null hypothesis for single sample t test

Statistics Made Easy

One Sample t-test: Definition, Formula, and Example

A  one sample t-test  is used to test whether or not the mean of a population is equal to some value.

This tutorial explains the following:

• The motivation for performing a one sample t-test.
• The formula to perform a one sample t-test.
• The assumptions that should be met to perform a one sample t-test.
• An example of how to perform a one sample t-test.

One Sample t-test: Motivation

Suppose we want to know whether or not the mean weight of a certain species of turtle in Florida is equal to 310 pounds. Since there are thousands of turtles in Florida, it would be extremely time-consuming and costly to go around and weigh each individual turtle.

Instead, we might take a simple random sample of 40 turtles and use the mean weight of the turtles in this sample to estimate the true population mean:

However, it’s virtually guaranteed that the mean weight of turtles in our sample will differ from 310 pounds. The question is whether or not this difference is statistically significant . Fortunately, a one sample t-test allows us to answer this question.

One Sample t-test: Formula

A one-sample t-test always uses the following null hypothesis:

• H 0 :  μ = μ 0 (population mean is equal to some hypothesized value μ 0 )

The alternative hypothesis can be either two-tailed, left-tailed, or right-tailed:

• H 1 (two-tailed):  μ ≠ μ 0 (population mean is not equal to some hypothesized value μ 0 )
• H 1 (left-tailed):  μ < μ 0 (population mean is less than some hypothesized value μ 0 )
• H 1 (right-tailed):  μ > μ 0 (population mean is greater than some hypothesized value μ 0 )

We use the following formula to calculate the test statistic t:

t = ( x  – μ) / (s/√ n )

• x : sample mean
• μ 0 : hypothesized population mean
• s:  sample standard deviation
• n:  sample size

If the p-value that corresponds to the test statistic t with (n-1) degrees of freedom is less than your chosen significance level (common choices are 0.10, 0.05, and 0.01) then you can reject the null hypothesis.

One Sample t-test: Assumptions

For the results of a one sample t-test to be valid, the following assumptions should be met:

• The variable under study should be either an interval or ratio variable .
• The observations in the sample should be independent .
• The variable under study should be approximately normally distributed.  You can check this assumption by creating a histogram and visually checking if the distribution has roughly a “bell shape.”
• The variable under study should have no outliers. You can check this assumption by creating a boxplot and visually checking for outliers.

One Sample t-test : Example

Suppose we want to know whether or not the mean weight of a certain species of turtle is equal to 310 pounds. To test this, will perform a one-sample t-test at significance level α = 0.05 using the following steps:

Step 1: Gather the sample data.

Suppose  we collect a random sample of turtles with the following information:

• Sample size n = 40
• Sample mean weight  x  = 300
• Sample standard deviation s = 18.5

Step 2: Define the hypotheses.

We will perform the one sample t-test with the following hypotheses:

• H 0 :  μ = 310 (population mean is equal to 310 pounds)
• H 1 :  μ ≠ 310 (population mean is not equal to 310 pounds)

Step 3: Calculate the test statistic  t .

t = ( x  – μ) / (s/√ n ) = (300-310) / (18.5/√ 40 ) =  -3.4187

Step 4: Calculate the p-value of the test statistic  t .

According to the T Score to P Value Calculator , the p-value associated with t = -3.4817 and degrees of freedom = n-1 = 40-1 = 39 is  0.00149 .

Step 5: Draw a conclusion.

Since this p-value is less than our significance level α = 0.05, we reject the null hypothesis. We have sufficient evidence to say that the mean weight of this species of turtle is not equal to 310 pounds.

Note:  You can also perform this entire one sample t-test by simply using the One Sample t-test calculator .

Additional Resources

The following tutorials explain how to perform a one-sample t-test using different statistical programs:

How to Perform a One Sample t-test in Excel How to Perform a One Sample t-test in SPSS How to Perform a One Sample t-test in Stata How to Perform a One Sample t-test in R How to Conduct a One Sample t-test in Python How to Perform a One Sample t-test on a TI-84 Calculator

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The One-Sample t -Test

What is the one-sample t -test.

The one-sample t-test is a statistical hypothesis test used to determine whether an unknown population mean is different from a specific value.

When can I use the test?

You can use the test for continuous data. Your data should be a random sample from a normal population.

What if my data isn’t nearly normally distributed?

If your sample sizes are very small, you might not be able to test for normality. You might need to rely on your understanding of the data. When you cannot safely assume normality, you can perform a nonparametric test that doesn’t assume normality.

Using the one-sample t -test

See how to perform a one-sample t -test using statistical software.

• Download JMP to follow along using the sample data included with the software.
• To see more JMP tutorials, visit the JMP Learning Library .

The sections below discuss what we need for the test, checking our data, performing the test, understanding test results and statistical details.

What do we need?

For the one-sample t -test, we need one variable.

We also have an idea, or hypothesis, that the mean of the population has some value. Here are two examples:

• A hospital has a random sample of cholesterol measurements for men. These patients were seen for issues other than cholesterol. They were not taking any medications for high cholesterol. The hospital wants to know if the unknown mean cholesterol for patients is different from a goal level of 200 mg.
• We measure the grams of protein for a sample of energy bars. The label claims that the bars have 20 grams of protein. We want to know if the labels are correct or not.

One-sample t -test assumptions

For a valid test, we need data values that are:

• Independent (values are not related to one another).
• Continuous.
• Obtained via a simple random sample from the population.

Also, the population is assumed to be normally distributed .

One-sample t -test example

Imagine we have collected a random sample of 31 energy bars from a number of different stores to represent the population of energy bars available to the general consumer. The labels on the bars claim that each bar contains 20 grams of protein.

Table 1: Grams of protein in random sample of energy bars

If you look at the table above, you see that some bars have less than 20 grams of protein. Other bars have more. You might think that the data support the idea that the labels are correct. Others might disagree. The statistical test provides a sound method to make a decision, so that everyone makes the same decision on the same set of data values. 

Checking the data

Let’s start by answering: Is the t -test an appropriate method to test that the energy bars have 20 grams of protein ? The list below checks the requirements for the test.

• The data values are independent. The grams of protein in one energy bar do not depend on the grams in any other energy bar. An example of dependent values would be if you collected energy bars from a single production lot. A sample from a single lot is representative of that lot, not energy bars in general.
• The data values are grams of protein. The measurements are continuous.
• We assume the energy bars are a simple random sample from the population of energy bars available to the general consumer (i.e., a mix of lots of bars).
• We assume the population from which we are collecting our sample is normally distributed, and for large samples, we can check this assumption.

We decide that the t -test is an appropriate method.

Before jumping into analysis, we should take a quick look at the data. The figure below shows a histogram and summary statistics for the energy bars.

From a quick look at the histogram, we see that there are no unusual points, or outliers . The data look roughly bell-shaped, so our assumption of a normal distribution seems reasonable.

From a quick look at the statistics, we see that the average is 21.40, above 20. Does this  average from our sample of 31 bars invalidate the label's claim of 20 grams of protein for the unknown entire population mean? Or not?

How to perform the one-sample t -test

For the t -test calculations we need the mean, standard deviation and sample size. These are shown in the summary statistics section of Figure 1 above.

We round the statistics to two decimal places. Software will show more decimal places, and use them in calculations. (Note that Table 1 shows only two decimal places; the actual data used to calculate the summary statistics has more.)

We start by finding the difference between the sample mean and 20:

$ 21.40-20\ =\ 1.40$

Next, we calculate the standard error for the mean. The calculation is:

Standard Error for the mean = $ \frac{s}{\sqrt{n}}= \frac{2.54}{\sqrt{31}}=0.456 $

This matches the value in Figure 1 above.

We now have the pieces for our test statistic. We calculate our test statistic as:

$ t =  \frac{\text{Difference}}{\text{Standard Error}}= \frac{1.40}{0.456}=3.07 $

To make our decision, we compare the test statistic to a value from the t- distribution. This activity involves four steps.

• We calculate a test statistic. Our test statistic is 3.07.
• We decide on the risk we are willing to take for declaring a difference when there is not a difference. For the energy bar data, we decide that we are willing to take a 5% risk of saying that the unknown population mean is different from 20 when in fact it is not. In statistics-speak, we set α = 0.05. In practice, setting your risk level (α) should be made before collecting the data.

We find the value from the t- distribution based on our decision. For a t -test, we need the degrees of freedom to find this value. The degrees of freedom are based on the sample size. For the energy bar data:

degrees of freedom = $ n - 1 = 31 - 1 = 30 $

The critical value of t with α = 0.05 and 30 degrees of freedom is +/- 2.043. Most statistics books have look-up tables for the distribution. You can also find tables online. The most likely situation is that you will use software and will not use printed tables.

We compare the value of our statistic (3.07) to the t value. Since 3.07 > 2.043, we reject the null hypothesis that the mean grams of protein is equal to 20. We make a practical conclusion that the labels are incorrect, and the population mean grams of protein is greater than 20.

Statistical details

Let’s look at the energy bar data and the 1-sample t -test using statistical terms.

Our null hypothesis is that the underlying population mean is equal to 20. The null hypothesis is written as:

$ H_o:  \mathrm{\mu} = 20 $

The alternative hypothesis is that the underlying population mean is not equal to 20. The labels claiming 20 grams of protein would be incorrect. This is written as:

$ H_a:  \mathrm{\mu} ≠ 20 $

This is a two-sided test. We are testing if the population mean is different from 20 grams in either direction. If we can reject the null hypothesis that the mean is equal to 20 grams, then we make a practical conclusion that the labels for the bars are incorrect. If we cannot reject the null hypothesis, then we make a practical conclusion that the labels for the bars may be correct.

We calculate the average for the sample and then calculate the difference with the population mean, mu:

$  \overline{x} - \mathrm{\mu} $

We calculate the standard error as:

$ \frac{s}{ \sqrt{n}} $

The formula shows the sample standard deviation as s and the sample size as n .  

The test statistic uses the formula shown below:

$  \dfrac{\overline{x} - \mathrm{\mu}} {s / \sqrt{n}} $

We compare the test statistic to a t value with our chosen alpha value and the degrees of freedom for our data. Using the energy bar data as an example, we set α = 0.05. The degrees of freedom ( df ) are based on the sample size and are calculated as:

$ df = n - 1 = 31 - 1 = 30 $

Statisticians write the t value with α = 0.05 and 30 degrees of freedom as:

$ t_{0.05,30} $

The t value for a two-sided test with α = 0.05 and 30 degrees of freedom is +/- 2.042. There are two possible results from our comparison:

• The test statistic is less extreme than the critical  t  values; in other words, the test statistic is not less than -2.042, or is not greater than +2.042. You fail to reject the null hypothesis that the mean is equal to the specified value. In our example, you would be unable to conclude that the label for the protein bars should be changed.
• The test statistic is more extreme than the critical  t  values; in other words, the test statistic is less than -2.042, or is greater than +2.042. You reject the null hypothesis that the mean is equal to the specified value. In our example, you conclude that either the label should be updated or the production process should be improved to produce, on average, bars with 20 grams of protein.

Testing for normality

The normality assumption is more important for small sample sizes than for larger sample sizes.

Normal distributions are symmetric, which means they are “even” on both sides of the center. Normal distributions do not have extreme values, or outliers. You can check these two features of a normal distribution with graphs. Earlier, we decided that the energy bar data was “close enough” to normal to go ahead with the assumption of normality. The figure below shows a normal quantile plot for the data, and supports our decision.

You can also perform a formal test for normality using software. The figure below shows results of testing for normality with JMP software. We cannot reject the hypothesis of a normal distribution. 

We can go ahead with the assumption that the energy bar data is normally distributed.

What if my data are not from a Normal distribution?

If your sample size is very small, it is hard to test for normality. In this situation, you might need to use your understanding of the measurements. For example, for the energy bar data, the company knows that the underlying distribution of grams of protein is normally distributed. Even for a very small sample, the company would likely go ahead with the t -test and assume normality.

What if you know the underlying measurements are not normally distributed? Or what if your sample size is large and the test for normality is rejected? In this situation, you can use a nonparametric test. Nonparametric  analyses do not depend on an assumption that the data values are from a specific distribution. For the one-sample t ­-test, the one possible nonparametric test is the Wilcoxon Signed Rank test. 

Understanding p-values

Using a visual, you can check to see if your test statistic is more extreme than a specified value in the distribution. The figure below shows a t- distribution with 30 degrees of freedom.

Since our test is two-sided and we set α = 0.05, the figure shows that the value of 2.042 “cuts off” 5% of the data in the tails combined.

The next figure shows our results. You can see the test statistic falls above the specified critical value. It is far enough “out in the tail” to reject the hypothesis that the mean is equal to 20.

Putting it all together with Software

You are likely to use software to perform a t -test. The figure below shows results for the 1-sample t -test for the energy bar data from JMP software.  

The software shows the null hypothesis value of 20 and the average and standard deviation from the data. The test statistic is 3.07. This matches the calculations above.

The software shows results for a two-sided test and for one-sided tests. We want the two-sided test. Our null hypothesis is that the mean grams of protein is equal to 20. Our alternative hypothesis is that the mean grams of protein is not equal to 20.  The software shows a p- value of 0.0046 for the two-sided test. This p- value describes the likelihood of seeing a sample average as extreme as 21.4, or more extreme, when the underlying population mean is actually 20; in other words, the probability of observing a sample mean as different, or even more different from 20, than the mean we observed in our sample. A p -value of 0.0046 means there is about 46 chances out of 10,000. We feel confident in rejecting the null hypothesis that the population mean is equal to 20.

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One Sample T Test – Clearly Explained with Examples | ML+

• October 8, 2020
• Selva Prabhakaran

One sample T-Test tests if the given sample of observations could have been generated from a population with a specified mean.

If it is found from the test that the means are statistically different, we infer that the sample is unlikely to have come from the population.

For example: If you want to test a car manufacturer’s claim that their cars give a highway mileage of 20kmpl on an average. You sample 10 cars from the dealership, measure their mileage and use the T-test to determine if the manufacturer’s claim is true.

By end of this, you will know when and how to do the T-Test, the concept, math, how to set the null and alternate hypothesis, how to use the T-tables, how to understand the one-tailed and two-tailed T-Test and see how to implement in R and Python using a practical example.

Introduction

Purpose of one sample t test, how to set the null and alternate hypothesis, procedure to do one sample t test, one sample t test example, one sample t test implementation, how to decide which t test to perform two tailed, upper tailed or lower tailed.

• Related Posts

The ‘One sample T Test’ is one of the 3 types of T Tests . It is used when you want to test if the mean of the population from which the sample is drawn is of a hypothesized value. You will understand this statement better (and all of about One Sample T test) better by the end of this post.

T Test was first invented by William Sealy Gosset, in 1908. Since he used the pseudo name as ‘Student’ when publishing his method in the paper titled ‘Biometrika’, the test came to be know as Student’s T Test.

Since it assumes that the test statistic, typically the sample mean, follows the sampling distribution, the Student’s T Test is considered as a Parametric test.

The purpose of the One Sample T Test is to determine if a sample observations could have come from a process that follows a specific parameter (like the mean).

It is typically implemented on small samples.

For example, given a sample of 15 items, you want to test if the sample mean is the same as a hypothesized mean (population). That is, essentially you want to know if the sample came from the given population or not.

Let’s suppose, you want to test if the mean weight of a manufactured component (from a sample size 15) is of a particular value (55 grams), with a 99% confidence.

How did we determine One sample T-test is the right test for this?

Because, there is only one sample involved and you want to compare the mean of this sample against a particular (hypothesized) value..

To do this, you need to set up a null hypothesis and an alternate hypothesis .

The null hypothesis usually assumes that there is no difference in the sample means and the hypothesized mean (comparison mean). The purpose of the T Test is to test if the null hypothesis can be rejected or not.

Depending on the how the problem is stated, the alternate hypothesis can be one of the following 3 cases:

• Case 1: H1 : x̅ != µ. Used when the true sample mean is not equal to the comparison mean. Use Two Tailed T Test.
• Case 2: H1 : x̅ > µ. Used when the true sample mean is greater than the comparison mean. Use Upper Tailed T Test.
• Case 3: H1 : x̅ < µ. Used when the true sample mean is lesser than the comparison mean. Use Lower Tailed T Test.

Where x̅ is the sample mean and µ is the population mean for comparison. We will go more into the detail of these three cases after solving some practical examples.

Example 1: A customer service company wants to know if their support agents are performing on par with industry standards.

According to a report the standard mean resolution time is 20 minutes per ticket. The sample group has a mean at 21 minutes per ticket with a standard deviation of 7 minutes.

Can you tell if the company’s support performance is better than the industry standard or not?

Example 2: A farming company wants to know if a new fertilizer has improved crop yield or not.

Historic data shows the average yield of the farm is 20 tonne per acre. They decide to test a new organic fertilizer on a smaller sample of farms and observe the new yield is 20.175 tonne per acre with a standard deviation of 3.02 tonne for 12 different farms.

Did the new fertilizer work?

Step 1: Define the Null Hypothesis (H0) and Alternate Hypothesis (H1)

H0: Sample mean (x̅) = Hypothesized Population mean (µ)

H1: Sample mean (x̅) != Hypothesized Population mean (µ)

The alternate hypothesis can also state that the sample mean is greater than or less than the comparison mean.

Step 2: Compute the test statistic (T)

$$t = \frac{Z}{s} = \frac{\bar{X} – \mu}{\frac{\hat{\sigma}}{\sqrt{n}}}$$

where s is the standard error .

Step 3: Find the T-critical from the T-Table

Use the degree of freedom and the alpha level (0.05) to find the T-critical.

Step 4: Determine if the computed test statistic falls in the rejection region.

Alternately, simply compute the P-value. If it is less than the significance level (0.05 or 0.01), reject the null hypothesis.

Problem Statement:

We have the potato yield from 12 different farms. We know that the standard potato yield for the given variety is µ=20.

x = [21.5, 24.5, 18.5, 17.2, 14.5, 23.2, 22.1, 20.5, 19.4, 18.1, 24.1, 18.5]

Test if the potato yield from these farms is significantly better than the standard yield.

Step 1: Define the Null and Alternate Hypothesis

H0: x̅ = 20

H1: x̅ > 20

n = 12. Since this is one sample T test, the degree of freedom = n-1 = 12-1 = 11.

Let’s set alpha = 0.05, to meet 95% confidence level.

Step 2: Calculate the Test Statistic (T) 1. Calculate sample mean

$$\bar{X} = \frac{x_1 + x_2 + x_3 + . . + x_n}{n}$$

$$\bar{x} = 20.175$$

• Calculate sample standard deviation

$$\bar{\sigma} = \frac{(x_1 – \bar{x})^2 + (x_2 – \bar{x})^2 + (x_3 – \bar{x})^2 + . . + (x_n – \bar{x})^2}{n-1}$$

$$\sigma = 3.0211$$

• Substitute in the T Statistic formula

$$T = \frac{\bar{x} – \mu}{se} = \frac{\bar{x} – \mu}{\frac{\sigma}{\sqrt{n}}}$$

$$T = (20.175 – 20)/(3.0211/\sqrt{12}) = 0.2006$$

Step 3: Find the T-Critical

Confidence level = 0.95, alpha=0.05. For one tailed test, look under 0.05 column. For d.o.f = 12 – 1 = 11, T-Critical = 1.796 .

Now you might wonder why ‘One Tailed test’ was chosen. This is because of the way you define the alternate hypothesis. Had the null hypothesis simply stated that the sample means is not equal to 20, then we would have gone for a two tailed test. More details about this topic in the next section.

Step 4: Does it fall in rejection region?

Since the computed T Statistic is less than the T-critical, it does not fall in the rejection region.

Clearly, the calculated T statistic does not fall in the rejection region. So, we do not reject the null hypothesis.

Since you want to perform a ‘One Tailed Greater than’ test (that is, the sample mean is greater than the comparison mean), you need to specify alternative='greater' in the t.test() function. Because, by default, the t.test() does a two tailed test (which is what you do when your alternate hypothesis simply states sample mean != comparison mean).

The P-value computed here is nothing but p = Pr(T > t) (upper-tailed), where t is the calculated T statistic.

In Python, One sample T Test is implemented in ttest_1samp() function in the scipy package. However, it does a Two tailed test by default , and reports a signed T statistic. That means, the reported P-value will always be computed for a Two-tailed test. To calculate the correct P value, you need to divide the output P-value by 2.

Apply the following logic if you are performing a one tailed test:

For greater than test: Reject H0 if p/2 < alpha (0.05). In this case, t will be greater than 0. For lesser than test: Reject H0 if p/2 < alpha (0.05). In this case, t will be less than 0.

Since it is one tailed test, the real p-value is 0.8446/2 = 0.4223. We do not rejecting the Null Hypothesis anyway.

The decision of whether the computed test statistic falls in the rejection region depends on how the alternate hypothesis is defined.

We know the Null Hypothesis is H0: µD = 0. Where, µD is the difference in the means, that is sample mean minus the comparison mean.

You can also write H0 as: x̅ = µ , where x̅ is sample mean and ‘µ’ is the comparison mean.

Case 1: If H1 : x̅ != µ , then rejection region lies on both tails of the T-Distribution (two-tailed). This means the alternate hypothesis just states the difference in means is not equal. There is no comparison if one of the means is greater or lesser than the other.

In this case, use Two Tailed T Test .

Here, P value = 2 . Pr(T > | t |)

Case 2: If H1: x̅ > µ , then rejection region lies on upper tail of the T-Distribution (upper-tailed). If the mean of the sample of interest is greater than the comparison mean. Example: If Component A has a longer time-to-failure than Component B.

In such case, use Upper Tailed based test.

Here, P-value = Pr(T > t)

Case 3: If H1: x̅ < µ , then rejection region lies on lower tail of the T-Distribution (lower-tailed). If the mean of the sample of interest is lesser than the comparison mean.

In such case, use lower tailed test.

Here, P-value = Pr(T < t)

Hope you are now familiar and clear about with the One Sample T Test. If some thing is still not clear, write in comment. Next, topic is Two sample T test . Stay tuned.

More Articles

Correlation – connecting the dots, the role of correlation in data analysis, hypothesis testing – a deep dive into hypothesis testing, the backbone of statistical inference, sampling and sampling distributions – a comprehensive guide on sampling and sampling distributions, law of large numbers – a deep dive into the world of statistics, central limit theorem – a deep dive into central limit theorem and its significance in statistics, skewness and kurtosis – peaks and tails, understanding data through skewness and kurtosis”, similar articles, complete introduction to linear regression in r, how to implement common statistical significance tests and find the p value, logistic regression – a complete tutorial with examples in r.

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One Sample t Test

Spss tutorials: one sample t test.

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Our tutorials reference a dataset called "sample" in many examples. If you'd like to download the sample dataset to work through the examples, choose one of the files below:

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The One Sample t Test examines whether the mean of a population is statistically different from a known or hypothesized value. The One Sample t Test is a parametric test.

This test is also known as:

• Single Sample t Test

The variable used in this test is known as:

• Test variable

In a One Sample t Test, the test variable's mean is compared against a "test value", which is a known or hypothesized value of the mean in the population. Test values may come from a literature review, a trusted research organization, legal requirements, or industry standards. For example:

• A particular factory's machines are supposed to fill bottles with 150 milliliters of product. A plant manager wants to test a random sample of bottles to ensure that the machines are not under- or over-filling the bottles.
• The United States Environmental Protection Agency (EPA) sets clearance levels for the amount of lead present in homes: no more than 10 micrograms per square foot on floors and no more than 100 micrograms per square foot on window sills ( as of December 2020 ). An inspector wants to test if samples taken from units in an apartment building exceed the clearance level.

Common Uses

The One Sample  t  Test is commonly used to test the following:

• Statistical difference between a mean and a known or hypothesized value of the mean in the population.
• This approach involves creating a change score from two variables, and then comparing the mean change score to zero, which will indicate whether any change occurred between the two time points for the original measures. If the mean change score is not significantly different from zero, no significant change occurred.

Note: The One Sample t Test can only compare a single sample mean to a specified constant. It can not compare sample means between two or more groups. If you wish to compare the means of multiple groups to each other, you will likely want to run an Independent Samples t Test (to compare the means of two groups) or a One-Way ANOVA (to compare the means of two or more groups).

Data Requirements

Your data must meet the following requirements:

• Test variable that is continuous (i.e., interval or ratio level)
• There is no relationship between scores on the test variable
• Violation of this assumption will yield an inaccurate p value
• Random sample of data from the population
• Non-normal population distributions, especially those that are thick-tailed or heavily skewed, considerably reduce the power of the test
• Among moderate or large samples, a violation of normality may still yield accurate p values
• Homogeneity of variances (i.e., variances approximately equal in both the sample and population)
• No outliers

The null hypothesis ( H 0 ) and (two-tailed) alternative hypothesis ( H 1 ) of the one sample T test can be expressed as:

H 0 : µ =  µ 0   ("the population mean is equal to the [proposed] population mean") H 1 : µ ≠  µ 0   ("the population mean is not equal to the [proposed] population mean")

where µ is the "true" population mean and µ 0 is the proposed value of the population mean.

Test Statistic

The test statistic for a One Sample t Test is denoted t , which is calculated using the following formula:

$$ t = \frac{\overline{x}-\mu{}_{0}}{s_{\overline{x}}} $$

$$ s_{\overline{x}} = \frac{s}{\sqrt{n}} $$

\(\mu_{0}\) = The test value -- the proposed constant for the population mean \(\bar{x}\) = Sample mean \(n\) = Sample size (i.e., number of observations) \(s\) = Sample standard deviation \(s_{\bar{x}}\) = Estimated standard error of the mean ( s /sqrt( n ))

The calculated t value is then compared to the critical t value from the t distribution table with degrees of freedom df = n - 1 and chosen confidence level. If the calculated t value > critical t value, then we reject the null hypothesis.

Data Set-Up

Your data should include one continuous, numeric variable (represented in a column) that will be used in the analysis. The variable's measurement level should be defined as Scale in the Variable View window.

Run a One Sample t Test

To run a One Sample t Test in SPSS, click  Analyze > Compare Means > One-Sample T Test .

The One-Sample T Test window opens where you will specify the variables to be used in the analysis. All of the variables in your dataset appear in the list on the left side. Move variables to the Test Variable(s) area by selecting them in the list and clicking the arrow button.

A Test Variable(s): The variable whose mean will be compared to the hypothesized population mean (i.e., Test Value). You may run multiple One Sample t Tests simultaneously by selecting more than one test variable. Each variable will be compared to the same Test Value. 

B Test Value: The hypothesized population mean against which your test variable(s) will be compared.

C Estimate effect sizes: Optional. If checked, will print effect size statistics -- namely, Cohen's d -- for the test(s). (Note: Effect sizes calculations for t tests were first added to SPSS Statistics in version 27, making them a relatively recent addition. If you do not see this option when you use SPSS, check what version of SPSS you're using.)

D Options: Clicking Options will open a window where you can specify the Confidence Interval Percentage and how the analysis will address Missing Values (i.e., Exclude cases analysis by analysis or Exclude cases listwise ). Click Continue when you are finished making specifications.

Click OK to run the One Sample t Test.

Problem Statement

According to the CDC , the mean height of U.S. adults ages 20 and older is about 66.5 inches (69.3 inches for males, 63.8 inches for females).

In our sample data, we have a sample of 435 college students from a single college. Let's test if the mean height of students at this college is significantly different than 66.5 inches using a one-sample t test. The null and alternative hypotheses of this test will be:

H 0 : µ Height = 66.5  ("the mean height is equal to 66.5") H 1 : µ Height ≠ 66.5  ("the mean height is not equal to 66.5")

Before the Test

In the sample data, we will use the variable Height , which a continuous variable representing each respondent’s height in inches. The heights exhibit a range of values from 55.00 to 88.41 ( Analyze > Descriptive Statistics > Descriptives ).

Let's create a histogram of the data to get an idea of the distribution, and to see if  our hypothesized mean is near our sample mean. Click Graphs > Legacy Dialogs > Histogram . Move variable Height to the Variable box, then click OK .

To add vertical reference lines at the mean (or another location), double-click on the plot to open the Chart Editor, then click Options > X Axis Reference Line . In the Properties window, you can enter a specific location on the x-axis for the vertical line, or you can choose to have the reference line at the mean or median of the sample data (using the sample data). Click Apply to make sure your new line is added to the chart. Here, we have added two reference lines: one at the sample mean (the solid black line), and the other at 66.5 (the dashed red line).

From the histogram, we can see that height is relatively symmetrically distributed about the mean, though there is a slightly longer right tail. The reference lines indicate that sample mean is slightly greater than the hypothesized mean, but not by a huge amount. It's possible that our test result could come back significant.

Running the Test

To run the One Sample t Test, click  Analyze > Compare Means > One-Sample T Test.  Move the variable Height to the Test Variable(s) area. In the Test Value field, enter 66.5.

If you are using SPSS Statistics 27 or later :

If you are using SPSS Statistics 26 or earlier :

Two sections (boxes) appear in the output: One-Sample Statistics and One-Sample Test . The first section, One-Sample Statistics , provides basic information about the selected variable, Height , including the valid (nonmissing) sample size ( n ), mean, standard deviation, and standard error. In this example, the mean height of the sample is 68.03 inches, which is based on 408 nonmissing observations.

The second section, One-Sample Test , displays the results most relevant to the One Sample t Test. 

A Test Value : The number we entered as the test value in the One-Sample T Test window.

B t Statistic : The test statistic of the one-sample t test, denoted t . In this example, t = 5.810. Note that t is calculated by dividing the mean difference (E) by the standard error mean (from the One-Sample Statistics box).

C df : The degrees of freedom for the test. For a one-sample t test, df = n - 1; so here, df = 408 - 1 = 407.

D Significance (One-Sided p and Two-Sided p): The p-values corresponding to one of the possible one-sided alternative hypotheses (in this case, µ Height > 66.5) and two-sided alternative hypothesis (µ Height ≠ 66.5), respectively. In our problem statement above, we were only interested in the two-sided alternative hypothesis.

E Mean Difference : The difference between the "observed" sample mean (from the One Sample Statistics box) and the "expected" mean (the specified test value (A)). The sign of the mean difference corresponds to the sign of the t value (B). The positive t value in this example indicates that the mean height of the sample is greater than the hypothesized value (66.5).

F Confidence Interval for the Difference : The confidence interval for the difference between the specified test value and the sample mean.

Decision and Conclusions

Recall that our hypothesized population value was 66.5 inches, the [approximate] average height of the overall adult population in the U.S. Since p < 0.001, we reject the null hypothesis that the mean height of students at this college is equal to the hypothesized population mean of 66.5 inches and conclude that the mean height is significantly different than 66.5 inches.

Based on the results, we can state the following:

• There is a significant difference in the mean height of the students at this college and the overall adult population in the U.S. ( p < .001).
• The average height of students at this college is about 1.5 inches taller than the U.S. adult population average (95% CI [1.013, 2.050]).
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• Next: Paired Samples t Test >>
• Last Updated: Dec 18, 2023 12:59 PM
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One-Sample T-Test – Quick Tutorial & Example

Null hypothesis, assumptions, effect size, confidence intervals for means, apa style reporting.

A one-sample t-test evaluates if a population mean is likely to be x : some hypothesized value.

One-Sample T-Test Example

A school director thinks his students perform poorly due to low IQ scores. Now, most IQ tests have been calibrated to have a mean of 100 points in the general population. So the question is does the student population have a mean IQ score of 100? Now, our school has 1,114 students and the IQ tests are somewhat costly to administer. Our director therefore draws a simple random sample of N = 38 students and tests them on 4 IQ components:

• verb (Verbal Intelligence )
• math (Mathematical Ability )
• clas (Classification Skills )
• logi (Logical Reasoning Skills)

The raw data thus collected are in this Googlesheet , partly shown below. Note that a couple of scores are missing due to illness and unknown reasons.

We'll try to demonstrate that our students have low IQ scores by rejecting the null hypothesis that the mean IQ score for the entire student population is 100 for each of the 4 IQ components measured. Our main challenge is that we only have data on a sample of 38 students from a population of N = 1,114. But let's first just look at some descriptive statistics for each component:

• N - sample size;
• M - sample mean and
• SD - sample standard deviation.

Descriptive Statistics

Our first basic conclusion is that our 38 students score lower than 100 points on all 4 IQ components. The differences for verb (99.29) and math (97.97) are small. Those for clas (93.91) and logi (94.74) seem somewhat more serious. Now, our sample of 38 students may obviously come up with slightly different means than our population of N = 1,114. So what can we (not) conclude regarding our population? We'll try to generalize these sample results to our population with 2 different approaches:

• Statistical significance : how likely are these sample means if the population means are really all 100 points?
• Confidence intervals : given the sample results, what are likely ranges for the population means?

Both approaches require some assumptions so let's first look into those.

The assumptions required for our one-sample t-tests are

• independent observations and
• normality : the IQ scores must be normally distributed in the entire population.

Do our data meet these assumptions? First off, 1. our students didn't interact during their tests. Therefore, our observations are likely to be independent. 2. Normality is only needed for small sample sizes, say N < 25 or so. For the data at hand, normality is no issue. For smaller sample sizes, you could evaluate the normality assumption by

• inspecting if the histograms roughly follow normal curves,
• inspecting if both skewness and kurtosis are close to 0 and
• running a Shapiro-Wilk test or a Kolmogorov-Smirnov test .

However, the data at hand meet all assumptions so let's now look into the actual tests.

If we'd draw many samples of students, such samples would come up with different means. We can compute the standard deviation of those means over hypothesized samples: the standard error of the mean or \(SE_{mean}\) $$SE_{mean} = \frac{SD}{\sqrt{N}}$$ for our first IQ component, this results in $$SE_{mean} = \frac{12.45}{\sqrt{38}} = 2.02$$ Our null hypothesis is that the population mean, \(\mu_0 = 100\). If this is true, then the average sample mean should also be 100. We now basically compute the z-score for our sample mean: the test statistic \(t\) $$t = \frac{M - \mu_0}{SE_{mean}}$$ for our first IQ component, this results in $$t = \frac{99.29 - 100}{2.02} = -0.35$$ If the assumptions are met, \(t\) follows a t distribution with the degrees of freedom or \(df\) given by $$df = N - 1$$ For a sample of 38 respondents, this results in $$df = 38 - 1 = 37$$ Given \(t\) and \(df\), we can simply look up that the 2-tailed significance level \(p\) = 0.73 in this Googlesheet , partly shown below.

Interpretation

As a rule of thumb, we reject the null hypothesis if p < 0.05. We just found that p = 0.73 so we don't reject our null hypothesis: given our sample data, the population mean being 100 is a credible statement. So precisely what does p = 0.73 mean? Well, it means there's a 0.73 (or 73%) probability that t < -0.35 or t > 0.35. The figure below illustrates how this probability results from the sampling distribution , t(37).

Next, remember that t is just a standardized mean difference. For our data, t = -0.35 corresponds to a difference of -0.71 IQ points. Therefore, p = 0.73 means that there's a 0.73 probability of finding an absolute mean difference of at least 0.71 points. Roughly speaking, the sample mean we found is likely to occur if the null hypothesis is true.

The only effect size measure for a one-sample t-test is Cohen’s D defined as $$Cohen's\;D = \frac{M - \mu_0}{SD}$$ For our first IQ test component, this results in $$Cohen's\;D = \frac{99.29 - 100}{12.45} = -0.06$$ Some general conventions are that

• | Cohen’s D | = 0.20 indicates a small effect size;
• | Cohen’s D | = 0.50 indicates a medium effect size;
• | Cohen’s D | = 0.80 indicates a large effect size.

This means that Cohen’s D = -0.06 indicates a negligible effect size for our first test component. Cohen’s D is completely absent from SPSS except for SPSS 27 . However, we can easily obtain it from JASP . The JASP output below shows the effect sizes for all 4 IQ test components.

Note that the last 2 IQ components -clas and logi- almost have medium effect sizes. These are also the 2 components whose means differ significantly from 100: p < 0.05 for both means (third table column).

Our data came up with sample means for our 4 IQ test components. Now, we know that sample means typically differ somewhat from their population counterparts. So what are likely ranges for the population means we're after? This is often answered by computing 95% confidence intervals . We'll demonstrate the procedure for our last IQ component, logical reasoning. Since we've 34 observations, t follows a t-distribution with df = 33. We'll first look up which t-values enclose the most likely 95% from the inverse t-distribution. We'll do so by typing =T.INV(0.025,33) into any cell of a Googlesheet , which returns -2.03. Note that 0.025 is 2.5%. This is because the 5% most un likely values are divided over both tails of the distribution as shown below.

Now, our t-value of -2.03 estimates that our 95% of our sample means fluctuate between ± 2.03 standard errors denoted by \(SE_{mean}\) For our last IQ component, $$SE_{mean} = \frac{12.57}{\sqrt34} = 2.16 $$ We now know that 95% of our sample means are estimated to fluctuate between ± 2.03 · 2.16 = 4.39 IQ test points. Last, we combine this fluctuation with our observed sample mean of 94.74: $$CI_{95\%} = [94.74 - 4.39,94.74 + 4.39] = [90.35,99.12]$$ Note that our 95% confidence interval does not enclose our hypothesized population mean of 100. This implies that we'll reject this null hypothesis at α = 0.05. We don't even need to run the actual t-test for drawing this conclusion.

A single t-test is usually reported in text as in “The mean for verbal skills did not differ from 100, t(37) = -0.35, p = 0.73, Cohen’s D = 0.06.” For multiple tests, a simple overview table as shown below is recommended. We feel that confidence intervals for means (not mean differences ) should also be included. Since the APA does not mention these, we left them out for now.

Right. Well, I can't think of anything else that is relevant regarding the one-sample t-test. If you do, don't be shy. Just write us a comment below. We're always happy to hear from you!

Thanks for reading!

Tell us what you think!

This tutorial has 3 comments:.

By YY Ma on February 23rd, 2021

An excellent introduction! Cohen's D is a useful statistic. I think, if the sample size of each study is identical, | t | can be used as the effect size. And | t (0.05,df) | is the threshold for assessing whether a effect size is significantly large.

By SHAMSUDDEEN IDRIS RIMINGADO on January 9th, 2022

In accordance with your explanation, does a one sample t test be use to test this hypothesis : There is significant difference between male and female exposed to error analysis in student with handwriting difficulties

By Ruben Geert van den Berg on January 10th, 2022

For your question, you'd typically use an independent samples t-test , which is a bit more complicated than the one-sample t-test discussed in this tutorial.

Hope that helps!

SPSS tutorials

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Single Sample T-Test

The StatsTest Flow: Difference >> Continuous Variable of Interest >> One Sample Tests (single group) >> Normal Variable of Interest

Not sure this is the right statistical method? Use the Choose Your StatsTest workflow to select the right method.

What is a Single Sample T-Test?

The Single Sample T-Test is a statistical test used to determine if a single group is significantly different from a known or hypothesized population value on your variable of interest. Your variable of interest should be continuous and normally distributed and you should have enough data (more than 5 values).

The Single Sample T-Test is also called a One-Sample T-Test, Single Sample Student T-Test, or One-Sample Test of Means.

Assumptions for a Single Sample T-Test

Every statistical method has assumptions. Assumptions mean that your data must satisfy certain properties in order for statistical method results to be accurate.

The assumptions for the Single Sample T-Test include:

Normally Distributed

Random sample, enough data.

Let’s dive in to each one of these separately.

The variable that you care about (and want to see if it is different between your group and the population) must be continuous. Continuous means that the variable can take on any reasonable value.

Some good examples of continuous variables include age, weight, height, test scores, survey scores, yearly salary, etc.

If the variable that you care about is a proportion (48% of males voted vs 56% of females voted) and you have more than 5 in each group then you should use the One-Proportion Z-Test . If your variable of interest is a proportion and you have less than 5 in a group, you should use the Exact Test of Goodness of Fit .

Normally Distributed Variable of Interest

The variable that you care about must be spread out in a normal way. In statistics, this is called being normally distributed (aka it must look like a bell curve when you graph the data). Only use a single sample t-test with your data if the variable you care about is normally distributed.

If your variable is not normally distributed, you should use Single-Sample Wilcoxon Signed-Rank Test instead.

The data points for each group in your analysis must have come from a simple random sample. This means that if you wanted to see if drinking sugary soda makes you gain weight, you would need to randomly select a group of soda drinkers for your soda drinker group, and then you would compare that to a known population weight for non-sugary-soda drinkers.

The key here is that the data points for each group were randomly selected. This is important because if your group is not randomly determined then your analysis will be incorrect. In statistical terms this is called bias, or a tendency to have incorrect results because of bad data.

If you do not have a random sample, the conclusions you can draw from your results are very limited. You should try to get a simple random sample. If you have paired samples (2 measurements from the same group of subjects) then you should use a Paired Samples T-Test instead. If you want to compare 2 groups of subjects instead of a single group with a population mean, then you should use an Independent Samples T-Test instead

The sample size (or data set size) should be greater than 5 in your group. Some people argue for more than 15 or even 30, but more than 5 is probably sufficient.

It also depends on the expected size of the difference between groups. If you expect a large difference between groups, then you can get away with a smaller sample size. If you expect a small difference between groups, then you likely need a larger sample (30+).

If your sample size is greater than 30 (and you know the average and standard deviation or spread of the population values), you should run a Single Sample Z-Test instead.

When to use a Single Sample T-Test?

You should use a Single Sample T-Test in the following scenario:

• You want to know if one group is different from a known or hypothesized population value on your variable of interest
• Your variable of interest is continuous
• You have one group
• Your variable of interest is normally distributed

Let’s clarify these to help you know when to use a Single Sample T-Test.

You are looking for a statistical test to see whether a single group is significantly different from a population value on your variable of interest. This is a difference question. Other types of analyses include examining the relationship between two variables (correlation) or predicting one variable using another variable (prediction).

Continuous Data

Your variable of interest must be continuous. Continuous means that your variable of interest can basically take on any value, such as heart rate, height, weight, number of ice cream bars you can eat in 1 minute, etc.

Types of data that are NOT continuous include ordered data (such as finishing place in a race, best business rankings, etc.), categorical data (gender, eye color, race, etc.), or binary data (purchased the product or not, has the disease or not, etc.).

A Single Sample T-Test can only be used to compare a single group with a known population value on your variable of interest.

If you have three or more groups, you should use a One Way Anova analysis instead. If you have two groups to compare, you should use an Independent Samples T-Test instead.

Normally distributed was covered earlier and means that your variable of interest should look like a bell curve when you graph it as a histogram.

If you get a group of students to take a pre-test and the same students to take a post-test, you have two different variables for the same group of students, which would be paired data, in which case you would need to use a Paired Samples T-Test instead.

Single Sample T-Test Example

Group 1 : Received the experimental medical treatment. Population Value : On average in the population, it takes 12 days to recover from the disease Variable of interest : Time to recover from the disease in days.

In this example, group 1 is our treatment group because they received the experimental medical treatment. The population value is essentially our control group because they did not receive the treatment.

The null hypothesis, which is statistical lingo for what would happen if the treatment does nothing, is that group 1 and our population will recover from the disease in about the same number of days, on average. We are trying to determine if receiving the experimental medical treatment will shorten the number of days it takes for patients to recover from the disease.

As we run the experiment, we track how long it takes for each patient to fully recover from the disease. In order to use a Single Sample T-Test on our data, our variable of interest has to be normally distributed (bell curve shaped). In this case, recovery from the disease in days is normal for our treatment group.

After the experiment is over, we compare our treatment group to the population value on our variable of interest (days to fully recover) using a Single Sample T-Test. When we run the analysis, we get a t-statistic and a p-value. The t-statistic is a measure of how different our group is from the population value on our recovery variable of interest. A p-value is the chance of seeing our results assuming the treatment actually doesn’t do anything. A p-value less than or equal to 0.05 means that our result is statistically significant and we can trust that the difference is not due to chance alone.

Frequently Asked Questions

Q: What is the difference between a single sample t-test and a one sample t-test? A: Nothing. They are two names for the same analysis.

Q: What if I don’t know the population average for my variable of interest? A: You cannot run a single sample t-test without a comparison group or value. You either need to collect data for a control group or find data on what the population average is.

Q: How do I run a single sample t-test in SPSS, R, SAS, or STATA? A: This resource is focused on helping you pick the right statistical method every time. There are many resources available to help you figure out how to run this method with your data: SPSS article: https://libguides.library.kent.edu/SPSS/OneSampletTest SPSS video: https://www.youtube.com/watch?v=2zVeV1ohGCU R article: http://www.sthda.com/english/wiki/one-sample-t-test-in-r R video: https://www.youtube.com/watch?v=kvmSAXhX9Hs

If you still can’t figure something out, feel free to reach out .

t-test Calculator

When to use a t-test, which t-test, how to do a t-test, p-value from t-test, t-test critical values, how to use our t-test calculator, one-sample t-test, two-sample t-test, paired t-test, t-test vs z-test.

Welcome to our t-test calculator! Here you can not only easily perform one-sample t-tests , but also two-sample t-tests , as well as paired t-tests .

Do you prefer to find the p-value from t-test, or would you rather find the t-test critical values? Well, this t-test calculator can do both! 😊

What does a t-test tell you? Take a look at the text below, where we explain what actually gets tested when various types of t-tests are performed. Also, we explain when to use t-tests (in particular, whether to use the z-test vs. t-test) and what assumptions your data should satisfy for the results of a t-test to be valid. If you've ever wanted to know how to do a t-test by hand, we provide the necessary t-test formula, as well as tell you how to determine the number of degrees of freedom in a t-test.

A t-test is one of the most popular statistical tests for location , i.e., it deals with the population(s) mean value(s).

There are different types of t-tests that you can perform:

• A one-sample t-test;
• A two-sample t-test; and
• A paired t-test.

In the next section , we explain when to use which. Remember that a t-test can only be used for one or two groups . If you need to compare three (or more) means, use the analysis of variance ( ANOVA ) method.

The t-test is a parametric test, meaning that your data has to fulfill some assumptions :

• The data points are independent; AND
• The data, at least approximately, follow a normal distribution .

If your sample doesn't fit these assumptions, you can resort to nonparametric alternatives. Visit our Mann–Whitney U test calculator or the Wilcoxon rank-sum test calculator to learn more. Other possibilities include the Wilcoxon signed-rank test or the sign test.

Your choice of t-test depends on whether you are studying one group or two groups:

One sample t-test

Choose the one-sample t-test to check if the mean of a population is equal to some pre-set hypothesized value .

The average volume of a drink sold in 0.33 l cans — is it really equal to 330 ml?

The average weight of people from a specific city — is it different from the national average?

Choose the two-sample t-test to check if the difference between the means of two populations is equal to some pre-determined value when the two samples have been chosen independently of each other.

In particular, you can use this test to check whether the two groups are different from one another .

The average difference in weight gain in two groups of people: one group was on a high-carb diet and the other on a high-fat diet.

The average difference in the results of a math test from students at two different universities.

This test is sometimes referred to as an independent samples t-test , or an unpaired samples t-test .

A paired t-test is used to investigate the change in the mean of a population before and after some experimental intervention , based on a paired sample, i.e., when each subject has been measured twice: before and after treatment.

In particular, you can use this test to check whether, on average, the treatment has had any effect on the population .

The change in student test performance before and after taking a course.

The change in blood pressure in patients before and after administering some drug.

So, you've decided which t-test to perform. These next steps will tell you how to calculate the p-value from t-test or its critical values, and then which decision to make about the null hypothesis.

Decide on the alternative hypothesis :

Use a two-tailed t-test if you only care whether the population's mean (or, in the case of two populations, the difference between the populations' means) agrees or disagrees with the pre-set value.

Use a one-tailed t-test if you want to test whether this mean (or difference in means) is greater/less than the pre-set value.

Compute your T-score value :

Formulas for the test statistic in t-tests include the sample size , as well as its mean and standard deviation . The exact formula depends on the t-test type — check the sections dedicated to each particular test for more details.

Determine the degrees of freedom for the t-test:

The degrees of freedom are the number of observations in a sample that are free to vary as we estimate statistical parameters. In the simplest case, the number of degrees of freedom equals your sample size minus the number of parameters you need to estimate . Again, the exact formula depends on the t-test you want to perform — check the sections below for details.

The degrees of freedom are essential, as they determine the distribution followed by your T-score (under the null hypothesis). If there are d degrees of freedom, then the distribution of the test statistics is the t-Student distribution with d degrees of freedom . This distribution has a shape similar to N(0,1) (bell-shaped and symmetric) but has heavier tails . If the number of degrees of freedom is large (>30), which generically happens for large samples, the t-Student distribution is practically indistinguishable from N(0,1).

💡 The t-Student distribution owes its name to William Sealy Gosset, who, in 1908, published his paper on the t-test under the pseudonym "Student". Gosset worked at the famous Guinness Brewery in Dublin, Ireland, and devised the t-test as an economical way to monitor the quality of beer. Cheers! 🍺🍺🍺

Recall that the p-value is the probability (calculated under the assumption that the null hypothesis is true) that the test statistic will produce values at least as extreme as the T-score produced for your sample . As probabilities correspond to areas under the density function, p-value from t-test can be nicely illustrated with the help of the following pictures:

The following formulae say how to calculate p-value from t-test. By cdf t,d we denote the cumulative distribution function of the t-Student distribution with d degrees of freedom:

p-value from left-tailed t-test:

p-value = cdf t,d (t score )

p-value from right-tailed t-test:

p-value = 1 − cdf t,d (t score )

p-value from two-tailed t-test:

p-value = 2 × cdf t,d (−|t score |)

or, equivalently: p-value = 2 − 2 × cdf t,d (|t score |)

However, the cdf of the t-distribution is given by a somewhat complicated formula. To find the p-value by hand, you would need to resort to statistical tables, where approximate cdf values are collected, or to specialized statistical software. Fortunately, our t-test calculator determines the p-value from t-test for you in the blink of an eye!

Recall, that in the critical values approach to hypothesis testing, you need to set a significance level, α, before computing the critical values , which in turn give rise to critical regions (a.k.a. rejection regions).

Formulas for critical values employ the quantile function of t-distribution, i.e., the inverse of the cdf :

Critical value for left-tailed t-test: cdf t,d -1 (α)

critical region:

(-∞, cdf t,d -1 (α)]

Critical value for right-tailed t-test: cdf t,d -1 (1-α)

[cdf t,d -1 (1-α), ∞)

Critical values for two-tailed t-test: ±cdf t,d -1 (1-α/2)

(-∞, -cdf t,d -1 (1-α/2)] ∪ [cdf t,d -1 (1-α/2), ∞)

To decide the fate of the null hypothesis, just check if your T-score lies within the critical region:

If your T-score belongs to the critical region , reject the null hypothesis and accept the alternative hypothesis.

If your T-score is outside the critical region , then you don't have enough evidence to reject the null hypothesis.

Choose the type of t-test you wish to perform:

A one-sample t-test (to test the mean of a single group against a hypothesized mean);

A two-sample t-test (to compare the means for two groups); or

A paired t-test (to check how the mean from the same group changes after some intervention).

Two-tailed;

Left-tailed; or

Right-tailed.

This t-test calculator allows you to use either the p-value approach or the critical regions approach to hypothesis testing!

Enter your T-score and the number of degrees of freedom . If you don't know them, provide some data about your sample(s): sample size, mean, and standard deviation, and our t-test calculator will compute the T-score and degrees of freedom for you .

Once all the parameters are present, the p-value, or critical region, will immediately appear underneath the t-test calculator, along with an interpretation!

The null hypothesis is that the population mean is equal to some value μ 0 \mu_0 μ 0 ​ .

The alternative hypothesis is that the population mean is:

• different from μ 0 \mu_0 μ 0 ​ ;
• smaller than μ 0 \mu_0 μ 0 ​ ; or
• greater than μ 0 \mu_0 μ 0 ​ .

One-sample t-test formula :

• μ 0 \mu_0 μ 0 ​ — Mean postulated in the null hypothesis;
• n n n — Sample size;
• x ˉ \bar{x} x ˉ — Sample mean; and
• s s s — Sample standard deviation.

Number of degrees of freedom in t-test (one-sample) = n − 1 n-1 n − 1 .

The null hypothesis is that the actual difference between these groups' means, μ 1 \mu_1 μ 1 ​ , and μ 2 \mu_2 μ 2 ​ , is equal to some pre-set value, Δ \Delta Δ .

The alternative hypothesis is that the difference μ 1 − μ 2 \mu_1 - \mu_2 μ 1 ​ − μ 2 ​ is:

• Different from Δ \Delta Δ ;
• Smaller than Δ \Delta Δ ; or
• Greater than Δ \Delta Δ .

In particular, if this pre-determined difference is zero ( Δ = 0 \Delta = 0 Δ = 0 ):

The null hypothesis is that the population means are equal.

The alternate hypothesis is that the population means are:

• μ 1 \mu_1 μ 1 ​ and μ 2 \mu_2 μ 2 ​ are different from one another;
• μ 1 \mu_1 μ 1 ​ is smaller than μ 2 \mu_2 μ 2 ​ ; and
• μ 1 \mu_1 μ 1 ​ is greater than μ 2 \mu_2 μ 2 ​ .

Formally, to perform a t-test, we should additionally assume that the variances of the two populations are equal (this assumption is called the homogeneity of variance ).

There is a version of a t-test that can be applied without the assumption of homogeneity of variance: it is called a Welch's t-test . For your convenience, we describe both versions.

Two-sample t-test if variances are equal

Use this test if you know that the two populations' variances are the same (or very similar).

Two-sample t-test formula (with equal variances) :

where s p s_p s p ​ is the so-called pooled standard deviation , which we compute as:

• Δ \Delta Δ — Mean difference postulated in the null hypothesis;
• n 1 n_1 n 1 ​ — First sample size;
• x ˉ 1 \bar{x}_1 x ˉ 1 ​ — Mean for the first sample;
• s 1 s_1 s 1 ​ — Standard deviation in the first sample;
• n 2 n_2 n 2 ​ — Second sample size;
• x ˉ 2 \bar{x}_2 x ˉ 2 ​ — Mean for the second sample; and
• s 2 s_2 s 2 ​ — Standard deviation in the second sample.

Number of degrees of freedom in t-test (two samples, equal variances) = n 1 + n 2 − 2 n_1 + n_2 - 2 n 1 ​ + n 2 ​ − 2 .

Two-sample t-test if variances are unequal (Welch's t-test)

Use this test if the variances of your populations are different.

Two-sample Welch's t-test formula if variances are unequal:

• s 1 s_1 s 1 ​ — Standard deviation in the first sample;
• s 2 s_2 s 2 ​ — Standard deviation in the second sample.

The number of degrees of freedom in a Welch's t-test (two-sample t-test with unequal variances) is very difficult to count. We can approximate it with the help of the following Satterthwaite formula :

Alternatively, you can take the smaller of n 1 − 1 n_1 - 1 n 1 ​ − 1 and n 2 − 1 n_2 - 1 n 2 ​ − 1 as a conservative estimate for the number of degrees of freedom.

🔎 The Satterthwaite formula for the degrees of freedom can be rewritten as a scaled weighted harmonic mean of the degrees of freedom of the respective samples: n 1 − 1 n_1 - 1 n 1 ​ − 1 and n 2 − 1 n_2 - 1 n 2 ​ − 1 , and the weights are proportional to the standard deviations of the corresponding samples.

As we commonly perform a paired t-test when we have data about the same subjects measured twice (before and after some treatment), let us adopt the convention of referring to the samples as the pre-group and post-group.

The null hypothesis is that the true difference between the means of pre- and post-populations is equal to some pre-set value, Δ \Delta Δ .

The alternative hypothesis is that the actual difference between these means is:

Typically, this pre-determined difference is zero. We can then reformulate the hypotheses as follows:

The null hypothesis is that the pre- and post-means are the same, i.e., the treatment has no impact on the population .

The alternative hypothesis:

• The pre- and post-means are different from one another (treatment has some effect);
• The pre-mean is smaller than the post-mean (treatment increases the result); or
• The pre-mean is greater than the post-mean (treatment decreases the result).

Paired t-test formula

In fact, a paired t-test is technically the same as a one-sample t-test! Let us see why it is so. Let x 1 , . . . , x n x_1, ... , x_n x 1 ​ , ... , x n ​ be the pre observations and y 1 , . . . , y n y_1, ... , y_n y 1 ​ , ... , y n ​ the respective post observations. That is, x i , y i x_i, y_i x i ​ , y i ​ are the before and after measurements of the i -th subject.

For each subject, compute the difference, d i : = x i − y i d_i := x_i - y_i d i ​ := x i ​ − y i ​ . All that happens next is just a one-sample t-test performed on the sample of differences d 1 , . . . , d n d_1, ... , d_n d 1 ​ , ... , d n ​ . Take a look at the formula for the T-score :

Δ \Delta Δ — Mean difference postulated in the null hypothesis;

n n n — Size of the sample of differences, i.e., the number of pairs;

x ˉ \bar{x} x ˉ — Mean of the sample of differences; and

s s s  — Standard deviation of the sample of differences.

Number of degrees of freedom in t-test (paired): n − 1 n - 1 n − 1

We use a Z-test when we want to test the population mean of a normally distributed dataset, which has a known population variance . If the number of degrees of freedom is large, then the t-Student distribution is very close to N(0,1).

Hence, if there are many data points (at least 30), you may swap a t-test for a Z-test, and the results will be almost identical. However, for small samples with unknown variance, remember to use the t-test because, in such cases, the t-Student distribution differs significantly from the N(0,1)!

🙋 Have you concluded you need to perform the z-test? Head straight to our z-test calculator !

What is a t-test?

A t-test is a widely used statistical test that analyzes the means of one or two groups of data. For instance, a t-test is performed on medical data to determine whether a new drug really helps.

What are different types of t-tests?

Different types of t-tests are:

• One-sample t-test;
• Two-sample t-test; and
• Paired t-test.

How to find the t value in a one sample t-test?

To find the t-value:

• Subtract the null hypothesis mean from the sample mean value.
• Divide the difference by the standard deviation of the sample.
• Multiply the resultant with the square root of the sample size.

Expected value

Ideal egg boiling, relative risk, schwarzschild radius.

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One sample t-test

The t-test is one of the most common hypothesis tests in statistics. The t-test determines either whether the sample mean and the mean of the population differ or if two sample means differ statistically. The t-test distinguishes between

• one sample t-test
• independent sample t-test
• paired samples t-test

The choice of which t-test to use depends on whether one or two samples are available. If two samples are available, a distinction is made between dependent and independent samples. In this tutorial you will find everything about the one sample t-test .

Tip: Do you want to calculate the t-value? You can easily calculate it for all three t-tests online in the t-test calculator on DATAtab

The one sample t-test is used to test whether the population differs from a fixed value. So, the question is: Are there statistically significant differences between a sample mean and the fixed value? The set value may, for example, reflect the remaining population percentage or a set quality target that is to be controlled.

Social science example:

You want to find out whether the health perception of managers in Canada differs from that of the population as a whole. For this purpose you ask 50 managers about their perception of health.

Technical example:

You want to find out if the screws your company produces really weigh 10 grams on average. To test this, weigh 50 screws and compare the actual weight with the weight they should have (10 grams).

Medical example:

A pharmaceutical company promises that its new drug lowers blood pressure by 10 mmHg in one week. You want to find out if this is correct. To do this, compare the observed reduction in blood pressure of 75 test subjects with the expected reduction of 10 mmHg.

Assumptions

In a one sample t-test, the data under consideration must be from a random sample, have metric scale of measurement , and be normally distributed.

One tailed and two tailed t-test

So if you want to know whether a sample differs from the population, you have to calculate a one sample t-test . But before the t-test can be calculated, a question and the hypotheses must first be defined. This determines whether a one tailed (directional) or a two tailed (non-directional) t-test must be calculated.

The question helps you to define the object of investigation. In the case of the one sample t-test the question is:

Two tailed (non-directional)

Is there a statistically significant difference between the mean value of the sample and the population?

One tailed (directional)

Is the mean value of the sample significantly larger (or smaller) than the mean value of the population?

For the examples above, this gives us the following questions:

• Does the health perception of managers in Canada differ from that of the overall population in Canada?
• Does the production plant produce screws with a weight of 10 grams?
• Does the new drug lower blood pressure by 10 mmHg within one week?

Hypotheses t-Test

In order to perform a one sample t-test, the following hypotheses are formulated:

• Null hypothesis H 0 : The mean value of the population is equal to the specified value.
• Alternative hypothesis H 1 : The mean value of the population is not equal to the specified value.
• Null hypothesis H 0 : The mean value of the population is equal to or greater than (or less than) that of the specified value.
• Alternative hypothesis H 1 : The mean value of the population is smaller (or larger) than the specified values.

One sample t-test equation

You can calculate the t-test either with a statistics software like DATAtab or by hand. For the calculation by hand you first need the test statistics t , which can be calculated for the one sample t-test with the equation

In order to check whether the mean sample value differs significantly from that of the population, the critical t-value must be calculated. First the number of degrees of freedom, abbreviated df , is required, which is calculated by taking the number of samples minus one.

where the standard deviation is the population standard deviation estimated using the sample.

If the number of degrees of freedom is known, the critical t-value can be determined using the table of t-values . For a sample of 12 people, the degree of freedom is 11, and the significance level is assumed to be 5 %. The table below shows the t values for a one tailed open distribution. Depending on whether you want to calculate a one tailed (directional) or two tailed (non-directional) t-test, you must read the t value at either 0.95 or 0.975. For the non-directional hypothesis and an significance level of 5%, the critical t-value is 2.201.

If the calculated t value is below the critical t value, there is no significant difference between the sample and the population; if it is above the critical t value, there is a significant difference.

Interpret t-value

The t-value is calculated by dividing the measured difference by the scatter in the sample data. The larger the magnitude of t, the more this argues against the null hypothesis. If the calculated t-value is larger than the critical t-value, the null hypothesis is rejected.

Number of degrees of freedom - df

The number of degrees of freedom indicates how many values are allowed to vary freely. The degrees of freedom are therefore the number of independent individual pieces of information.

One sample t-test example

As an example for the t-test for one sample, we examine whether an online statistics tutorial newly introduced at the university has an effect on the students' examination results.

The average score in the statistics test at a university has been 28 points for years. This semester a new online statistics tutorial was introduced. Now the course management would like to know whether the success of the studies has changed since the introduction of the statistics tutorial: Does the online statistics tutorial have a positive effect on exam results?

The population considered is all students who have written the statistics exam since the new statistics tutorial was introduced. The reference value to be compared is 28.

Null hypothesis H0

The mean value from the sample and the predefined value does not differ significantly. The online statistics tutorial has no significant effect on exam results.

Here's how it goes on DATAtab:

Do you want to calculate a t-test independently? Calculate the example in the Statistics Calculator. Just copy the upper table including the first row into the t-Test Calculator . Datatab will then provide you with the tables below.

The following results are obtained with DATAtab: The mean value is 32.33 and the standard deviation 5.46. This leads to a standard error of the mean value of 1.57. The t-statistic thus gives 2.75

You would now like to know whether your hypothesis (the score is 28) is significant or not. To do this, you first specify a significance level in Datatab, usually 5% is used, which is preselected. Now you will get the table below in Datatab.

One sample t-test (Test Value = 28)

95% confidence interval of the difference.

To interpret whether your hypothesis is significant one of the two values can be used:

• p-value (2-tailed)
• lower and upper confidence interval of the difference

In this example p-value (2-tailed) is equal to 0.02, i.e. 2 %. Put into words this means: The probability that a sample with a mean difference of 4.33 or more will be drawn from the population is 2%. The significance level was set at 5%, which is greater than 2%. For this reason, a significant difference between the sample and the population is assumed.

Whether or not there is a significant difference can also be read from the confidence interval of the difference. If the lower and upper limits go throw zero, there is no significant difference. If this is not the case, there is a significant difference. In this example, the lower value is 0.86 and the upper value is 7.81. Since the lower and upper values do not touch zero, there is a significant difference.

APA Style | One sample t-test

If we were to write the top results for publication in an APA journal, that is, in an APA format, we would write it that way:

A t-test showed a statistically reliable difference between the score of students who attended the online course and the average score of students who did not attend an online course. (M = 32.33, s = 5.47) and 28, t(11) = 2.75, p < 0.02, α = 0.05.

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Cite DATAtab: DATAtab Team (2024). DATAtab: Online Statistics Calculator. DATAtab e.U. Graz, Austria. URL https://datatab.net

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SPSS Annotated Output T-test

The t-test procedure performs t-tests for one sample, two samples and paired observations.  The single-sample t-test compares the mean of the sample to a given number (which you supply).  The independent samples t-test compares the difference in the means from the two groups to a given value (usually 0). In other words, it tests whether the difference in the means is 0.  The dependent-sample or paired t-test compares the difference in the means from the two variables measured on the same set of subjects to a given number (usually 0), while taking into account the fact that the scores are not independent.  In our examples, we will use the hsb2 data set.

Single sample t-test

The single sample t-test tests the null hypothesis that the population mean is equal to the number specified by the user.  SPSS calculates the t-statistic and its p-value under the assumption that the sample comes from an approximately normal distribution. If the p-value associated with the t-test is small (0.05 is often used as the threshold), there is evidence that the mean is different from the hypothesized value.  If the p-value associated with the t-test is not small (p > 0.05), then the null hypothesis is not rejected and you can conclude that the mean is not different from the hypothesized value.

In this example, the t-statistic is 4.140 with 199 degrees of freedom.  The corresponding two-tailed p-value is .000, which is less than 0.05.  We conclude that the mean of variable write is different from 50.

One-Sample Statistics

a.  – This is the list of variables.  Each variable that was listed on the variables= statement in the above code will have its own line in this part of the output.

b.  N – This is the number of valid (i.e., non-missing) observations used in calculating the t-test.

c.  Mean – This is the mean of the variable.

d.  Std. Deviation – This is the standard deviation of the variable.

e.  Std. Error Mean – This is the estimated standard deviation of the sample mean.  If we drew repeated samples of size 200, we would expect the standard deviation of the sample means to be close to the standard error.  The standard deviation of the distribution of sample mean is estimated as the standard deviation of the sample divided by the square root of sample size: 9.47859/(sqrt(200)) = .67024.

Test statistics

f. – This identifies the variables.  Each variable that was listed on the variables= statement will have its own line in this part of the output.  If a variables= statement is not specified, t-test will conduct a t-test on all numerical variables in the dataset.

g.  t – This is the Student t-statistic.  It is the ratio of the difference between the sample mean and the given number to the standard error of the mean: (52.775 – 50) / .6702372 = 4.1403. Since the standard error of the mean measures the variability of the sample mean, the smaller the standard error of the mean, the more likely that our sample mean is close to the true population mean.  This is illustrated by the following three figures.

In all three cases, the difference between the population means is the same. But with large variability of sample means, second graph, two populations overlap a great deal.  Therefore, the difference may well come by chance.  On the other hand, with small variability, the difference is more clear as in the third graph.  The smaller the standard error of the mean, the larger the magnitude of the t-value and therefore, the smaller the p-value.

h.  df – The degrees of freedom for the single sample t-test is simply the number of valid observations minus 1.  We lose one degree of freedom because we have estimated the mean from the sample.  We have used some of the information from the data to estimate the mean, therefore it is not available to use for the test and the degrees of freedom accounts for this.

i.   Sig (2-tailed) – This is the two-tailed p-value evaluating the null against an alternative that the mean is not equal to 50. It is equal to the probability of observing a greater absolute value of t under the null hypothesis.  If the p-value is less than the pre-specified alpha level (usually .05 or .01) we will conclude that mean is statistically significantly different from zero.  For example, the p-value is smaller than 0.05. So we conclude that the mean for  write is different from 50.

j.   Mean Difference – This is the difference between the sample mean and the test value.

k.  95% Confidence Interval of the Difference – These are the lower and upper bound of the confidence interval for the mean. A confidence interval for the mean specifies a range of values within which the unknown population parameter, in this case the mean, may lie.  It is given by

where s is the sample deviation of the observations and N is the number of valid observations.  The t-value in the formula can be computed or found in any statistics book with the degrees of freedom being N-1 and the p-value being 1- alpha /2, where alpha is the confidence level and by default is .95.

Paired t-test

A paired (or “dependent”) t-test is used when the observations are not independent of one another. In the example below, the same students took both the writing and the reading test. Hence, you would expect there to be a relationship between the scores provided by each student.  The paired t-test accounts for this.  For each student, we are essentially looking at the differences in the values of the two variables and testing if the mean of these differences is equal to zero.

In this example, the t-statistic is 0.8673 with 199 degrees of freedom.  The corresponding two-tailed p-value is 0.3868, which is greater than 0.05.  We conclude that the mean difference of write and read is not different from 0.

Summary statistics

a.  – This is the list of variables.

b.  Mean – These are the respective means of the variables.

c.  N – This is the number of valid (i.e., non-missing) observations used in calculating the t-test.

d.  Std. Deviation – This is the standard deviations of the variables.

e.  Std Error Mean – Standard Error Mean is the estimated standard deviation of the sample mean.  This value is estimated as the standard deviation of one sample divided by the square root of sample size: 9.47859/sqrt(200) = .67024, 10.25294/sqrt(200) = .72499. This provides a measure of the variability of the sample mean.

f. Correlation – This is the correlation coefficient of the pair of variables indicated.  This is a measure of the strength and direction of the linear relationship between the two variables.  The correlation coefficient can range from -1 to +1, with -1 indicating a perfect negative correlation, +1 indicating a perfect positive correlation, and 0 indicating no correlation at all.  (A variable correlated with itself will always have a correlation coefficient of 1.)  You can think of the correlation coefficient as telling you the extent to which you can guess the value of one variable given a value of the other variable. The .597 is the numerical description of how tightly around the imaginary line the points lie. If the correlation was higher, the points would tend to be closer to the line; if it was smaller, they would tend to be further away from the line.

g. Sig – This is the p-value associated with the correlation. Here, correlation is significant at the .05 level.

g.  writing score-reading score – This is the value measured within each subject: the difference between the writing and reading scores.  The paired t-test forms a single random sample of the paired difference. The mean of these values among all subjects is compared to 0 in a paired t-test.

h.  Mean – This is the mean within-subject difference between the two variables.

i.   Std. Deviation – This is the standard deviation of the mean paired difference.

j.   Std Error Mean – This is the estimated standard deviation of the sample mean.  This value is estimated as the standard deviation of one sample divided by the square root of sample size: 8.88667/sqrt(200) = .62838.  This provides a measure of the variability of the sample mean.

k.  95% Confidence Interval of the Difference – These are the lower and upper bound of the confidence interval for the mean difference. A confidence interval for the mean specifies a range of values within which the unknown population parameter, in this case the mean, may lie.  It is given by

l.   t – This is the t-statistic.  It is the ratio of the mean of the difference to the standard error of the difference: (.545/.62838).

m. degrees of freedom – The degrees of freedom for the paired observations is simply the number of observations minus 1. This is because the test is conducted on the one sample of the paired differences.

n.  Sig. (2-tailed) – This is the two-tailed p-value computed using the t distribution.  It is the probability of observing a greater absolute value of t under the null hypothesis.  If the p-value is less than the pre-specified alpha level (usually .05 or .01, here the former) we will conclude that mean difference between writing score and reading score is statistically significantly different from zero.  For example, the p-value for the difference between the two variables is greater than 0.05 so we conclude that the mean difference is not statistically significantly different from 0.

Independent group t-test

This t-test is designed to compare means of same variable between two groups. In our example, we compare the mean writing score between the group of female students and the group of male students. Ideally, these subjects are randomly selected from a larger population of subjects. The test assumes that variances for the two populations are the same.  The interpretation for p-value is the same as in other type of t-tests.

In this example, the t-statistic is -3.7341 with 198 degrees of freedom.  The corresponding two-tailed p-value is 0.0002, which is less than 0.05.  We conclude that the difference of means in write between males and females is different from 0.

a.  female – This column gives categories of the independent variable female . This variable is necessary for doing the independent group t-test and is specified by the t-test groups=  statement.

b.  N – This is the number of valid (i.e., non-missing) observations in each group.

c.  Mean – This is the mean of the dependent variable for each level of the independent variable.

d.  Std. Deviation – This is the standard deviation of the dependent variable for each of the levels of the independent variable.

e.  Std. Error Mean – This is the standard error of the mean, the ratio of the standard deviation to the square root of the respective number of observations.

f. – This column lists the dependent variable(s).  In our example, the dependent variable is write (labeled “writing score”).

g. – This column specifies the method for computing the standard error of the difference of the means.  The method of computing this value is based on the assumption regarding the variances of the two groups. If we assume that the two populations have the same variance, then the first method, called pooled variance estimator, is used. Otherwise, when the variances are not assumed to be equal, the Satterthwaite’s method is used.

h.  F – This column lists Levene’s test statistic. Assume \(k\) is the number of groups, \(N\) is the total number of observations, and \(N_i\) is the number of observations in each \(i\)-th group for dependent variable \(Y_{ij}\). Then Levene’s test statistic is defined as

\begin{equation} W = \frac{(N-k)}{(k-1)} \frac{\sum_{i=1}^{k} N_i (\bar{Z}_{i.}-\bar{Z}_{..})^2}{\sum_{i=1}^{k}\sum_{j=1}^{N_i}(Z_{ij}-\bar{Z}_{i.})^2} \end{equation}

\begin{equation} Z_{ij} = |Y_{ij}-\bar{Y}_{i.}| \end{equation}

where \(\bar{Y}_{i.}\) is the mean of the dependent variable and \(\bar{Z}_{i.}\) is the mean of \(Z_{ij}\) for each \(i\)-th group respectively, and \(\bar{Z}_{..}\) is the grand mean of \(Z_{ij}\).

i.  Sig. – This is the two-tailed p-value associated with the null that the two groups have the same variance. In our example, the probability is less than 0.05. So there is evidence that the variances for the two groups, female students and male students, are different. Therefore, we may want to use the second method (Satterthwaite variance estimator) for our t-test.

j.  t – These are the t-statistics under the two different assumptions: equal variances and unequal variances.  These are the ratios of the mean of the differences to the standard errors of the difference under the two different assumptions: (-4.86995 / 1.30419) =  -3.734, (-4.86995/1.33189) = -3.656.

k.  df – The degrees of freedom when we assume equal variances is simply the sum of the two sample sizes (109 and 91) minus 2. The degrees of freedom when we assume unequal variances is calculated using the Satterthwaite formula.

l.   Sig. (2-tailed) – The p-value is the two-tailed probability computed using the t distribution.  It is the probability of observing a t-value of equal or greater absolute value under the null hypothesis.  For a one-tailed test, halve this probability.  If the p-value is less than our pre-specified alpha level, usually 0.05, we will conclude that the difference is significantly different from zero.  For example, the p-value for the difference between females and males is less than 0.05 in both cases, so we conclude that the difference in means is statistically significantly different from 0.

m. Mean Difference – This is the difference between the means.

n. Std Error Difference – Standard Error difference is the estimated standard deviation of the difference between the sample means.  If we drew repeated samples of size 200, we would expect the standard deviation of the sample means to be close to the standard error. This provides a measure of the variability of the sample mean.  The Central Limit Theorem tells us that the sample means are approximately normally distributed when the sample size is 30 or greater.  Note that the standard error difference is calculated differently under the two different assumptions.

o.   95% Confidence Interval of the Difference – These are the lower and upper bound of the confidence interval for the mean difference. A confidence interval for the mean specifies a range of values within which the unknown population parameter, in this case the mean, may lie.  It is given by

where s is the sample deviation of the observations and N is the number of valid observations.  The t-value in the formula can be computed or found in any statistics book with the degrees of freedom being N-1 and the p-value being 1- width /2, where width is the confidence level and by default is .95.

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• One sample t test

A one sample t test compares the mean with a hypothetical value. In most cases, the hypothetical value comes from theory. For example, if you express your data as 'percent of control', you can test whether the average differs significantly from 100. The hypothetical value can also come from previous data. For example, compare whether the mean systolic blood pressure differs from 135, a value determined in a previous study.

1. Choose data entry format

Caution: Changing format will erase your data.

2. Specify the hypothetical mean value

3. enter data, 4. view the results, learn more about the one sample t test.

In this article you will learn the requirements and assumptions of a one sample t test, how to format and interpret the results of a one sample t test, and when to use different types of t tests.

One sample t test: Overview

The one sample t test, also referred to as a single sample t test, is a statistical hypothesis test used to determine whether the mean calculated from sample data collected from a single group is different from a designated value specified by the researcher. This designated value does not come from the data itself, but is an external value chosen for scientific reasons. Often, this designated value is a mean previously established in a population, a standard value of interest, or a mean concluded from other studies. Like all hypothesis testing, the one sample t test determines if there is enough evidence reject the null hypothesis (H0) in favor of an alternative hypothesis (H1). The null hypothesis for a one sample t test can be stated as: "The population mean equals the specified mean value." The alternative hypothesis for a one sample t test can be stated as: "The population mean is different from the specified mean value."

The one sample t test differs from most statistical hypothesis tests because it does not compare two separate groups or look at a relationship between two variables. It is a straightforward comparison between data gathered on a single variable from one population and a specified value defined by the researcher. The one sample t test can be used to look for a difference in only one direction from the standard value (a one-tailed t test ) or can be used to look for a difference in either direction from the standard value (a two-tailed t test ).

Requirements and Assumptions for a one sample t test

A one sample t test should be used only when data has been collected on one variable for a single population and there is no comparison being made between groups. For a valid one sample t test analysis, data values must be all of the following:

The one sample t test assumes that all "errors" in the data are independent. The term "error" refers to the difference between each value and the group mean. The results of a t test only make sense when the scatter is random - that whatever factor caused a value to be too high or too low affects only that one value. Prism cannot test this assumption, but there are graphical ways to explore data to verify this assumption is met.

A t test is only appropriate to apply in situations where data represent variables that are continuous measurements. As they rely on the calculation of a mean value, variables that are categorical should not be analyzed using a t test.

The results of a t test should be based on a random sample and only be generalized to the larger population from which samples were drawn.

As with all parametric hypothesis testing, the one sample t test assumes that you have sampled your data from a population that follows a normal (or Gaussian) distribution. While this assumption is not as important with large samples, it is important with small sample sizes, especially less than 10. If your data do not come from a Gaussian distribution , there are three options to accommodate this. One option is to transform the values to make the distribution more Gaussian, perhaps by transforming all values to their reciprocals or logarithms. Another choice is to use the Wilcoxon signed rank nonparametric test instead of the t test. A final option is to use the t test anyway, knowing that the t test is fairly robust to departures from a Gaussian distribution with large samples.

How to format a one sample t test

Ideally, data for a one sample t test should be collected and entered as a single column from which a mean value can be easily calculated. If data is entered on a table with multiple subcolumns, Prism requires one of the following choices to be selected to perform the analysis:

• Each subcolumn of data can be analyzed separately
• An average of the values in the columns across each row can be calculated, and the analysis conducted on this new stack of means, or
• All values in all columns can be treated as one sample of data (paying no attention to which row or column any values are in).

How the one sample t test calculator works

Prism calculates the t ratio by dividing the difference between the actual and hypothetical means by the standard error of the actual mean. The equation is written as follows, where x is the calculated mean, μ is the hypothetical mean (specified value), S is the standard deviation of the sample, and n is the sample size:

A p value is computed based on the calculated t ratio and the numbers of degrees of freedom present (which equals sample size minus 1). The one sample t test calculator assumes it is a two-tailed one sample t test, meaning you are testing for a difference in either direction from the specified value.

How to interpret results of a one sample t test

As discussed, a one sample t test compares the mean of a single column of numbers against a hypothetical mean. This hypothetical mean can be based upon a specific standard or other external prediction. The test produces a P value which requires careful interpretation.

The p value answers this question: If the data were sampled from a Gaussian population with a mean equal to the hypothetical value you entered, what is the chance of randomly selecting N data points and finding a mean as far (or further) from the hypothetical value as observed here?

If the p value is large (usually defined to mean greater than 0.05), the data do not give you any reason to conclude that the population mean differs from the designated value to which it has been compared. This is not the same as saying that the true mean equals the hypothetical value, but rather states that there is no evidence of a difference. Thus, we cannot reject the null hypothesis (H0).

If the p value is small (usually defined to mean less than or equal to 0.05), then it is unlikely that the discrepancy observed between the sample mean and hypothetical mean is due to a coincidence arising from random sampling. There is evidence to reject the idea that the difference is coincidental and conclude instead that the population has a mean that is different from the hypothetical value to which it has been compared. The difference is statistically significant, and the null hypothesis is therefore rejected.

If the null hypothesis is rejected, the question of whether the difference is scientifically important still remains. The confidence interval can be a useful tool in answering this question. Prism reports the 95% confidence interval for the difference between the actual and hypothetical mean. In interpreting these results, one can be 95% sure that this range includes the true difference. It requires scientific judgment to determine if this difference is truly meaningful.

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When to use different types of t tests

There are three types of t tests which can be used for hypothesis testing:

• Independent two-sample (or unpaired) t test
• Paired sample t test

As described, a one sample t test should be used only when data has been collected on one variable for a single population and there is no comparison being made between groups. It only applies when the mean value for data is intended to be compared to a fixed and defined number.

In most cases involving data analysis, however, there are multiple groups of data either representing different populations being compared, or the same population being compared at different times or conditions. For these situations, it is not appropriate to use a one sample t test. Other types of t tests are appropriate for these specific circumstances:

Independent Two-Sample t test (Unpaired t test)

The independent sample t test, also referred to as the unpaired t test, is used to compare the means of two different samples. The independent two-sample t test comes in two different forms:

• the standard Student's t test, which assumes that the variance of the two groups are equal.
• the Welch's t test , which is less restrictive compared to the original Student's test. This is the test where you do not assume that the variance is the same in the two groups, which results in fractional degrees of freedom.

The two methods give very similar results when the sample sizes are equal and the variances are similar.

Paired Sample t test

The paired sample t test is used to compare the means of two related groups of samples. Put into other words, it is used in a situation where you have two values (i.e., a pair of values) for the same group of samples. Often these two values are measured from the same samples either at two different times, under two different conditions, or after a specific intervention.

You can perform multiple independent two-sample comparison tests simultaneously in Prism. Select from parametric and nonparametric tests and specify if the data are unpaired or paired. Try performing a t test with a 30-day free trial of Prism .

Watch this video to learn how to choose between a paired and unpaired t test.

Example of how to apply the appropriate t test

"Alkaline" labeled bottled drinking water has become fashionable over the past several years. Imagine we have collected a random sample of 30 bottles of "alkaline" drinking water from a number of different stores to represent the population of "alkaline" bottled water for a particular brand available to the general consumer. The labels on each of the bottles claim that the pH of the "alkaline" water is 8.5. A laboratory then proceeds to measure the exact pH of the water in each bottle.

Table 1: pH of water in random sample of "alkaline bottled water"

If you look at the table above, you see that some bottles have a pH measured to be lower than 8.5, while other bottles have a pH measured to be higher. What can the data tell us about the actual pH levels found in this brand of "alkaline" water bottles marketed to the public as having a pH of 8.5? Statistical hypothesis testing provides a sound method to evaluate this question. Which specific test to use, however, depends on the specific question being asked.

Is a t test appropriate to apply to this data?

Let's start by asking: Is a t test an appropriate method to analyze this set of pH data? The following list reviews the requirements and assumptions for using a t test:

• Independent sampling : In an independent sample t test, the data values are independent. The pH of one bottle of water does not depend on the pH of any other water bottle. (An example of dependent values would be if you collected water bottles from a single production lot. A sample from a single lot is representative only of that lot, not of alkaline bottled water in general).
• Continuous variable : The data values are pH levels, which are numerical measurements that are continuous.
• Random sample : We assume the water bottles are a simple random sample from the population of "alkaline" water bottles produced by this brand as they are a mix of many production lots.
• Normal distribution : We assume the population from which we collected our samples has pH levels that are normally distributed. To verify this, we should visualize the data graphically. The figure below shows a histogram for the pH measurements of the water bottles. From a quick look at the histogram, we see that there are no unusual points, or outliers. The data look roughly bell-shaped, so our assumption of a normal distribution seems reasonable. The QQ plot can also be used to graphically assess normality and is the preferred choice when the sample size is small.

Based upon these features and assumptions being met, we can conclude that a t test is an appropriate method to be applied to this set of data.

Which t test is appropriate to use?

The next decision is which t test to apply, and this depends on the exact question we would like our analysis to answer. This example illustrates how each type of t test could be chosen for a specific analysis, and why the one sample t test is the correct choice to determine if the measured pH of the bottled water samples match the advertised pH of 8.5.

We could be interested in determining whether a certain characteristic of a water bottle is associated with having a higher or lower pH, such as whether bottles are glass or plastic. For this questions, we would effectively be dividing the bottles into 2 separate groups and comparing the means of the pH between the 2 groups. For this analysis, we would elect to use a two sample t test because we are comparing the means of two independent groups.

We could also be interested in learning if pH is affected by a water bottle being opened and exposed to the air for a week. In this case, each original sample would be tested for pH level after a week had elapsed and the water had been exposed to the air, creating a second set of sample data. To evaluate whether this exposure affected pH, we would again be comparing two different groups of data, but this time the data are in paired samples each having an original pH measurement and a second measurement from after the week of exposure to the open air. For this analysis, it is appropriate to use a paired t test so that data for each bottle is assembled in rows, and the change in pH is considered bottle by bottle.

Returning to the original question we set out to answer-whether bottled water that is advertised to have a pH of 8.5 actually meets this claim-it is now clear that neither an independent two sample t test or a paired t test would be appropriate. In this case, all 30 pH measurements are sampled from one group representing bottled drinking water labeled "alkaline" available to the general consumer. We wish to compare this measured mean with an expected advertised value of 8.5. This is the exact situation for which one should employ a one sample t test!

From a quick look at the descriptive statistics, we see that the mean of the sample measurements is 8.513, slightly above 8.5. Does this average from our sample of 30 bottles validate the advertised claim of pH 8.5? By applying Prism's one sample t test analysis to this data set, we will get results by which we can evaluate whether the null hypothesis (that there is no difference between the mean pH level in the water bottles and the pH level advertised on the bottles) should be accepted or rejected.

How to Perform a One Sample T Test in Prism

In prior versions of Prism, the one sample t test and the Wilcoxon rank sum tests were computed as part of Prism's Column Statistics analysis. Now, starting with Prism 8, performing one sample t tests is even easier with a separate analysis in Prism.

Steps to perform a one sample t test in Prism

• Create a Column data table.
• Enter each data set in a single Y column so all values from each group are stacked into a column. Prism will perform a one sample t test (or Wilcoxon rank sum test) on each column you enter.
• Click Analyze, look in the list of Column analyses, and choose one sample t test and Wilcoxon test.

It's that simple! Prism streamlines your t test analysis so you can make more accurate and more informed data interpretations. Start your 30-day free trial of Prism and try performing your first one sample t test in Prism.

Watch this video for a step-by-step tutorial on how to perform a t test in Prism.

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• Null and Alternative Hypotheses | Definitions & Examples

Null & Alternative Hypotheses | Definitions, Templates & Examples

Published on May 6, 2022 by Shaun Turney . Revised on June 22, 2023.

The null and alternative hypotheses are two competing claims that researchers weigh evidence for and against using a statistical test :

• Null hypothesis ( H 0 ): There’s no effect in the population .
• Alternative hypothesis ( H a or H 1 ) : There’s an effect in the population.

Table of contents

Answering your research question with hypotheses, what is a null hypothesis, what is an alternative hypothesis, similarities and differences between null and alternative hypotheses, how to write null and alternative hypotheses, other interesting articles, frequently asked questions.

The null and alternative hypotheses offer competing answers to your research question . When the research question asks “Does the independent variable affect the dependent variable?”:

• The null hypothesis ( H 0 ) answers “No, there’s no effect in the population.”
• The alternative hypothesis ( H a ) answers “Yes, there is an effect in the population.”

The null and alternative are always claims about the population. That’s because the goal of hypothesis testing is to make inferences about a population based on a sample . Often, we infer whether there’s an effect in the population by looking at differences between groups or relationships between variables in the sample. It’s critical for your research to write strong hypotheses .

You can use a statistical test to decide whether the evidence favors the null or alternative hypothesis. Each type of statistical test comes with a specific way of phrasing the null and alternative hypothesis. However, the hypotheses can also be phrased in a general way that applies to any test.

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The null hypothesis is the claim that there’s no effect in the population.

If the sample provides enough evidence against the claim that there’s no effect in the population ( p ≤ α), then we can reject the null hypothesis . Otherwise, we fail to reject the null hypothesis.

Although “fail to reject” may sound awkward, it’s the only wording that statisticians accept . Be careful not to say you “prove” or “accept” the null hypothesis.

Null hypotheses often include phrases such as “no effect,” “no difference,” or “no relationship.” When written in mathematical terms, they always include an equality (usually =, but sometimes ≥ or ≤).

You can never know with complete certainty whether there is an effect in the population. Some percentage of the time, your inference about the population will be incorrect. When you incorrectly reject the null hypothesis, it’s called a type I error . When you incorrectly fail to reject it, it’s a type II error.

Examples of null hypotheses

The table below gives examples of research questions and null hypotheses. There’s always more than one way to answer a research question, but these null hypotheses can help you get started.

*Note that some researchers prefer to always write the null hypothesis in terms of “no effect” and “=”. It would be fine to say that daily meditation has no effect on the incidence of depression and p 1 = p 2 .

The alternative hypothesis ( H a ) is the other answer to your research question . It claims that there’s an effect in the population.

Often, your alternative hypothesis is the same as your research hypothesis. In other words, it’s the claim that you expect or hope will be true.

The alternative hypothesis is the complement to the null hypothesis. Null and alternative hypotheses are exhaustive, meaning that together they cover every possible outcome. They are also mutually exclusive, meaning that only one can be true at a time.

Alternative hypotheses often include phrases such as “an effect,” “a difference,” or “a relationship.” When alternative hypotheses are written in mathematical terms, they always include an inequality (usually ≠, but sometimes < or >). As with null hypotheses, there are many acceptable ways to phrase an alternative hypothesis.

Examples of alternative hypotheses

The table below gives examples of research questions and alternative hypotheses to help you get started with formulating your own.

Null and alternative hypotheses are similar in some ways:

• They’re both answers to the research question.
• They both make claims about the population.
• They’re both evaluated by statistical tests.

However, there are important differences between the two types of hypotheses, summarized in the following table.

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To help you write your hypotheses, you can use the template sentences below. If you know which statistical test you’re going to use, you can use the test-specific template sentences. Otherwise, you can use the general template sentences.

General template sentences

The only thing you need to know to use these general template sentences are your dependent and independent variables. To write your research question, null hypothesis, and alternative hypothesis, fill in the following sentences with your variables:

Does independent variable affect dependent variable ?

• Null hypothesis ( H 0 ): Independent variable does not affect dependent variable.
• Alternative hypothesis ( H a ): Independent variable affects dependent variable.

Test-specific template sentences

Once you know the statistical test you’ll be using, you can write your hypotheses in a more precise and mathematical way specific to the test you chose. The table below provides template sentences for common statistical tests.

Note: The template sentences above assume that you’re performing one-tailed tests . One-tailed tests are appropriate for most studies.

If you want to know more about statistics , methodology , or research bias , make sure to check out some of our other articles with explanations and examples.

• Normal distribution
• Descriptive statistics
• Measures of central tendency
• Correlation coefficient

Methodology

• Cluster sampling
• Stratified sampling
• Types of interviews
• Cohort study
• Thematic analysis

Research bias

• Implicit bias
• Cognitive bias
• Survivorship bias
• Availability heuristic
• Nonresponse bias
• Regression to the mean

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics. It is used by scientists to test specific predictions, called hypotheses , by calculating how likely it is that a pattern or relationship between variables could have arisen by chance.

Null and alternative hypotheses are used in statistical hypothesis testing . The null hypothesis of a test always predicts no effect or no relationship between variables, while the alternative hypothesis states your research prediction of an effect or relationship.

The null hypothesis is often abbreviated as H 0 . When the null hypothesis is written using mathematical symbols, it always includes an equality symbol (usually =, but sometimes ≥ or ≤).

The alternative hypothesis is often abbreviated as H a or H 1 . When the alternative hypothesis is written using mathematical symbols, it always includes an inequality symbol (usually ≠, but sometimes < or >).

A research hypothesis is your proposed answer to your research question. The research hypothesis usually includes an explanation (“ x affects y because …”).

A statistical hypothesis, on the other hand, is a mathematical statement about a population parameter. Statistical hypotheses always come in pairs: the null and alternative hypotheses . In a well-designed study , the statistical hypotheses correspond logically to the research hypothesis.

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One-sample and paired-sample t -test

Description

h = ttest( x ) returns a test decision for the null hypothesis that the data in x comes from a normal distribution with mean equal to zero and unknown variance, using the one-sample t -test . The alternative hypothesis is that the population distribution does not have a mean equal to zero. The result h is 1 if the test rejects the null hypothesis at the 5% significance level, and 0 otherwise.

h = ttest( x , y ) returns a test decision for the null hypothesis that the data in x – y comes from a normal distribution with mean equal to zero and unknown variance, using the paired-sample t -test.

h = ttest( x , y , Name,Value ) returns a test decision for the paired-sample t -test with additional options specified by one or more name-value pair arguments. For example, you can change the significance level or conduct a one-sided test.

h = ttest( x , m ) returns a test decision for the null hypothesis that the data in x comes from a normal distribution with mean m and unknown variance. The alternative hypothesis is that the mean is not m .

h = ttest( x , m , Name,Value ) returns a test decision for the one-sample t -test with additional options specified by one or more name-value pair arguments. For example, you can change the significance level or conduct a one-sided test.

[ h , p ] = ttest( ___ ) also returns the p -value, p , of the test, using any of the input arguments from the previous syntax groups.

[ h , p , ci , stats ] = ttest( ___ ) also returns the confidence interval ci for the mean of x , or of x – y for the paired t -test, and the structure stats containing information about the test statistic.

collapse all

t -Test for Mean Equal to Zero

Load the sample data. Create a vector containing the third column of the stock returns data.

Test the null hypothesis that the sample data comes from a population with mean equal to zero.

The returned value h = 1 indicates that ttest rejects the null hypothesis at the 5% significance level.

t -Test at Different Significance Level

Test the null hypothesis that the sample data are from a population with mean equal to zero at the 1% significance level.

The returned value h = 0 indicates that ttest does not reject the null hypothesis at the 1% significance level.

Paired-Sample t -Test

Load the sample data. Create vectors containing the first and second columns of the data matrix to represent students’ grades on two exams.

Test the null hypothesis that the pairwise difference between data vectors x and y has a mean equal to zero.

The returned value of h = 0 indicates that ttest does not reject the null hypothesis at the default 5% significance level.

Paired-Sample t -Test at Different Significance Level

Test the null hypothesis that the pairwise difference between data vectors x and y has a mean equal to zero at the 1% significance level.

The returned value of h = 0 indicates that ttest does not reject the null hypothesis at the 1% significance level.

t -Test for a Hypothesized Mean

Load the sample data. Create a vector containing the first column of the students' exam grades data.

Test the null hypothesis that sample data comes from a distribution with mean m = 75 .

The returned value of h = 0 indicates that ttest does not reject the null hypothesis at the 5% significance level.

One-Sided t -Test

Load the sample data. Create a vector containing the first column of the students’ exam grades data.

Plot a histogram of the exam grades data and fit a normal density function.

Use a right-tailed t -test to test the null hypothesis that the data comes from a population with mean equal to 65, against the alternative that the mean is greater than 65.

The returned value of h = 1 indicates that ttest rejects the null hypothesis at the default significance level of 5%, in favor of the alternative hypothesis that the data comes from a population with a mean greater than 65.

Plot the corresponding Student's t- distribution, the returned t -statistic, and the critical t -value. Calculate the critical t -value for the default confidence level of 95% by using tinv .

The orange dot represents the t -statistic and is located to the right of the dashed black line that represents the critical t -value.

Input Arguments

X — sample data vector | matrix | multidimensional array.

Sample data, specified as a vector, matrix, or multidimensional array . ttest performs a separate t -test along each column and returns a vector of results. If y sample data is specified, x and y must be the same size.

Data Types: single | double

y — Sample data vector | matrix | multidimensional array

Sample data, specified as a vector, matrix, or multidimensional array . If y sample data is specified, x and y must be the same size.

m — Hypothesized population mean 0 (default) | scalar value

Hypothesized population mean, specified as a scalar value.

Name-Value Arguments

Specify optional pairs of arguments as Name1=Value1,...,NameN=ValueN , where Name is the argument name and Value is the corresponding value. Name-value arguments must appear after other arguments, but the order of the pairs does not matter.

Before R2021a, use commas to separate each name and value, and enclose Name in quotes.

Example: 'Tail','right','Alpha',0.01 conducts a right-tailed hypothesis test at the 1% significance level.

Alpha — Significance level 0.05 (default) | scalar value in the range (0,1)

Significance level of the hypothesis test, specified as the comma-separated pair consisting of 'Alpha' and a scalar value in the range (0,1).

Example: 'Alpha',0.01

Dim — Dimension first nonsingleton dimension (default) | positive integer value

Dimension of the input matrix along which to test the means, specified as the comma-separated pair consisting of 'Dim' and a positive integer value. For example, specifying 'Dim',1 tests the column means, while 'Dim',2 tests the row means.

Example: 'Dim',2

Tail — Type of alternative hypothesis 'both' (default) | 'right' | 'left'

Type of alternative hypothesis to evaluate, specified as the comma-separated pair consisting of 'Tail' and one of:

'both' — Test against the alternative hypothesis that the population mean is not m .

'right' — Test against the alternative hypothesis that the population mean is greater than m .

'left' — Test against the alternative hypothesis that the population mean is less than m .

ttest tests the null hypothesis that the population mean is m against the specified alternative hypothesis.

Example: 'Tail','right'

Output Arguments

H — hypothesis test result 1 | 0.

Hypothesis test result, returned as 1 or 0 .

If h = 1 , this indicates the rejection of the null hypothesis at the Alpha significance level.

If h = 0 , this indicates a failure to reject the null hypothesis at the Alpha significance level.

p — p -value scalar value in the range [0,1]

p -value of the test, returned as a scalar value in the range [0,1]. p is the probability of observing a test statistic as extreme as, or more extreme than, the observed value under the null hypothesis. Small values of p cast doubt on the validity of the null hypothesis.

ci — Confidence interval vector

Confidence interval for the true population mean, returned as a two-element vector containing the lower and upper boundaries of the 100 × (1 – Alpha )% confidence interval.

stats — Test statistics structure

Test statistics, returned as a structure containing the following:

tstat — Value of the test statistic.

df — Degrees of freedom of the test.

sd — Estimated population standard deviation. For a paired t -test, sd is the standard deviation of x – y .

One-Sample t-Test

The one-sample t -test is a parametric test of the location parameter when the population standard deviation is unknown.

The test statistic is

t = x ¯ − μ s / n ,

where x ¯ is the sample mean, μ is the hypothesized population mean, s is the sample standard deviation, and n is the sample size. Under the null hypothesis, the test statistic has Student’s t distribution with n – 1 degrees of freedom.

Multidimensional Array

A multidimensional array has more than two dimensions. For example, if x is a 1-by-3-by-4 array, then x is a three-dimensional array.

First Nonsingleton Dimension

The first nonsingleton dimension is the first dimension of an array whose size is not equal to 1. For example, if x is a 1-by-2-by-3-by-4 array, then the second dimension is the first nonsingleton dimension of x .

Use sampsizepwr to calculate:

The sample size that corresponds to specified power and parameter values;

The power achieved for a particular sample size, given the true parameter value;

The parameter value detectable with the specified sample size and power.

Extended Capabilities

Gpu arrays accelerate code by running on a graphics processing unit (gpu) using parallel computing toolbox™..

This function fully supports GPU arrays. For more information, see Run MATLAB Functions on a GPU (Parallel Computing Toolbox) .

Version History

Introduced before R2006a

ztest | ttest2 | sampsizepwr

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Single Sample T-Test Calculator

A single sample t-test (or one sample t-test) is used to compare the mean of a single sample of scores to a known or hypothetical population mean. So, for example, it could be used to determine whether the mean diastolic blood pressure of a particular group differs from 85, a value determined by a previous study.

Requirements

• The data is normally distributed
• Scale of measurement should be interval or ratio
• A randomized sample from a defined population

Null Hypothesis

H 0 : M - μ = 0, where M is the sample mean and μ is the population or hypothesized mean.

As above, the null hypothesis is that there is no difference between the sample mean and the known or hypothesized population mean.

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11.3: The Independent Samples t-test (Student Test)

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• Danielle Navarro
• University of New South Wales

Although the one sample t-test has its uses, it’s not the most typical example of a t-test 189 . A much more common situation arises when you’ve got two different groups of observations. In psychology, this tends to correspond to two different groups of participants, where each group corresponds to a different condition in your study. For each person in the study, you measure some outcome variable of interest, and the research question that you’re asking is whether or not the two groups have the same population mean. This is the situation that the independent samples t-test is designed for.

Suppose we have 33 students taking Dr Harpo’s statistics lectures, and Dr Harpo doesn’t grade to a curve. Actually, Dr Harpo’s grading is a bit of a mystery, so we don’t really know anything about what the average grade is for the class as a whole. There are two tutors for the class, Anastasia and Bernadette. There are N 1 =15 students in Anastasia’s tutorials, and N 2 =18 in Bernadette’s tutorials. The research question I’m interested in is whether Anastasia or Bernadette is a better tutor, or if it doesn’t make much of a difference. Dr Harpo emails me the course grades, in the harpo.Rdata file. As usual, I’ll load the file and have a look at what variables it contains:

As we can see, there’s a single data frame with two variables, grade and tutor . The grade variable is a numeric vector, containing the grades for all N=33 students taking Dr Harpo’s class; the tutor variable is a factor that indicates who each student’s tutor was. The first six observations in this data set are shown below:

We can calculate means and standard deviations, using the mean() and sd() functions. Rather than show the R output, here’s a nice little summary table:

To give you a more detailed sense of what’s going on here, I’ve plotted histograms showing the distribution of grades for both tutors (Figure 13.6 and 13.7). Inspection of these histograms suggests that the students in Anastasia’s class may be getting slightly better grades on average, though they also seem a little more variable.

Here is a simpler plot showing the means and corresponding confidence intervals for both groups of students (Figure 13.8).

Introducing the test

The independent samples t-test comes in two different forms, Student’s and Welch’s. The original Student t-test – which is the one I’ll describe in this section – is the simpler of the two, but relies on much more restrictive assumptions than the Welch t-test. Assuming for the moment that you want to run a two-sided test, the goal is to determine whether two “independent samples” of data are drawn from populations with the same mean (the null hypothesis) or different means (the alternative hypothesis). When we say “independent” samples, what we really mean here is that there’s no special relationship between observations in the two samples. This probably doesn’t make a lot of sense right now, but it will be clearer when we come to talk about the paired samples t-test later on. For now, let’s just point out that if we have an experimental design where participants are randomly allocated to one of two groups, and we want to compare the two groups’ mean performance on some outcome measure, then an independent samples t-test (rather than a paired samples t-test) is what we’re after.

Okay, so let’s let μ 1 denote the true population mean for group 1 (e.g., Anastasia’s students), and μ 2 will be the true population mean for group 2 (e.g., Bernadette’s students), 190 and as usual we’ll let \(\bar{X}_{1}\) and \(\bar{X}_{2}\) denote the observed sample means for both of these groups. Our null hypothesis states that the two population means are identical (μ 1 =μ 2 ) and the alternative to this is that they are not (μ 1 ≠μ 2 ). Written in mathematical-ese, this is…

H 0 :μ 1 =μ 2

H 1 :μ 1 ≠μ 2

To construct a hypothesis test that handles this scenario, we start by noting that if the null hypothesis is true, then the difference between the population means is exactly zero, μ 1 −μ 2 =0 As a consequence, a diagnostic test statistic will be based on the difference between the two sample means. Because if the null hypothesis is true, then we’d expect

\(\bar{X}_{1}\) - \(\bar{X}_{2}\)

to be pretty close to zero. However, just like we saw with our one-sample tests (i.e., the one-sample z-test and the one-sample t-test) we have to be precise about exactly how close to zero this difference

\(\ t ={\bar{X}_1 - \bar{X}_2 \over SE}\)

We just need to figure out what this standard error estimate actually is. This is a bit trickier than was the case for either of the two tests we’ve looked at so far, so we need to go through it a lot more carefully to understand how it works.

“pooled estimate” of the standard deviation

In the original “Student t-test”, we make the assumption that the two groups have the same population standard deviation: that is, regardless of whether the population means are the same, we assume that the population standard deviations are identical, σ 1 =σ 2 . Since we’re assuming that the two standard deviations are the same, we drop the subscripts and refer to both of them as σ. How should we estimate this? How should we construct a single estimate of a standard deviation when we have two samples? The answer is, basically, we average them. Well, sort of. Actually, what we do is take a weighed average of the variance estimates, which we use as our pooled estimate of the variance . The weight assigned to each sample is equal to the number of observations in that sample, minus 1. Mathematically, we can write this as

\(\ \omega_{1}\)=N 1 −1

\(\ \omega_{2}\)=N 2 −1

Now that we’ve assigned weights to each sample, we calculate the pooled estimate of the variance by taking the weighted average of the two variance estimates, \(\ \hat{\sigma_1}^2\) and \(\ \hat{\sigma_2}^2\)

\(\ \hat{\sigma_p}^2 ={ \omega_{1}\hat{\sigma_1}^2+\omega_{2}\hat{\sigma_2}^2 \over \omega_{1}+\omega_{2}}\)

Finally, we convert the pooled variance estimate to a pooled standard deviation estimate, by taking the square root. This gives us the following formula for \(\ \hat{\sigma_p}\),

\(\ \hat{\sigma_p} =\sqrt{\omega_1\hat{\sigma_1}^2+\omega_2\hat{\sigma_2}^2\over \omega_1+\omega_2} \)

And if you mentally substitute \(\ \omega_1\)=N1−1 and \(\ \omega_2\)=N2−1 into this equation you get a very ugly looking formula; a very ugly formula that actually seems to be the “standard” way of describing the pooled standard deviation estimate. It’s not my favourite way of thinking about pooled standard deviations, however. 191

same pooled estimate, described differently

I prefer to think about it like this. Our data set actually corresponds to a set of N observations, which are sorted into two groups. So let’s use the notation X ik to refer to the grade received by the i-th student in the k-th tutorial group: that is, X 11 is the grade received by the first student in Anastasia’s class, X 21 is her second student, and so on. And we have two separate group means \(\ \bar{X_1}\) and \(\ \bar{X_2}\), which we could “generically” refer to using the notation \(\ \bar{X_k}\), i.e., the mean grade for the k-th tutorial group. So far, so good. Now, since every single student falls into one of the two tutorials, and so we can describe their deviation from the group mean as the difference

\(\ X_{ik} - \bar{X_k}\)

So why not just use these deviations (i.e., the extent to which each student’s grade differs from the mean grade in their tutorial?) Remember, a variance is just the average of a bunch of squared deviations, so let’s do that. Mathematically, we could write it like this:

\(\ ∑_{ik} (X_{ik}-\bar{X}_k)^2 \over N \)

where the notation “∑ ik ” is a lazy way of saying “calculate a sum by looking at all students in all tutorials”, since each “ik” corresponds to one student. 192 But, as we saw in Chapter 10, calculating the variance by dividing by N produces a biased estimate of the population variance. And previously, we needed to divide by N−1 to fix this. However, as I mentioned at the time, the reason why this bias exists is because the variance estimate relies on the sample mean; and to the extent that the sample mean isn’t equal to the population mean, it can systematically bias our estimate of the variance. But this time we’re relying on two sample means! Does this mean that we’ve got more bias? Yes, yes it does. And does this mean we now need to divide by N−2 instead of N−1, in order to calculate our pooled variance estimate? Why, yes…

\(\hat{\sigma}_{p}\ ^{2}=\dfrac{\sum_{i k}\left(X_{i k}-X_{k}\right)^{2}}{N-2}\)

Oh, and if you take the square root of this then you get \(\ \hat{\sigma_{P}}\), the pooled standard deviation estimate. In other words, the pooled standard deviation calculation is nothing special: it’s not terribly different to the regular standard deviation calculation.

Completing the test

Regardless of which way you want to think about it, we now have our pooled estimate of the standard deviation. From now on, I’ll drop the silly p subscript, and just refer to this estimate as \(\ \hat{\sigma}\). Great. Let’s now go back to thinking about the bloody hypothesis test, shall we? Our whole reason for calculating this pooled estimate was that we knew it would be helpful when calculating our standard error estimate. But, standard error of what ? In the one-sample t-test, it was the standard error of the sample mean, SE (\(\ \bar{X}\)), and since SE (\(\ \bar{X}=\sigma/ \sqrt{N}\) that’s what the denominator of our t-statistic looked like. This time around, however, we have two sample means. And what we’re interested in, specifically, is the the difference between the two \(\ \bar{X_1}\) - \(\ \bar{X_2}\). As a consequence, the standard error that we need to divide by is in fact the standard error of the difference between means. As long as the two variables really do have the same standard deviation, then our estimate for the standard error is

\(\operatorname{SE}\left(\bar{X}_{1}-\bar{X}_{2}\right)=\hat{\sigma} \sqrt{\dfrac{1}{N_{1}}+\dfrac{1}{N_{2}}}\)

and our t-statistic is therefore

\(t=\dfrac{\bar{X}_{1}-\bar{X}_{2}}{\operatorname{SE}\left(\bar{X}_{1}-\bar{X}_{2}\right)}\)

(shocking, isn’t it?) as long as the null hypothesis is true, and all of the assumptions of the test are met. The degrees of freedom, however, is slightly different. As usual, we can think of the degrees of freedom to be equal to the number of data points minus the number of constraints. In this case, we have N observations (N1 in sample 1, and N2 in sample 2), and 2 constraints (the sample means). So the total degrees of freedom for this test are N−2.

Doing the test in R

Not surprisingly, you can run an independent samples t-test using the t.test() function (Section 13.7), but once again I’m going to start with a somewhat simpler function in the lsr package. That function is unimaginatively called independentSamplesTTest() . First, recall that our data look like this:

The outcome variable for our test is the student grade , and the groups are defined in terms of the tutor for each class. So you probably won’t be too surprised to see that we’re going to describe the test that we want in terms of an R formula that reads like this grade ~ tutor . The specific command that we need is:

The first two arguments should be familiar to you. The first one is the formula that tells R what variables to use and the second one tells R the name of the data frame that stores those variables. The third argument is not so obvious. By saying var.equal = TRUE , what we’re really doing is telling R to use the Student independent samples t-test. More on this later. For now, lets ignore that bit and look at the output:

The output has a very familiar form. First, it tells you what test was run, and it tells you the names of the variables that you used. The second part of the output reports the sample means and standard deviations for both groups (i.e., both tutorial groups). The third section of the output states the null hypothesis and the alternative hypothesis in a fairly explicit form. It then reports the test results: just like last time, the test results consist of a t-statistic, the degrees of freedom, and the p-value. The final section reports two things: it gives you a confidence interval, and an effect size. I’ll talk about effect sizes later. The confidence interval, however, I should talk about now.

It’s pretty important to be clear on what this confidence interval actually refers to: it is a confidence interval for the difference between the group means. In our example, Anastasia’s students had an average grade of 74.5, and Bernadette’s students had an average grade of 69.1, so the difference between the two sample means is 5.4. But of course the difference between population means might be bigger or smaller than this. The confidence interval reported by the independentSamplesTTest() function tells you that there’s a 95% chance that the true difference between means lies between 0.2 and 10.8.

In any case, the difference between the two groups is significant (just barely), so we might write up the result using text like this:

The mean grade in Anastasia’s class was 74.5% (std dev = 9.0), whereas the mean in Bernadette’s class was 69.1% (std dev = 5.8). A Student’s independent samples t-test showed that this 5.4% difference was significant (t(31)=2.1, p<.05, CI 95 =[0.2,10.8], d=.74), suggesting that a genuine difference in learning outcomes has occurred.

Notice that I’ve included the confidence interval and the effect size in the stat block. People don’t always do this. At a bare minimum, you’d expect to see the t-statistic, the degrees of freedom and the p value. So you should include something like this at a minimum: t(31)=2.1, p<.05. If statisticians had their way, everyone would also report the confidence interval and probably the effect size measure too, because they are useful things to know. But real life doesn’t always work the way statisticians want it to: you should make a judgment based on whether you think it will help your readers, and (if you’re writing a scientific paper) the editorial standard for the journal in question. Some journals expect you to report effect sizes, others don’t. Within some scientific communities it is standard practice to report confidence intervals, in other it is not. You’ll need to figure out what your audience expects. But, just for the sake of clarity, if you’re taking my class: my default position is that it’s usually worth includng the effect size, but don’t worry about the confidence interval unless the assignment asks you to or implies that you should.

Positive and negative t values

Before moving on to talk about the assumptions of the t-test, there’s one additional point I want to make about the use of t-tests in practice. The first one relates to the sign of the t-statistic (that is, whether it is a positive number or a negative one). One very common worry that students have when they start running their first t-test is that they often end up with negative values for the t-statistic, and don’t know how to interpret it. In fact, it’s not at all uncommon for two people working independently to end up with R outputs that are almost identical, except that one person has a negative t values and the other one has a positive t value. Assuming that you’re running a two-sided test, then the p-values will be identical. On closer inspection, the students will notice that the confidence intervals also have the opposite signs. This is perfectly okay: whenever this happens, what you’ll find is that the two versions of the R output arise from slightly different ways of running the t-test. What’s happening here is very simple. The t-statistic that R is calculating here is always of the form

\(t=\dfrac{(\text { mean } 1)-(\text { mean } 2)}{(\mathrm{SE})}\)

If “mean 1” is larger than “mean 2” the t statistic will be positive, whereas if “mean 2” is larger then the t statistic will be negative. Similarly, the confidence interval that R reports is the confidence interval for the difference “(mean 1) minus (mean 2)”, which will be the reverse of what you’d get if you were calculating the confidence interval for the difference “(mean 2) minus (mean 1)”.

Okay, that’s pretty straightforward when you think about it, but now consider our t-test comparing Anastasia’s class to Bernadette’s class. Which one should we call “mean 1” and which one should we call “mean 2”. It’s arbitrary. However, you really do need to designate one of them as “mean 1” and the other one as “mean 2”. Not surprisingly, the way that R handles this is also pretty arbitrary. In earlier versions of the book I used to try to explain it, but after a while I gave up, because it’s not really all that important, and to be honest I can never remember myself. Whenever I get a significant t-test result, and I want to figure out which mean is the larger one, I don’t try to figure it out by looking at the t-statistic. Why would I bother doing that? It’s foolish. It’s easier just look at the actual group means, since the R output actually shows them!

Here’s the important thing. Because it really doesn’t matter what R printed out, I usually try to report the t-statistic in such a way that the numbers match up with the text. Here’s what I mean… suppose that what I want to write in my report is “Anastasia’s class had higher grades than Bernadette’s class”. The phrasing here implies that Anastasia’s group comes first, so it makes sense to report the t-statistic as if Anastasia’s class corresponded to group 1. If so, I would write

Anastasia’s class had higher grades than Bernadette’s class (t(31)=2.1,p=.04).

(I wouldn’t actually emphasise the word “higher” in real life, I’m just doing it to emphasise the point that “higher” corresponds to positive t values). On the other hand, suppose the phrasing I wanted to use has Bernadette’s class listed first. If so, it makes more sense to treat her class as group 1, and if so, the write up looks like this:

Bernadette’s class had lower grades than Anastasia’s class (t(31)=−2.1,p=.04).

Because I’m talking about one group having “lower” scores this time around, it is more sensible to use the negative form of the t-statistic. It just makes it read more cleanly.

One last thing: please note that you can’t do this for other types of test statistics. It works for t-tests, but it wouldn’t be meaningful for chi-square testsm F-tests or indeed for most of the tests I talk about in this book. So don’t overgeneralise this advice! I’m really just talking about t-tests here and nothing else!

Assumptions of the test

As always, our hypothesis test relies on some assumptions. So what are they? For the Student t-test there are three assumptions, some of which we saw previously in the context of the one sample t-test (see Section 13.2.3):

• Normality . Like the one-sample t-test, it is assumed that the data are normally distributed. Specifically, we assume that both groups are normally distributed. In Section 13.9 we’ll discuss how to test for normality, and in Section 13.10 we’ll discuss possible solutions.
• Independence . Once again, it is assumed that the observations are independently sampled. In the context of the Student test this has two aspects to it. Firstly, we assume that the observations within each sample are independent of one another (exactly the same as for the one-sample test). However, we also assume that there are no cross-sample dependencies. If, for instance, it turns out that you included some participants in both experimental conditions of your study (e.g., by accidentally allowing the same person to sign up to different conditions), then there are some cross sample dependencies that you’d need to take into account.
• Homogeneity of variance (also called “homoscedasticity”). The third assumption is that the population standard deviation is the same in both groups. You can test this assumption using the Levene test, which I’ll talk about later on in the book (Section 14.7). However, there’s a very simple remedy for this assumption, which I’ll talk about in the next section.