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10.26: Hypothesis Test for a Population Mean (5 of 5)

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Learning Objectives

• Interpret the P-value as a conditional probability.

We finish our discussion of the hypothesis test for a population mean with a review of the meaning of the P-value, along with a review of type I and type II errors.

Review of the Meaning of the P-value

At this point, we assume you know how to use a P-value to make a decision in a hypothesis test. The logic is always the same. If we pick a level of significance (α), then we compare the P-value to α.

• If the P-value ≤ α, reject the null hypothesis. The data supports the alternative hypothesis.
• If the P-value > α, do not reject the null hypothesis. The data is not strong enough to support the alternative hypothesis.

In fact, we find that we treat these as “rules” and apply them without thinking about what the P-value means. So let’s pause here and review the meaning of the P-value, since it is the connection between probability and decision-making in inference.

Birth Weights in a Town

Let’s return to the familiar context of birth weights for babies in a town. Suppose that babies in the town had a mean birth weight of 3,500 grams in 2010. This year, a random sample of 50 babies has a mean weight of about 3,400 grams with a standard deviation of about 500 grams. Here is the distribution of birth weights in the sample.

Obviously, this sample weighs less on average than the population of babies in the town in 2010. A decrease in the town’s mean birth weight could indicate a decline in overall health of the town. But does this sample give strong evidence that the town’s mean birth weight is less than 3,500 grams this year?

We now know how to answer this question with a hypothesis test. Let’s use a significance level of 5%.

Let μ = mean birth weight in the town this year. The null hypothesis says there is “no change from 2010.”

• H 0 : μ < 3,500
• H a : μ = 3,500

Since the sample is large, we can conduct the T-test (without worrying about the shape of the distribution of birth weights for individual babies.)

Statistical software tells us the P-value is 0.082 = 8.2%. Since the P-value is greater than 0.05, we fail to reject the null hypothesis.

Our conclusion: This sample does not suggest that the mean birth weight this year is less than 3,500 grams ( P -value = 0.082). The sample from this year has a mean of 3,400 grams, which is 100 grams lower than the mean in 2010. But this difference is not statistically significant. It can be explained by the chance fluctuation we expect to see in random sampling.

What Does the P-Value of 0.082 Tell Us?

A simulation can help us understand the P-value. In a simulation, we assume that the population mean is 3,500 grams. This is the null hypothesis. We assume the null hypothesis is true and select 1,000 random samples from a population with a mean of 3,500 grams. The mean of the sampling distribution is at 3,500 (as predicted by the null hypothesis.) We see this in the simulated sampling distribution.

In the simulation, we can see that about 8.6% of the samples have a mean less than 3,400. Since probability is the relative frequency of an event in the long run, we say there is an 8.6% chance that a random sample of 500 babies has a mean less than 3,400 if the population mean is 3,500. We can see that the corresponding area to the left of T = −1.41 in the T-model (with df = 49) also gives us a good estimate of the probability. This area is the P-value, about 8.2%.

If we generalize this statement, we say the P-value is the probability that random samples have results more extreme than the data if the null hypothesis is true. (By more extreme, we mean further from value of the parameter, in the direction of the alternative hypothesis.) We can also describe the P-value in terms of T-scores. The P-value is the probability that the test statistic from a random sample has a value more extreme than that associated with the data if the null hypothesis is true.

What Does a P-Value Mean?

Do women who smoke run the risk of shorter pregnancy and premature birth? The mean pregnancy length is 266 days. We test the following hypotheses.

• H 0 : μ = 266
• H a : μ < 266

Suppose a random sample of 40 women who smoke during their pregnancy have a mean pregnancy length of 260 days with a standard deviation of 21 days. The P-value is 0.04.

What probability does the P-value of 0.04 describe? Label each of the following interpretations as valid or invalid.

https://assessments.lumenlearning.co...sessments/3654

https://assessments.lumenlearning.co...sessments/3655

https://assessments.lumenlearning.co...sessments/3656

Review of Type I and Type II Errors

We know that statistical inference is based on probability, so there is always some chance of making a wrong decision. Recall that there are two types of wrong decisions that can be made in hypothesis testing. When we reject a null hypothesis that is true, we commit a type I error. When we fail to reject a null hypothesis that is false, we commit a type II error.

The following table summarizes the logic behind type I and type II errors.

It is possible to have some influence over the likelihoods of committing these errors. But decreasing the chance of a type I error increases the chance of a type II error. We have to decide which error is more serious for a given situation. Sometimes a type I error is more serious. Other times a type II error is more serious. Sometimes neither is serious.

Recall that if the null hypothesis is true, the probability of committing a type I error is α. Why is this? Well, when we choose a level of significance (α), we are choosing a benchmark for rejecting the null hypothesis. If the null hypothesis is true, then the probability that we will reject a true null hypothesis is α. So the smaller α is, the smaller the probability of a type I error.

It is more complicated to calculate the probability of a type II error. The best way to reduce the probability of a type II error is to increase the sample size. But once the sample size is set, larger values of α will decrease the probability of a type II error (while increasing the probability of a type I error).

General Guidelines for Choosing a Level of Significance

• If the consequences of a type I error are more serious, choose a small level of significance (α).
• If the consequences of a type II error are more serious, choose a larger level of significance (α). But remember that the level of significance is the probability of committing a type I error.
• In general, we pick the largest level of significance that we can tolerate as the chance of a type I error.

Let’s return to the investigation of the impact of smoking on pregnancy length.

Recap of the hypothesis test: The mean human pregnancy length is 266 days. We test the following hypotheses.

https://assessments.lumenlearning.co...sessments/3778

https://assessments.lumenlearning.co...sessments/3779

https://assessments.lumenlearning.co...sessments/3780

Let’s Summarize

In this “Hypothesis Test for a Population Mean,” we looked at the four steps of a hypothesis test as they relate to a claim about a population mean.

Step 1: Determine the hypotheses.

• The hypotheses are claims about the population mean, µ.
• The null hypothesis is a hypothesis that the mean equals a specific value, µ 0 .

Step 2: Collect the data.

Since the hypothesis test is based on probability, random selection or assignment is essential in data production. Additionally, we need to check whether the t-model is a good fit for the sampling distribution of sample means. To use the t-model, the variable must be normally distributed in the population or the sample size must be more than 30. In practice, it is often impossible to verify that the variable is normally distributed in the population. If this is the case and the sample size is not more than 30, researchers often use the t-model if the sample is not strongly skewed and does not have outliers.

Step 3: Assess the evidence.

• If a t-model is appropriate, determine the t-test statistic for the data’s sample mean.
• Use the test statistic, together with the alternative hypothesis, to determine the P-value.
• The P-value is the probability of finding a random sample with a mean at least as extreme as our sample mean, assuming that the null hypothesis is true.
• As in all hypothesis tests, if the alternative hypothesis is greater than, the P-value is the area to the right of the test statistic. If the alternative hypothesis is less than, the P-value is the area to the left of the test statistic. If the alternative hypothesis is not equal to, the P-value is equal to double the tail area beyond the test statistic.

Step 4: Give the conclusion.

The logic of the hypothesis test is always the same. To state a conclusion about H 0 , we compare the P-value to the significance level, α.

• If P ≤ α, we reject H 0 . We conclude there is significant evidence in favor of H a .
• If P > α, we fail to reject H 0 . We conclude the sample does not provide significant evidence in favor of H a .
• We write the conclusion in the context of the research question. Our conclusion is usually a statement about the alternative hypothesis (we accept H a or fail to acceptH a ) and should include the P-value.

Other Hypothesis Testing Notes

• Remember that the P-value is the probability of seeing a sample mean at least as extreme as the one from the data if the null hypothesis is true. The probability is about the random sample; it is not a “chance” statement about the null or alternative hypothesis.
• If our test results in rejecting a null hypothesis that is actually true, then it is called a type I error.
• If our test results in failing to reject a null hypothesis that is actually false, then it is called a type II error.
• If rejecting a null hypothesis would be very expensive, controversial, or dangerous, then we really want to avoid a type I error. In this case, we would set a strict significance level (a small value of α, such as 0.01).
• Finally, remember the phrase “garbage in, garbage out.” If the data collection methods are poor, then the results of a hypothesis test are meaningless.

Contributors and Attributions

• Concepts in Statistics. Provided by : Open Learning Initiative. Located at : http://oli.cmu.edu . License : CC BY: Attribution

9.1 Null and Alternative Hypotheses

The actual test begins by considering two hypotheses . They are called the null hypothesis and the alternative hypothesis . These hypotheses contain opposing viewpoints.

H 0 , the — null hypothesis: a statement of no difference between sample means or proportions or no difference between a sample mean or proportion and a population mean or proportion. In other words, the difference equals 0.

H a —, the alternative hypothesis: a claim about the population that is contradictory to H 0 and what we conclude when we reject H 0 .

Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data.

After you have determined which hypothesis the sample supports, you make a decision. There are two options for a decision. They are reject H 0 if the sample information favors the alternative hypothesis or do not reject H 0 or decline to reject H 0 if the sample information is insufficient to reject the null hypothesis.

Mathematical Symbols Used in H 0 and H a :

H 0 always has a symbol with an equal in it. H a never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test. However, be aware that many researchers use = in the null hypothesis, even with > or < as the symbol in the alternative hypothesis. This practice is acceptable because we only make the decision to reject or not reject the null hypothesis.

Example 9.1

H 0 : No more than 30 percent of the registered voters in Santa Clara County voted in the primary election. p ≤ 30 H a : More than 30 percent of the registered voters in Santa Clara County voted in the primary election. p > 30

A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25 percent. State the null and alternative hypotheses.

Example 9.2

We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are the following: H 0 : μ = 2.0 H a : μ ≠ 2.0

We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : μ __ 66
• H a : μ __ 66

Example 9.3

We want to test if college students take fewer than five years to graduate from college, on the average. The null and alternative hypotheses are the following: H 0 : μ ≥ 5 H a : μ < 5

We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : μ __ 45
• H a : μ __ 45

Example 9.4

An article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third of the students pass. The same article stated that 6.6 percent of U.S. students take advanced placement exams and 4.4 percent pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6 percent. State the null and alternative hypotheses. H 0 : p ≤ 0.066 H a : p > 0.066

On a state driver’s test, about 40 percent pass the test on the first try. We want to test if more than 40 percent pass on the first try. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : p __ 0.40
• H a : p __ 0.40

Collaborative Exercise

Bring to class a newspaper, some news magazines, and some internet articles. In groups, find articles from which your group can write null and alternative hypotheses. Discuss your hypotheses with the rest of the class.

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Module 10: Inference for Means

Hypothesis test for a population mean (1 of 5), learning outcomes.

• Recognize when to use a hypothesis test or a confidence interval to draw a conclusion about a population mean.
• Under appropriate conditions, conduct a hypothesis test about a population mean. State a conclusion in context.

Introduction

In Inference for Means , our focus is on inference when the variable is quantitative, so the parameters and statistics are means. In “Estimating a Population Mean,” we learned how to use a sample mean to calculate a confidence interval. The confidence interval estimates a population mean. In “Hypothesis Test for a Population Mean,” we learn to use a sample mean to test a hypothesis about a population mean.

We did hypothesis tests in earlier modules. In Inference for One Proportion , each claim involved a single population proportion. In Inference for Two Proportions , the claim was a statement about a treatment effect or a difference in population proportions. In “Hypothesis Test for a Population Mean,” the claims are statements about a population mean. But we will see that the steps and the logic of the hypothesis test are the same. Before we get into the details, let’s practice identifying research questions and studies that involve a population mean.

Cell Phone Data

Cell phones and cell phone plans can be very expensive, so consumers must think carefully when choosing a cell phone and service. This decision is as much about choosing the right cellular company as it is about choosing the right phone. Many people use the data/Internet capabilities of a phone as much as, if not more than, they use voice capability. The data service of a cell company is therefore an important factor in this decision. In the following example, a student named Melanie from Los Angeles applies what she learned in her statistics class to help her make a decision about buying a data plan for her smartphone.

Melanie read an advertisement from the Cell Phone Giants (CPG, for short, and yes, we’re using a fictitious company name) that she thinks is too good to be true. The CPG ad states that customers in Los Angeles get average data download speeds of 4 Mbps. With this speed, the ad claims, it takes, on average, only 12 seconds to download a typical 3-minute song from iTunes.

Only 12 seconds on average to download a 3-minute song from iTunes! Melanie has her doubts about this claim, so she gathers data to test it. She asks a friend who uses the CPG plan to download a song, and it takes 13 seconds to download a 3-minute song using the CPG network. Melanie decides to gather more evidence. She uses her friend’s phone and times the download of the same 3-minute song from various locations in Los Angeles. She gets a mean download time of 13.5 seconds for her sample of downloads.

What can Melanie conclude? Her sample has a mean download time that is greater than 12 seconds. Isn’t this evidence that the CPG claim is wrong? Why is a hypothesis test necessary? Isn’t the conclusion clear?

Let’s review the reason Melanie needs to do a hypothesis test before she can reach a conclusion.

Why should Melanie do a hypothesis test?

Melanie’s data (with a mean of 13.5 seconds) suggest that the average download time overall is greater than the 12 seconds claimed by the manufacturer. But wait. We know that samples will vary. If the CPG claim is correct, we don’t expect all samples to have a mean download time exactly equal to 12 seconds. There will be variability in the sample means. But if the overall average download time is 12 seconds, how much variability in sample means do we expect to see? We need to determine if the difference Melanie observed can be explained by chance.

We have to judge Melanie’s data against random samples that come from a population with a mean of 12. For this reason, we must do a simulation or use a mathematical model to examine the sampling distribution of sample means. Based on the sampling distribution, we ask, Is it likely that the samples will have mean download times that are greater than 13.5 seconds if the overall mean is 12 seconds? This probability (the P-value) determines whether Melanie’s data provides convincing evidence against the CPG claim.

Now let’s do the hypothesis test.

Step 1: Determine the hypotheses.

As always, hypotheses come from the research question. The null hypothesis is a hypothesis that the population mean equals a specific value. The alternative hypothesis reflects our claim. The alternative hypothesis says the population mean is “greater than” or “less than” or “not equal to” the value we assume is true in the null hypothesis.

Melanie’s hypotheses:

• H 0 : It takes 12 seconds on average to download Melanie’s song from iTunes with the CPG network in Los Angeles.
• H a : It takes more than 12 seconds on average to download Melanie’s song from iTunes using the CPG network in Los Angeles.

We can write the hypotheses in terms of µ. When we do so, we should always define µ. Here μ = the average number of seconds it takes to download Melanie’s song on the CPG network in Los Angeles.

• H 0 : μ = 12
• H a : μ > 12

Step 2: Collect the data.

To conduct a hypothesis test, Melanie knows she has to use a t-model of the sampling distribution. She thinks ahead to the conditions required, which helps her collect a useful sample.

Recall the conditions for use of a t-model.

• There is no reason to think the download times are normally distributed (they might be, but this isn’t something Melanie could know for sure). So the sample has to be large (more than 30).
• The sample has to be random. Melanie decides to use one phone but randomly selects days, times, and locations in Los Angeles.

Melanie collects a random sample of 45 downloads by using her friend’s phone to download her song from iTunes according to the randomly selected days, times, and locations.

Melanie’s sample of size 45 downloads has an average download time of 13.5 seconds. The standard deviation for the sample is 3.2 seconds. Now Melanie needs to determine how unlikely this data is if CPG’s claim is actually true.

Step 3: Assess the evidence.

Assuming the average download time for Melanie’s song is really 12 seconds, what is the probability that 45 random downloads of this song will have a mean of 13.5 seconds or more?

This is a question about sampling variability. Melanie must determine the standard error. She knows the standard error of random sample means is [latex]\sigma \text{}/\sqrt{n}[/latex]. Since she has no way of knowing the population standard deviation, σ, Melanie uses the sample standard deviation, s = 3.2, as an approximation. Therefore, Melanie approximates the standard error of all sample means ( n = 45) to be

[latex]s\text{}/\sqrt{n}\text{}=\text{}3.2\text{}/\sqrt{45}\text{}=\text{}0.48[/latex]

Now she can assess how far away her sample is from the claimed mean in terms of standard errors. That is, she can compute the t-score of her sample mean.

[latex]T\text{}=\text{}\frac{\mathrm{statistic}-\mathrm{parameter}}{\mathrm{standard}\text{}\mathrm{error}}\text{}=\text{}\frac{\stackrel{¯}{x}-μ}{s\text{}/\sqrt{n}}\text{}=\text{}\frac{13.5-12}{0.48}\text{}=\text{}3.14[/latex]

The sample mean for Melanie’s random sample is approximately 3.14 standard errors above the overall mean of 12. We know from previous experience that a sample mean this far above µ is very unlikely. With a t-score this large, the P-value is very small. We use a simulation of the t-model for 44 degrees of freedom to verify this.

We want the probability that the sample mean is greater than 13.5. This corresponds to the probability that T is greater than 3.14. The P-value is 0.0015.

Step 4: State a conclusion.

Here the logic is the same as for other hypothesis tests. We use the P-value to make a decision. The P-value helps us determine if the difference we see between the data and the hypothesized value of µ is statistically significant or due to chance. One of two outcomes can occur:

• One possibility is that results similar to the actual sample are extremely unlikely. This means the data does not fit with results from random samples selected from the population described by the null hypothesis. In this case, it is unlikely that the data came from this population. The probability as measured by the P-value is small, so we view this as strong evidence against the null hypothesis. We reject the null hypothesis in favor of the alternative hypothesis.
• The other possibility is that results similar to the actual sample are fairly likely (not unusual). This means the data fits with typical results from random samples selected from the population described by the null hypothesis. The probability as measured by the P-value is large. In this case, we do not have evidence against the null hypothesis, so we cannot reject it in favor of the alternative hypothesis.

Melanie’s data is very unlikely if µ = 12. The probability is essentially zero (P-value = 0.0015). This means we will rarely see sample means greater than 13.5 if µ = 12. So we reject the null and accept the alternative hypothesis. In other words, this sample provides strong evidence that CPG has overstated the speed of its data download capability.

The following activities give you an opportunity to practice parts of the hypothesis testing process for a population mean. Later you will have the opportunity to practice the hypothesis test from start to finish.

For the following scenarios, give the null and alternative hypotheses and state in words what µ represents in your hypotheses. A good definition of µ describes both the variable and the population.

In the previous example, Melanie did not state a significance level for her test. If she had, the logic is the same as we used for hypothesis tests in Modules 8 and 9. To come to a conclusion about H 0 , we compare the P-value to the significance level α.

• If P ≤ α, we reject H 0 . We conclude there is significant evidence in favor of H a .
• If P > α, we fail to reject H 0 . We conclude the sample does not provide significant evidence in favor of H a .

Use this simulation when needed to answer questions below.

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• Concepts in Statistics. Provided by : Open Learning Initiative. Located at : http://oli.cmu.edu . License : CC BY: Attribution

Teach yourself statistics

Hypothesis Test for a Mean

This lesson explains how to conduct a hypothesis test of a mean, when the following conditions are met:

• The sampling method is simple random sampling .
• The sampling distribution is normal or nearly normal.

Generally, the sampling distribution will be approximately normally distributed if any of the following conditions apply.

• The population distribution is normal.
• The population distribution is symmetric , unimodal , without outliers , and the sample size is 15 or less.
• The population distribution is moderately skewed , unimodal, without outliers, and the sample size is between 16 and 40.
• The sample size is greater than 40, without outliers.

This approach consists of four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results.

State the Hypotheses

Every hypothesis test requires the analyst to state a null hypothesis and an alternative hypothesis . The hypotheses are stated in such a way that they are mutually exclusive. That is, if one is true, the other must be false; and vice versa.

The table below shows three sets of hypotheses. Each makes a statement about how the population mean μ is related to a specified value M . (In the table, the symbol ≠ means " not equal to ".)

The first set of hypotheses (Set 1) is an example of a two-tailed test , since an extreme value on either side of the sampling distribution would cause a researcher to reject the null hypothesis. The other two sets of hypotheses (Sets 2 and 3) are one-tailed tests , since an extreme value on only one side of the sampling distribution would cause a researcher to reject the null hypothesis.

Formulate an Analysis Plan

The analysis plan describes how to use sample data to accept or reject the null hypothesis. It should specify the following elements.

• Significance level. Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10; but any value between 0 and 1 can be used.
• Test method. Use the one-sample t-test to determine whether the hypothesized mean differs significantly from the observed sample mean.

Analyze Sample Data

Using sample data, conduct a one-sample t-test. This involves finding the standard error, degrees of freedom, test statistic, and the P-value associated with the test statistic.

SE = s * sqrt{ ( 1/n ) * [ ( N - n ) / ( N - 1 ) ] }

SE = s / sqrt( n )

• Degrees of freedom. The degrees of freedom (DF) is equal to the sample size (n) minus one. Thus, DF = n - 1.

t = ( x - μ) / SE

• P-value. The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a t statistic, use the t Distribution Calculator to assess the probability associated with the t statistic, given the degrees of freedom computed above. (See sample problems at the end of this lesson for examples of how this is done.)

Sample Size Calculator

As you probably noticed, the process of hypothesis testing can be complex. When you need to test a hypothesis about a mean score, consider using the Sample Size Calculator. The calculator is fairly easy to use, and it is free. You can find the Sample Size Calculator in Stat Trek's main menu under the Stat Tools tab. Or you can tap the button below.

Interpret Results

If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, this involves comparing the P-value to the significance level , and rejecting the null hypothesis when the P-value is less than the significance level.

Test Your Understanding

In this section, two sample problems illustrate how to conduct a hypothesis test of a mean score. The first problem involves a two-tailed test; the second problem, a one-tailed test.

Problem 1: Two-Tailed Test

An inventor has developed a new, energy-efficient lawn mower engine. He claims that the engine will run continuously for 5 hours (300 minutes) on a single gallon of regular gasoline. From his stock of 2000 engines, the inventor selects a simple random sample of 50 engines for testing. The engines run for an average of 295 minutes, with a standard deviation of 20 minutes. Test the null hypothesis that the mean run time is 300 minutes against the alternative hypothesis that the mean run time is not 300 minutes. Use a 0.05 level of significance. (Assume that run times for the population of engines are normally distributed.)

Solution: The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

Null hypothesis: μ = 300

Alternative hypothesis: μ ≠ 300

• Formulate an analysis plan . For this analysis, the significance level is 0.05. The test method is a one-sample t-test .

SE = s / sqrt(n) = 20 / sqrt(50) = 20/7.07 = 2.83

DF = n - 1 = 50 - 1 = 49

t = ( x - μ) / SE = (295 - 300)/2.83 = -1.77

where s is the standard deviation of the sample, x is the sample mean, μ is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test , the P-value is the probability that the t statistic having 49 degrees of freedom is less than -1.77 or greater than 1.77. We use the t Distribution Calculator to find P(t < -1.77) is about 0.04.

• If you enter 1.77 as the sample mean in the t Distribution Calculator, you will find the that the P(t < 1.77) is about 0.04. Therefore, P(t >  1.77) is 1 minus 0.96 or 0.04. Thus, the P-value = 0.04 + 0.04 = 0.08.
• Interpret results . Since the P-value (0.08) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the population was normally distributed, and the sample size was small relative to the population size (less than 5%).

Problem 2: One-Tailed Test

Bon Air Elementary School has 1000 students. The principal of the school thinks that the average IQ of students at Bon Air is at least 110. To prove her point, she administers an IQ test to 20 randomly selected students. Among the sampled students, the average IQ is 108 with a standard deviation of 10. Based on these results, should the principal accept or reject her original hypothesis? Assume a significance level of 0.01. (Assume that test scores in the population of engines are normally distributed.)

Null hypothesis: μ >= 110

Alternative hypothesis: μ < 110

• Formulate an analysis plan . For this analysis, the significance level is 0.01. The test method is a one-sample t-test .

SE = s / sqrt(n) = 10 / sqrt(20) = 10/4.472 = 2.236

DF = n - 1 = 20 - 1 = 19

t = ( x - μ) / SE = (108 - 110)/2.236 = -0.894

Here is the logic of the analysis: Given the alternative hypothesis (μ < 110), we want to know whether the observed sample mean is small enough to cause us to reject the null hypothesis.

The observed sample mean produced a t statistic test statistic of -0.894. We use the t Distribution Calculator to find P(t < -0.894) is about 0.19.

• This means we would expect to find a sample mean of 108 or smaller in 19 percent of our samples, if the true population IQ were 110. Thus the P-value in this analysis is 0.19.
• Interpret results . Since the P-value (0.19) is greater than the significance level (0.01), we cannot reject the null hypothesis.

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8.7 Hypothesis Tests for a Population Mean with Unknown Population Standard Deviation

Learning objectives.

• Conduct and interpret hypothesis tests for a population mean with unknown population standard deviation.

Some notes about conducting a hypothesis test:

• The null hypothesis [latex]H_0[/latex] is always an “equal to.”  The null hypothesis is the original claim about the population parameter.
• The alternative hypothesis [latex]H_a[/latex] is a “less than,” “greater than,” or “not equal to.”  The form of the alternative hypothesis depends on the context of the question.
• If the alternative hypothesis is a “less than”,  then the test is left-tail.  The p -value is the area in the left-tail of the distribution.
• If the alternative hypothesis is a “greater than”, then the test is right-tail.  The p -value is the area in the right-tail of the distribution.
• If the alternative hypothesis is a “not equal to”, then the test is two-tail.  The p -value is the sum of the area in the two-tails of the distribution.  Each tail represents exactly half of the p -value.
• Think about the meaning of the p -value.  A data analyst (and anyone else) should have more confidence that they made the correct decision to reject the null hypothesis with a smaller p -value (for example, 0.001 as opposed to 0.04) even if using a significance level of  0.05.  Similarly, for a large p -value such as 0.4, as opposed to a p -value of 0.056 (a significance level of 0.05 is less than either number), a data analyst should have more confidence that they made the correct decision in not rejecting the null hypothesis.  This makes the data analyst use judgment rather than mindlessly applying rules.
• The significance level must be identified before collecting the sample data and conducting the test.  Generally, the significance level will be included in the question.  If no significance level is given, a common standard is to use a significance level of 5%.
• An alternative approach for hypothesis testing is to use what is called the critical value approach .  In this book, we will only use the p -value approach.  Some of the videos below may mention the critical value approach, but this approach will not be used in this book.

Steps to Conduct a Hypothesis Test for a Population Mean with Unknown Population Standard Deviation

• Write down the null and alternative hypotheses in terms of the population mean [latex]\mu[/latex].  Include appropriate units with the values of the mean.
• Use the form of the alternative hypothesis to determine if the test is left-tailed, right-tailed, or two-tailed.
• Collect the sample information for the test and identify the significance level [latex]\alpha[/latex].

[latex]\begin{eqnarray*} t & = & \frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}} \\ \\ df & = & n-1 \\ \\ \end{eqnarray*}[/latex]

• The results of the sample data are significant. There is sufficient evidence to conclude that the null hypothesis [latex]H_0[/latex] is an incorrect belief and that the alternative hypothesis [latex]H_a[/latex] is most likely correct.
• The results of the sample data are not significant. There is not sufficient evidence to conclude that the alternative hypothesis [latex]H_a[/latex] may be correct.
• Write down a concluding sentence specific to the context of the question.

USING EXCEL TO CALCULE THE P -VALUE FOR A HYPOTHESIS TEST ON A POPULATION MEAN WITH UNKNOWN POPULATION STANDARD DEVIATION

The p -value for a hypothesis test on a population mean is the area in the tail(s) of the distribution of the sample mean.  When the population standard deviation is unknown, use the [latex]t[/latex]-distribution to find the p -value.

If the p -value is the area in the left-tail:

• For t-score , enter the value of [latex]t[/latex] calculated from [latex]\displaystyle{t=\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}}[/latex].
• For degrees of freedom , enter the degrees of freedom for the [latex]t[/latex]-distribution [latex]n-1[/latex].
• For the logic operator , enter true .  Note:  Because we are calculating the area under the curve, we always enter true for the logic operator.
• The output from the t.dist function is the area under the [latex]t[/latex]-distribution to the left of the entered [latex]t[/latex]-score.
• Visit the Microsoft page for more information about the t.dist function.

If the p -value is the area in the right-tail:

• The output from the t.dist.rt function is the area under the [latex]t[/latex]-distribution to the right of the entered [latex]t[/latex]-score.
• Visit the Microsoft page for more information about the t.dist.rt function.

If the p -value is the sum of area in the tails:

• For t-score , enter the absolute value of [latex]t[/latex] calculated from [latex]\displaystyle{t=\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}}[/latex].  Note:  In the t.dist.2t function, the value of the [latex]t[/latex]-score must be a positive number.  If the [latex]t[/latex]-score is negative, enter the absolute value of the [latex]t[/latex]-score into the t.dist.2t function.
• The output from the t.dist.2t function is the sum of areas in the tails under the [latex]t[/latex]-distribution.
• Visit the Microsoft page for more information about the t.dist.2t function.

Statistics students believe that the mean score on the first statistics test is 65.  A statistics instructor thinks the mean score is higher than 65.  He samples ten statistics students and obtains the following scores:

The instructor performs a hypothesis test using a 1% level of significance. The test scores are assumed to be from a normal distribution.

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & \mu=65  \\ H_a: & & \mu \gt 65  \end{eqnarray*}[/latex]

From the question, we have [latex]n=10[/latex], [latex]\overline{x}=67[/latex], [latex]s=3.1972...[/latex] and [latex]\alpha=0.01[/latex].

This is a test on a population mean where the population standard deviation is unknown (we only know the sample standard deviation [latex]s=3.1972...[/latex]).  So we use a [latex]t[/latex]-distribution to calculate the p -value.  Because the alternative hypothesis is a [latex]\gt[/latex], the p -value is the area in the right-tail of the distribution.

To use the t.dist.rt function, we need to calculate out the [latex]t[/latex]-score:

[latex]\begin{eqnarray*} t & = & \frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}} \\ & = & \frac{67-65}{\frac{3.1972...}{\sqrt{10}}} \\ & = & 1.9781... \end{eqnarray*}[/latex]

The degrees of freedom for the [latex]t[/latex]-distribution is [latex]n-1=10-1=9[/latex].

So the p -value[latex]=0.0396[/latex].

Conclusion:

Because p -value[latex]=0.0396 \gt 0.01=\alpha[/latex], we do not reject the null hypothesis.  At the 1% significance level there is not enough evidence to suggest that mean score on the test is greater than 65.

• The null hypothesis [latex]\mu=65[/latex] is the claim that the mean test score is 65.
• The alternative hypothesis [latex]\mu \gt 65[/latex] is the claim that the mean test score is greater than 65.
• Keep all of the decimals throughout the calculation (i.e. in the sample standard deviation, the [latex]t[/latex]-score, etc.) to avoid any round-off error in the calculation of the p -value.  This ensures that we get the most accurate value for the p -value.
• The p -value is the area in the right-tail of the [latex]t[/latex]-distribution, to the right of [latex]t=1.9781...[/latex].
• The p -value of 0.0396 tells us that under the assumption that the mean test score is 65 (the null hypothesis), there is a 3.96% chance that the mean test score is 65 or more.  Compared to the 1% significance level, this is a large probability, and so is likely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely correct, and so the conclusion of the test is to not reject the null hypothesis.

A company claims that the average change in the value of their stock is $3.50 per week.  An investor believes this average is too high. The investor records the changes in the company’s stock price over 30 weeks and finds the average change in the stock price is $2.60 with a standard deviation of $1.80.  At the 5% significance level, is the average change in the company’s stock price lower than the company claims?

[latex]\begin{eqnarray*} H_0: & & \mu=$3.50  \\ H_a: & & \mu \lt $3.50  \end{eqnarray*}[/latex]

From the question, we have [latex]n=30[/latex], [latex]\overline{x}=2.6[/latex], [latex]s=1.8[/latex] and [latex]\alpha=0.05[/latex].

This is a test on a population mean where the population standard deviation is unknown (we only know the sample standard deviation [latex]s=1.8.[/latex]).  So we use a [latex]t[/latex]-distribution to calculate the p -value.  Because the alternative hypothesis is a [latex]\lt[/latex], the p -value is the area in the left-tail of the distribution.

To use the t.dist function, we need to calculate out the [latex]t[/latex]-score:

[latex]\begin{eqnarray*} t & = & \frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}} \\ & = & \frac{2.6-3.5}{\frac{1.8}{\sqrt{30}}} \\ & = & -1.5699... \end{eqnarray*}[/latex]

The degrees of freedom for the [latex]t[/latex]-distribution is [latex]n-1=30-1=29[/latex].

So the p -value[latex]=0.0636[/latex].

Because p -value[latex]=0.0636 \gt 0.05=\alpha[/latex], we do not reject the null hypothesis.  At the 5% significance level there is not enough evidence to suggest that average change in the stock price is lower than $3.50.

• The null hypothesis [latex]\mu=$3.50[/latex] is the claim that the average change in the company’s stock is $3.50 per week.
• The alternative hypothesis [latex]\mu \lt $3.50[/latex] is the claim that the average change in the company’s stock is less than $3.50 per week.
• The p -value is the area in the left-tail of the [latex]t[/latex]-distribution, to the left of [latex]t=-1.5699...[/latex].
• The p -value of 0.0636 tells us that under the assumption that the average change in the stock is $3.50 (the null hypothesis), there is a 6.36% chance that the average change is $3.50 or less.  Compared to the 5% significance level, this is a large probability, and so is likely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely correct, and so the conclusion of the test is to not reject the null hypothesis.  In other words, the company’s claim that the average change in their stock price is $3.50 per week is most likely correct.

A paint manufacturer has their production line set-up so that the average volume of paint in a can is 3.78 liters.  The quality control manager at the plant believes that something has happened with the production and the average volume of paint in the cans has changed.  The quality control department takes a sample of 100 cans and finds the average volume is 3.62 liters with a standard deviation of 0.7 liters.  At the 5% significance level, has the volume of paint in a can changed?

[latex]\begin{eqnarray*} H_0: & & \mu=3.78 \mbox{ liters}  \\ H_a: & & \mu \neq 3.78 \mbox{ liters}  \end{eqnarray*}[/latex]

From the question, we have [latex]n=100[/latex], [latex]\overline{x}=3.62[/latex], [latex]s=0.7[/latex] and [latex]\alpha=0.05[/latex].

This is a test on a population mean where the population standard deviation is unknown (we only know the sample standard deviation [latex]s=0.7[/latex]).  So we use a [latex]t[/latex]-distribution to calculate the p -value.  Because the alternative hypothesis is a [latex]\neq[/latex], the p -value is the sum of area in the tails of the distribution.

To use the t.dist.2t function, we need to calculate out the [latex]t[/latex]-score:

[latex]\begin{eqnarray*} t & = & \frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}} \\ & = & \frac{3.62-3.78}{\frac{0.07}{\sqrt{100}}} \\ & = & -2.2857... \end{eqnarray*}[/latex]

The degrees of freedom for the [latex]t[/latex]-distribution is [latex]n-1=100-1=99[/latex].

So the p -value[latex]=0.0244[/latex].

Because p -value[latex]=0.0244 \lt 0.05=\alpha[/latex], we reject the null hypothesis in favour of the alternative hypothesis.  At the 5% significance level there is enough evidence to suggest that average volume of paint in the cans has changed.

• The null hypothesis [latex]\mu=3.78[/latex] is the claim that the average volume of paint in the cans is 3.78.
• The alternative hypothesis [latex]\mu \neq 3.78[/latex] is the claim that the average volume of paint in the cans is not 3.78.
• Keep all of the decimals throughout the calculation (i.e. in the [latex]t[/latex]-score) to avoid any round-off error in the calculation of the p -value.  This ensures that we get the most accurate value for the p -value.
• The p -value is the sum of the area in the two tails.  The output from the t.dist.2t function is exactly the sum of the area in the two tails, and so is the p -value required for the test.  No additional calculations are required.
• The t.dist.2t function requires that the value entered for the [latex]t[/latex]-score is positive .  A negative [latex]t[/latex]-score entered into the t.dist.2t function generates an error in Excel.  In this case, the value of the [latex]t[/latex]-score is negative, so we must enter the absolute value of this [latex]t[/latex]-score into field 1.
• The p -value of 0.0244 is a small probability compared to the significance level, and so is unlikely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely incorrect, and so the conclusion of the test is to reject the null hypothesis in favour of the alternative hypothesis.  In other words, the average volume of paint in the cans has most likely changed from 3.78 liters.

Watch this video: Hypothesis Testing: t -test, right tail by ExcelIsFun [11:02]

Watch this video: Hypothesis Testing: t -test, left tail by ExcelIsFun [7:48]

Watch this video: Hypothesis Testing: t -test, two tail by ExcelIsFun [8:54]

Concept Review

The hypothesis test for a population mean is a well established process:

• Collect the sample information for the test and identify the significance level.
• When the population standard deviation is unknown, find the p -value (the area in the corresponding tail) for the test using the [latex]t[/latex]-distribution with [latex]\displaystyle{t=\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}}[/latex] and [latex]df=n-1[/latex].
• Compare the p -value to the significance level and state the outcome of the test.

Attribution

“ 9.6   Hypothesis Testing of a Single Mean and Single Proportion “ in Introductory Statistics by OpenStax  is licensed under a  Creative Commons Attribution 4.0 International License.

Introduction to Statistics Copyright © 2022 by Valerie Watts is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

Hypothesis Testing Calculator

Related: confidence interval calculator, type ii error.

The first step in hypothesis testing is to calculate the test statistic. The formula for the test statistic depends on whether the population standard deviation (σ) is known or unknown. If σ is known, our hypothesis test is known as a z test and we use the z distribution. If σ is unknown, our hypothesis test is known as a t test and we use the t distribution. Use of the t distribution relies on the degrees of freedom, which is equal to the sample size minus one. Furthermore, if the population standard deviation σ is unknown, the sample standard deviation s is used instead. To switch from σ known to σ unknown, click on $\boxed{\sigma}$ and select $\boxed{s}$ in the Hypothesis Testing Calculator.

Next, the test statistic is used to conduct the test using either the p-value approach or critical value approach. The particular steps taken in each approach largely depend on the form of the hypothesis test: lower tail, upper tail or two-tailed. The form can easily be identified by looking at the alternative hypothesis (H a ). If there is a less than sign in the alternative hypothesis then it is a lower tail test, greater than sign is an upper tail test and inequality is a two-tailed test. To switch from a lower tail test to an upper tail or two-tailed test, click on $\boxed{\geq}$ and select $\boxed{\leq}$ or $\boxed{=}$, respectively.

In the p-value approach, the test statistic is used to calculate a p-value. If the test is a lower tail test, the p-value is the probability of getting a value for the test statistic at least as small as the value from the sample. If the test is an upper tail test, the p-value is the probability of getting a value for the test statistic at least as large as the value from the sample. In a two-tailed test, the p-value is the probability of getting a value for the test statistic at least as unlikely as the value from the sample.

To test the hypothesis in the p-value approach, compare the p-value to the level of significance. If the p-value is less than or equal to the level of signifance, reject the null hypothesis. If the p-value is greater than the level of significance, do not reject the null hypothesis. This method remains unchanged regardless of whether it's a lower tail, upper tail or two-tailed test. To change the level of significance, click on $\boxed{.05}$. Note that if the test statistic is given, you can calculate the p-value from the test statistic by clicking on the switch symbol twice.

In the critical value approach, the level of significance ($\alpha$) is used to calculate the critical value. In a lower tail test, the critical value is the value of the test statistic providing an area of $\alpha$ in the lower tail of the sampling distribution of the test statistic. In an upper tail test, the critical value is the value of the test statistic providing an area of $\alpha$ in the upper tail of the sampling distribution of the test statistic. In a two-tailed test, the critical values are the values of the test statistic providing areas of $\alpha / 2$ in the lower and upper tail of the sampling distribution of the test statistic.

To test the hypothesis in the critical value approach, compare the critical value to the test statistic. Unlike the p-value approach, the method we use to decide whether to reject the null hypothesis depends on the form of the hypothesis test. In a lower tail test, if the test statistic is less than or equal to the critical value, reject the null hypothesis. In an upper tail test, if the test statistic is greater than or equal to the critical value, reject the null hypothesis. In a two-tailed test, if the test statistic is less than or equal the lower critical value or greater than or equal to the upper critical value, reject the null hypothesis.

When conducting a hypothesis test, there is always a chance that you come to the wrong conclusion. There are two types of errors you can make: Type I Error and Type II Error. A Type I Error is committed if you reject the null hypothesis when the null hypothesis is true. Ideally, we'd like to accept the null hypothesis when the null hypothesis is true. A Type II Error is committed if you accept the null hypothesis when the alternative hypothesis is true. Ideally, we'd like to reject the null hypothesis when the alternative hypothesis is true.

Hypothesis testing is closely related to the statistical area of confidence intervals. If the hypothesized value of the population mean is outside of the confidence interval, we can reject the null hypothesis. Confidence intervals can be found using the Confidence Interval Calculator . The calculator on this page does hypothesis tests for one population mean. Sometimes we're interest in hypothesis tests about two population means. These can be solved using the Two Population Calculator . The probability of a Type II Error can be calculated by clicking on the link at the bottom of the page.

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AP®︎/College Statistics

Course: ap®︎/college statistics   >   unit 10.

• Idea behind hypothesis testing

Examples of null and alternative hypotheses

• Writing null and alternative hypotheses
• P-values and significance tests
• Comparing P-values to different significance levels
• Estimating a P-value from a simulation
• Estimating P-values from simulations
• Using P-values to make conclusions

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Video transcript

7.3 - Comparing Two Population Means

Introduction.

In this section, we are going to approach constructing the confidence interval and developing the hypothesis test similarly to how we approached those of the difference in two proportions.

There are a few extra steps we need to take, however. First, we need to consider whether the two populations are independent. When considering the sample mean, there were two parameters we had to consider, \(\mu\) the population mean, and \(\sigma\) the population standard deviation. Therefore, the second step is to determine if we are in a situation where the population standard deviations are the same or if they are different.

Independent and Dependent Samples

It is important to be able to distinguish between an independent sample or a dependent sample.

The following are examples to illustrate the two types of samples.

Example 7-3: Gas Mileage

We want to compare the gas mileage of two brands of gasoline. Describe how to design a study involving...

• independent sample Answer: Randomly assign 12 cars to use Brand A and another 12 cars to use Brand B.
• dependent samples Answer: Using 12 cars, have each car use Brand A and Brand B. Compare the differences in mileage for each car.
• Design involving independent samples
• Design involving dependent samples
• Answer: Randomly assign half of the subjects to taste Coke and the other half to taste Pepsi.

Answer: Allow all the subjects to rate both Coke and Pepsi. The drinks should be given in random order. The same subject's ratings of the Coke and the Pepsi form a paired data set.

• We randomly select 20 males and 20 females and compare the average time they spend watching TV. Is this an independent sample or paired sample?
• We randomly select 20 couples and compare the time the husbands and wives spend watching TV. Is this an independent sample or paired sample?
• Answer: Independent Sample

Answer: Paired sample

The two types of samples require a different theory to construct a confidence interval and develop a hypothesis test. We consider each case separately, beginning with independent samples.

7.3.1 - Inference for Independent Means

Two-cases for independent means.

As with comparing two population proportions, when we compare two population means from independent populations, the interest is in the difference of the two means. In other words, if \(\mu_1\) is the population mean from population 1 and \(\mu_2\) is the population mean from population 2, then the difference is \(\mu_1-\mu_2\). If \(\mu_1-\mu_2=0\) then there is no difference between the two population parameters.

If each population is normal, then the sampling distribution of \(\bar{x}_i\) is normal with mean \(\mu_i\), standard error \(\dfrac{\sigma_i}{\sqrt{n_i}}\), and the estimated standard error \(\dfrac{s_i}{\sqrt{n_i}}\), for \(i=1, 2\).

Using the Central Limit Theorem, if the population is not normal, then with a large sample, the sampling distribution is approximately normal.

The theorem presented in this Lesson says that if either of the above are true, then \(\bar{x}_1-\bar{x}_2\) is approximately normal with mean \(\mu_1-\mu_2\), and standard error \(\sqrt{\dfrac{\sigma^2_1}{n_1}+\dfrac{\sigma^2_2}{n_2}}\).

However, in most cases, \(\sigma_1\) and \(\sigma_2\) are unknown, and they have to be estimated. It seems natural to estimate \(\sigma_1\) by \(s_1\) and \(\sigma_2\) by \(s_2\). When the sample sizes are small, the estimates may not be that accurate and one may get a better estimate for the common standard deviation by pooling the data from both populations if the standard deviations for the two populations are not that different.

Given this, there are two options for estimating the variances for the independent samples:

• Using pooled variances
• Using unpooled (or unequal) variances

When to use which? When we are reasonably sure that the two populations have nearly equal variances, then we use the pooled variances test. Otherwise, we use the unpooled (or separate) variance test.

7.3.1.1 - Pooled Variances

Confidence intervals for \(\boldsymbol{\mu_1-\mu_2}\): pooled variances.

When we have good reason to believe that the variance for population 1 is equal to that of population 2, we can estimate the common variance by pooling information from samples from population 1 and population 2.

An informal check for this is to compare the ratio of the two sample standard deviations. If the two are equal , the ratio would be 1, i.e. \(\frac{s_1}{s_2}=1\). However, since these are samples and therefore involve error, we cannot expect the ratio to be exactly 1. When the sample sizes are nearly equal (admittedly "nearly equal" is somewhat ambiguous, so often if sample sizes are small one requires they be equal), then a good Rule of Thumb to use is to see if the ratio falls from 0.5 to 2. That is, neither sample standard deviation is more than twice the other.

If this rule of thumb is satisfied, we can assume the variances are equal. Later in this lesson, we will examine a more formal test for equality of variances.

• Let \(n_1\) be the sample size from population 1 and let \(s_1\) be the sample standard deviation of population 1.
• Let \(n_2\) be the sample size from population 2 and \(s_2\) be the sample standard deviation of population 2.

Then the common standard deviation can be estimated by the pooled standard deviation:

\(s_p=\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s^2_2}{n_1+n_2-2}}\)

If we can assume the populations are independent, that each population is normal or has a large sample size, and that the population variances are the same, then it can be shown that...

\(t=\dfrac{\bar{x}_1-\bar{x_2}-0}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\)

follows a t-distribution with \(n_1+n_2-2\) degrees of freedom.

Now, we can construct a confidence interval for the difference of two means, \(\mu_1-\mu_2\).

where \(t_{\alpha/2}\) comes from a t-distribution with \(n_1+n_2-2\) degrees of freedom.

Hypothesis Tests for \(\boldsymbol{\mu_1-\mu_2}\): The Pooled t-test

Now let's consider the hypothesis test for the mean differences with pooled variances.

\(H_0\colon\mu_1-\mu_2=0\)

\(H_a\colon \mu_1-\mu_2\ne0\)

\(H_a\colon \mu_1-\mu_2>0\)

\(H_a\colon \mu_1-\mu_2<0\)

The assumptions/conditions are:

• The populations are independent
• The population variances are equal
• Each population is either normal or the sample size is large.

The test statistic is...

\(t^*=\dfrac{\bar{x}_1-\bar{x}_2-0}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\)

And \(t^*\) follows a t-distribution with degrees of freedom equal to \(df=n_1+n_2-2\).

The p-value, critical value, rejection region, and conclusion are found similarly to what we have done before.

Example 7-4: Comparing Packing Machines

In a packing plant, a machine packs cartons with jars. It is supposed that a new machine will pack faster on the average than the machine currently used. To test that hypothesis, the times it takes each machine to pack ten cartons are recorded. The results, ( machine.txt ), in seconds, are shown in the tables.

\(\bar{x}_1=42.14, \text{s}_1= 0.683\)

\(\bar{x}_2=43.23, \text{s}_2= 0.750\)

Do the data provide sufficient evidence to conclude that, on the average, the new machine packs faster?

• Hypothesis Test
• Confidence Interval

Are these independent samples? Yes, since the samples from the two machines are not related.

Are these large samples or a normal population?

We have \(n_1\lt 30\) and \(n_2\lt 30\). We do not have large enough samples, and thus we need to check the normality assumption from both populations. Let's take a look at the normality plots for this data:

From the normal probability plots, we conclude that both populations may come from normal distributions. Remember the plots do not indicate that they DO come from a normal distribution. It only shows if there are clear violations. We should proceed with caution.

Do the populations have equal variance? No information allows us to assume they are equal. We can use our rule of thumb to see if they are “close.” They are not that different as \(\dfrac{s_1}{s_2}=\dfrac{0.683}{0.750}=0.91\) is quite close to 1. This assumption does not seem to be violated.

We can thus proceed with the pooled t -test.

Let \(\mu_1\) denote the mean for the new machine and \(\mu_2\) denote the mean for the old machine.

The null hypothesis is that there is no difference in the two population means, i.e.

\(H_0\colon \mu_1-\mu_2=0\)

The alternative is that the new machine is faster, i.e.

The significance level is 5%. Since we may assume the population variances are equal, we first have to calculate the pooled standard deviation:

\begin{align} s_p&=\sqrt{\frac{(n_1-1)s^2_1+(n_2-1)s^2_2}{n_1+n_2-2}}\\ &=\sqrt{\frac{(10-1)(0.683)^2+(10-1)(0.750)^2}{10+10-2}}\\ &=\sqrt{\dfrac{9.261}{18}}\\ &=0.7173 \end{align}

The test statistic is:

\begin{align} t^*&=\dfrac{\bar{x}_1-\bar{x}_2-0}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\\ &=\dfrac{42.14-43.23}{0.7173\sqrt{\frac{1}{10}+\frac{1}{10}}}\\&=-3.398 \end{align}

The alternative is left-tailed so the critical value is the value \(a\) such that \(P(T<a)=0.05\), with \(10+10-2=18\) degrees of freedom. The critical value is -1.7341. The rejection region is \(t^*<-1.7341\).

Our test statistic, -3.3978, is in our rejection region, therefore, we reject the null hypothesis. With a significance level of 5%, we reject the null hypothesis and conclude there is enough evidence to suggest that the new machine is faster than the old machine.

To find the interval, we need all of the pieces. We calculated all but one when we conducted the hypothesis test. We only need the multiplier. For a 99% confidence interval, the multiplier is \(t_{0.01/2}\) with degrees of freedom equal to 18. This value is 2.878.

The interval is:

\(\bar{x}_1-\bar{x}_2\pm t_{\alpha/2}s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\)

\((42.14-43.23)\pm 2.878(0.7173)\sqrt{\frac{1}{10}+\frac{1}{10}}\)

\(-1.09\pm 0.9232\)

The 99% confidence interval is (-2.013, -0.167).

We are 99% confident that the difference between the two population mean times is between -2.012 and -0.167.

  Minitab: 2-Sample t-test - Pooled

The following steps are used to conduct a 2-sample t-test for pooled variances in Minitab.

• Choose Stat > Basic Statistics > 2-Sample t .

• Select the Options button and enter the desired 'confidence level', 'null hypothesis value' (again for our class this will be 0), and select the correct 'alternative hypothesis' from the drop-down menu. Finally, check the box for 'assume equal variances'. This latter selection should only be done when we have verified the two variances can be assumed equal.

The Minitab output for the packing time example:

Two-Sample T-Test and CI: New Machine, Old Machine

μ 1 : mean of New Machine

μ 2 : mean of Old Machine

Difference: μ 1 - μ 2

Equal variances are assumed for this analysis.

Descriptive Statistics

Estimation for difference.

Alternative hypothesis

H 1 : μ 1 - μ 2 < 0

7.3.1.2 - Unpooled Variances

When the assumption of equal variances is not valid, we need to use separate, or unpooled, variances. The mathematics and theory are complicated for this case and we intentionally leave out the details.

We still have the following assumptions:

If the assumptions are satisfied, then

\(t^*=\dfrac{\bar{x}_1-\bar{x_2}-0}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}\)

will have a t-distribution with degrees of freedom

\(df=\dfrac{(n_1-1)(n_2-1)}{(n_2-1)C^2+(1-C)^2(n_1-1)}\)

where \(C=\dfrac{\frac{s^2_1}{n_1}}{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}\).

Where \(t_{\alpha/2}\) comes from the t-distribution using the degrees of freedom above.

Minitab ®

Minitab: unpooled t-test.

To perform a separate variance 2-sample, t -procedure use the same commands as for the pooled procedure EXCEPT we do NOT check box for 'Use Equal Variances.'

• Choose Stat > Basic Statistics > 2-sample t
• Select the Options box and enter the desired 'Confidence level,' 'Null hypothesis value' (again for our class this will be 0), and select the correct 'Alternative hypothesis' from the drop-down menu.

For some examples, one can use both the pooled t-procedure and the separate variances (non-pooled) t -procedure and obtain results that are close to each other. However, when the sample standard deviations are very different from each other, and the sample sizes are different, the separate variances 2-sample t -procedure is more reliable.

Example 7-5: Grade Point Average

Independent random samples of 17 sophomores and 13 juniors attending a large university yield the following data on grade point averages ( student_gpa.txt ):

At the 5% significance level, do the data provide sufficient evidence to conclude that the mean GPAs of sophomores and juniors at the university differ?

There is no indication that there is a violation of the normal assumption for both samples. As before, we should proceed with caution.

Now, we need to determine whether to use the pooled t-test or the non-pooled (separate variances) t -test. The summary statistics are:

The standard deviations are 0.520 and 0.3093 respectively; both the sample sizes are small, and the standard deviations are quite different from each other. We, therefore, decide to use an unpooled t -test.

The null and alternative hypotheses are:

\(H_0\colon \mu_1-\mu_2=0\) vs \(H_a\colon \mu_1-\mu_2\ne0\)

The significance level is 5%. Perform the 2-sample t -test in Minitab with the appropriate alternative hypothesis.

Remember, the default for the 2-sample t-test in Minitab is the non-pooled one. Minitab generates the following output.

Two sample T for sophomores vs juniors

95% CI for mu sophomore - mu juniors: (-0.45, 0.173)

T-Test mu sophomore = mu juniors (Vs no =): T = -0.92

P = 0.36 DF = 26

Since the p-value of 0.36 is larger than \(\alpha=0.05\), we fail to reject the null hypothesis.

At 5% level of significance, the data does not provide sufficient evidence that the mean GPAs of sophomores and juniors at the university are different.

95% CI for mu sophomore- mu juniors is;

(-0.45, 0.173)

We are 95% confident that the difference between the mean GPA of sophomores and juniors is between -0.45 and 0.173.

7.3.2 - Inference for Paired Means

When we developed the inference for the independent samples, we depended on the statistical theory to help us. The theory, however, required the samples to be independent. What can we do when the two samples are not independent, i.e., the data is paired?

Consider an example where we are interested in a person’s weight before implementing a diet plan and after. Since the interest is focusing on the difference, it makes sense to “condense” these two measurements into one and consider the difference between the two measurements. For example, if instead of considering the two measures, we take the before diet weight and subtract the after diet weight. The difference makes sense too! It is the weight lost on the diet.

When we take the two measurements to make one measurement (i.e., the difference), we are now back to the one sample case! Now we can apply all we learned for the one sample mean to the difference (Cool!)

The Confidence Interval for the Difference of Paired Means, \(\mu_d\)

When we consider the difference of two measurements, the parameter of interest is the mean difference, denoted \(\mu_d\). The mean difference is the mean of the differences. We are still interested in comparing this difference to zero.

Suppose we have two paired samples of size \(n\):

\(x_1, x_2, …., x_n\) and \(y_1, y_2, … , y_n\)

Their difference can be denoted as:

\(d_1=x_1-y_1, d_2=x_2-y_2, …., d_n=x_n-y_n\)

The sample mean of the differences is:

\(\bar{d}=\frac{1}{n}\sum_{i=1}^n d_i\)

Denote the sample standard deviation of the differences as \(s_d\).

If \(\bar{d}\) is normal (or the sample size is large), the sampling distribution of \(\bar{d}\) is (approximately) normal with mean \(\mu_d\), standard error \(\dfrac{\sigma_d}{\sqrt{n}}\), and estimated standard error \(\dfrac{s_d}{\sqrt{n}}\).

At this point, the confidence interval will be the same as that of one sample.

\(\bar{d}\pm t_{\alpha/2}\frac{s_d}{\sqrt{n}}\)

where \(t_{\alpha/2}\) comes from \(t\)-distribution with \(n-1\) degrees of freedom

Example 7-6: Zinc Concentrations

Trace metals in drinking water affect the flavor and an unusually high concentration can pose a health hazard. Ten pairs of data were taken measuring zinc concentration in bottom water and surface water ( zinc_conc.txt ).

Does the data suggest that the true average concentration in the bottom water is different than that of surface water? Construct a confidence interval to address this question.

Zinc concentrations

In this example, the response variable is concentration and is a quantitative measurement. The explanatory variable is location (bottom or surface) and is categorical. The two populations (bottom or surface) are not independent. Therefore, we are in the paired data setting. The parameter of interest is \(\mu_d\).

Find the difference as the concentration of the bottom water minus the concentration of the surface water.

Since the problem did not provide a confidence level, we should use 5%.

To use the methods we developed previously, we need to check the conditions. The problem does not indicate that the differences come from a normal distribution and the sample size is small (n=10). We should check, using the Normal Probability Plot to see if there is any violation. First, we need to find the differences.

All of the differences fall within the boundaries, so there is no clear violation of the assumption. We can proceed with using our tools, but we should proceed with caution.

We need all of the pieces for the confidence interval. The sample mean difference is \(\bar{d}=0.0804\) and the standard deviation is \(s_d=0.0523\). For practice, you should find the sample mean of the differences and the standard deviation by hand. With \(n-1=10-1=9\) degrees of freedom, \(t_{0.05/2}=2.2622\).

The 95% confidence interval for the mean difference, \(\mu_d\) is:

\(\bar{d}\pm t_{\alpha/2}\dfrac{s_d}{\sqrt{n}}\)

\(0.0804\pm 2.2622\left( \dfrac{0.0523}{\sqrt{10}}\right)\)

(0.04299, 0.11781)

We are 95% confident that the population mean difference of bottom water and surface water zinc concentration is between 0.04299 and 0.11781.

If there is no difference between the means of the two measures, then the mean difference will be 0. Since 0 is not in our confidence interval, then the means are statistically different (or statistical significant or statistically different).

Note! Minitab will calculate the confidence interval and a hypothesis test simultaneously. We demonstrate how to find this interval using Minitab after presenting the hypothesis test.

Hypothesis Test for the Difference of Paired Means, \(\mu_d\)

In this section, we will develop the hypothesis test for the mean difference for paired samples. As we learned in the previous section, if we consider the difference rather than the two samples, then we are back in the one-sample mean scenario.

The possible null and alternative hypotheses are:

\(H_0\colon \mu_d=0\)

\(H_a\colon \mu_d\ne 0\)

\(H_a\colon \mu_d>0\)

\(H_a\colon \mu_d<0\)

We still need to check the conditions and at least one of the following need to be satisfied:

• The differences of the paired follow a normal distribution
• The sample size is large, \(n>30\).

If at least one is satisfied then...

\(t^*=\dfrac{\bar{d}-0}{\frac{s_d}{\sqrt{n}}}\)

Will follow a t-distribution with \(n-1\) degrees of freedom.

The same process for the hypothesis test for one mean can be applied. The test for the mean difference may be referred to as the paired t-test or the test for paired means.

Example 7-7: Zinc Concentrations - Hypothesis Test

Recall the zinc concentration example. Does the data suggest that the true average concentration in the bottom water exceeds that of surface water? Conduct this test using the rejection region approach. ( zinc_conc.txt ).

If we find the difference as the concentration of the bottom water minus the concentration of the surface water, then null and alternative hypotheses are:

\(H_0\colon \mu_d=0\) vs \(H_a\colon \mu_d>0\)

Note! If the difference was defined as surface - bottom, then the alternative would be left-tailed.

The desired significance level was not stated so we will use \(\alpha=0.05\).

The assumptions were discussed when we constructed the confidence interval for this example. Remember although the Normal Probability Plot for the differences showed no violation, we should still proceed with caution.

The next step is to find the critical value and the rejection region. The critical value is the value \(a\) such that \(P(T>a)=0.05\). Using the table or software, the value is 1.8331. For a right-tailed test, the rejection region is \(t^*>1.8331\).

Recall from the previous example, the sample mean difference is \(\bar{d}=0.0804\) and the sample standard deviation of the difference is \(s_d=0.0523\). Therefore, the test statistic is:

\(t^*=\dfrac{\bar{d}-0}{\frac{s_d}{\sqrt{n}}}=\dfrac{0.0804}{\frac{0.0523}{\sqrt{10}}}=4.86\)

The value of our test statistic falls in the rejection region. Therefore, we reject the null hypothesis. With a significance level of 5%, there is enough evidence in the data to suggest that the bottom water has higher concentrations of zinc than the surface level.

Minitab ®  – Paired t-Test

You can use a paired t -test in Minitab to perform the test. Alternatively, you can perform a 1-sample t -test on difference = bottom - surface.

• Choose Stat > Basic Statistics > Paired t
• Click Options to specify the confidence level for the interval and the alternative hypothesis you want to test. The default null hypothesis is 0.

Zinc Concentrations Example

The Minitab output for paired T for bottom - surface is as follows:

Paired T for bottom - surface

95% lower bound for mean difference: 0.0505

T-Test of mean difference = 0 (vs > 0): T-Value = 4.86 P-Value = 0.000

Note! In Minitab, if you choose a lower-tailed or an upper-tailed hypothesis test, an upper or lower confidence bound will be constructed, respectively, rather than a confidence interval.

Using the p -value to draw a conclusion about our example:

p -value = 0.000 < 0.05

Reject \(H_0\) and conclude that bottom zinc concentration is higher than surface zinc concentration.

Additional Notes

• For the zinc concentration problem, if you do not recognize the paired structure, but mistakenly use the 2-sample t -test treating them as independent samples, you will not be able to reject the null hypothesis. This demonstrates the importance of distinguishing the two types of samples. Also, it is wise to design an experiment efficiently whenever possible.
• What if the assumption of normality is not satisfied? Considering a nonparametric test would be wise.

Statistics Made Easy

How to Write a Null Hypothesis (5 Examples)

A hypothesis test uses sample data to determine whether or not some claim about a population parameter is true.

Whenever we perform a hypothesis test, we always write a null hypothesis and an alternative hypothesis, which take the following forms:

H 0 (Null Hypothesis): Population parameter =,  ≤, ≥ some value

H A  (Alternative Hypothesis): Population parameter <, >, ≠ some value

Note that the null hypothesis always contains the equal sign .

We interpret the hypotheses as follows:

Null hypothesis: The sample data provides no evidence to support some claim being made by an individual.

Alternative hypothesis: The sample data  does provide sufficient evidence to support the claim being made by an individual.

For example, suppose it’s assumed that the average height of a certain species of plant is 20 inches tall. However, one botanist claims the true average height is greater than 20 inches.

To test this claim, she may go out and collect a random sample of plants. She can then use this sample data to perform a hypothesis test using the following two hypotheses:

H 0 : μ ≤ 20 (the true mean height of plants is equal to or even less than 20 inches)

H A : μ > 20 (the true mean height of plants is greater than 20 inches)

If the sample data gathered by the botanist shows that the mean height of this species of plants is significantly greater than 20 inches, she can reject the null hypothesis and conclude that the mean height is greater than 20 inches.

Read through the following examples to gain a better understanding of how to write a null hypothesis in different situations.

Example 1: Weight of Turtles

A biologist wants to test whether or not the true mean weight of a certain species of turtles is 300 pounds. To test this, he goes out and measures the weight of a random sample of 40 turtles.

Here is how to write the null and alternative hypotheses for this scenario:

H 0 : μ = 300 (the true mean weight is equal to 300 pounds)

H A : μ ≠ 300 (the true mean weight is not equal to 300 pounds)

Example 2: Height of Males

It’s assumed that the mean height of males in a certain city is 68 inches. However, an independent researcher believes the true mean height is greater than 68 inches. To test this, he goes out and collects the height of 50 males in the city.

H 0 : μ ≤ 68 (the true mean height is equal to or even less than 68 inches)

H A : μ > 68 (the true mean height is greater than 68 inches)

Example 3: Graduation Rates

A university states that 80% of all students graduate on time. However, an independent researcher believes that less than 80% of all students graduate on time. To test this, she collects data on the proportion of students who graduated on time last year at the university.

H 0 : p ≥ 0.80 (the true proportion of students who graduate on time is 80% or higher)

H A : μ < 0.80 (the true proportion of students who graduate on time is less than 80%)

Example 4: Burger Weights

A food researcher wants to test whether or not the true mean weight of a burger at a certain restaurant is 7 ounces. To test this, he goes out and measures the weight of a random sample of 20 burgers from this restaurant.

H 0 : μ = 7 (the true mean weight is equal to 7 ounces)

H A : μ ≠ 7 (the true mean weight is not equal to 7 ounces)

Example 5: Citizen Support

A politician claims that less than 30% of citizens in a certain town support a certain law. To test this, he goes out and surveys 200 citizens on whether or not they support the law.

H 0 : p ≥ .30 (the true proportion of citizens who support the law is greater than or equal to 30%)

H A : μ < 0.30 (the true proportion of citizens who support the law is less than 30%)

Additional Resources

Introduction to Hypothesis Testing Introduction to Confidence Intervals An Explanation of P-Values and Statistical Significance

Hey there. My name is Zach Bobbitt. I have a Master of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike.  My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.

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10.1 Two Population Means with Unknown Standard Deviations

HYPOTHESIS TESTING FOR DIFFERENCE BETWEEN MEANS WITH TWO SAMPLES

When testing about claims that compare two groups (populations), we’ll need to obtain samples from each group to test those claims. Typically studies comparing two groups (such as the treatment and control groups in experiments/clinical trials or a company claiming that its product lasts longer than than its competitors, etc.) will require us to obtain sample from each group and use the difference between the two groups to test claim made about those populations.

Independent Samples: Two samples are independent if the selection of sample members from one population does not in any way affects the selection of sample members in the second population.

In this unit, we’ll perform testing using independent samples as well as depending samples. Keep in mind that the tests using independent samples follow a different procedure than that of the dependent samples.

All of hypothesis testing for means with two samples will fall into one of the following scenario:

Note that for scenarios 1 and 2, the null hypothesis may be written with [latex]=[/latex] sign instead of [latex]\ge[/latex] or [latex]\le[/latex].

When conducting a hypothesis test that compares two independent population means, the following characteristics should be present:

• The two independent samples are simple random samples that are independent from two distinct populations.
• If the sample sizes are small, the populations should be normally distributed. This requirement can be relaxed for larger sample sizes.

If the standard deviations of the two populations (or variances) are known, then we can conduct a [latex]z-[/latex]test to test the difference between the means. However, more often not, population standard deviations are not known. Therefore we normally do not perform this z-test. For sample sizes that are large, this test may be fine.

For the more realistic scenario when we don’t know the population standard deviations, we can estimate them using the two sample standard deviations from our independent samples. If we happen to know that the population standard deviations are the same (or similar), then we can combine sample information to compute a pooled estimate of the standard deviation and use that to compute the standard error (SE) , of the difference in sample means , [latex]\bar X_1-\bar X_2[/latex]: \[s_\text{pooled}=\sqrt{\frac{(n_1-1)s_1^2 +(n_2-1)s_2^2}{n_1+n_2-2}}\]

The standard error, SE: \[\text{SE}=s_\text{pooled}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}.\] The degrees of freedom, [latex]df[/latex], for the underlying [latex]t-[/latex]distribution is given by: [latex]df=n_1+n_2-2[/latex].

For situations where the population standard deviations are not known to be equal (unpooled) , we calculate the estimated standard deviation, or standard error (SE), of the difference in sample means, [latex]\bar X_1-\bar X_2[/latex] as follows: \[\text{SE}=\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}.\] The degrees of freedom, [latex]df[/latex], for the underlying [latex]t-[/latex]distribution is calculated either by using a conservative estimate [latex]df=\text{min}(n_1-1, n_2-1)[/latex] or by a more complicated formula, which our calculators can handle easily. \[df = \frac{\left( \frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1}\left(\frac{s_1^2 }{n_1} \right)^2+\frac{1}{n_2-1} \left(\frac{s_1^2 }{n_1} \right)^2}\]

You don’t need to compute they gnarly formula by hand, just use a calculator. When reporting the [latex]df[/latex] computed using this formula, it doesn’t always come out to be an whole number so you round down to the nearest whole number if required.

If we’re comparing two populations, we may find that the population means are different. It may be due to a difference in the populations or it may be due to chance . A hypothesis test can help determine if a difference in the estimated means reflects a difference in the population means.

Generally, the null hypothesis states that the two population means are the same. That is [latex]H_0:\mu_1=\mu_2[/latex]. Recall that we start hypothesis testing by assuming the equality in the null hypothesis is true. In this case, we’re assuming that [latex]\mu_1=\mu_2[/latex] which means that the two population means are the same (no difference). With no difference between the population means, we can compute the the standard deviation of the sampling distribution of the difference of the sample means as above (see both pooled and unpooled cases) and the test statistic is: \[ \frac{(\bar x_1-\bar x_2 )-(\mu_1-\mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}.\] These formulas for two sample test for the difference of means are shown here so that you can recognize them for what they are but when conducting two sample tests, we’ll be using calculators to speed up our calculations so that you won’t have to remember these formulas.

WORKED OUT EXAMPLE – Testing with Two Samples

College Students and Drinking Habits: A public health official is studying differences in drinking habits among students at two different universities. They collect a random sample of students independently from each of the two universities and ask each student how many alcoholic drinks they consumed in the previous week. The results are summarized in the table below.

The official wants to determine whether these data provide significant evidence that students at One State drink more than students at State Two .

First of all, there are two groups (or populations) that we are comparing here– the students at two universities. We’re testing to see if the average amount of drink is greater for One State . In order to do so, we’ll compare the average number of drinks for each of the two groups. Since we’re dealing with averages (or means), this test is going to be about means (more precisely the difference between population means). Note that since the data for each university is numerical, this points toward means as well.

Let’s label One State University students’ population as group 1 and University of State Two as group 2 . We don’t know the average number of drinks for all students at either university, so we label them as: [latex]\mu_1[/latex] = the mean number of drinks consumed by students from One State [latex]\mu_2[/latex] = the mean number of drinks consumed by students from State Two

Sample information Sample information is given in table in the question.

STEP 1: Write the claim (or what’s being tested) using mathematical symbols The claim we’re testing is that that students at One State drink more than students at State Two. If One State students drink less, their mean number of drinks will be lower than that of State Two students. That is, [latex]\mu_1\gt\mu_2[/latex].

STEP 2: Write the opposite of the claim using mathematical symbols Opposite of the claim: [latex]\mu_1\le\mu_2[/latex].

STEP 3: Write the null and alternative hypothesis The null hypothesis is the statement that contains the condition of equality which in this case is the opposite of the claim

STEP 4: Identify the tail-type of the test and the level of significance, α. If α is not given, assume it to be 5% or 0.05. This is a right tailed test. Why? Since the level of significance is not given, use [latex]\alpha=0.05[/latex] (or 5% level of significance)

STEP 5: Check if CLT conditions are valid. Sample sizes from both the groups are at least 30, so CLT conditions are met. We’ll perform a t-test for independent means since the population standard deviations are not known.

This is the stage where we assume the equality in the null hypothesis is true, that is, assume [latex]\mu_1=\mu_2[/latex]. We’re assuming that there’s no difference between the two universities when it comes to their students’ drinking habits. We know that the sampling distribution of the difference of sample means will be approximately normally distributed with mean of [latex]\mu_{\bar x_1-\bar x_2} = \mu_1 - \mu_2 = 0[/latex]. For the standard deviation of the sampling distribution, we need to know if the population variances are equal or unequal. If the variances are equal, then we pool their variances (POOLED) , and if they are not equal, then the variances can not be pooled (NOT POOLED) . In this question, we’re not told that the two universities have similar variances when it comes to student drinking, so we’ll proceed with an unpooled variance given by:\[\mu_{\bar x_1-\bar x_2} = \mu_1 – \mu_2 = 0 \]\[s_{\bar x_1-\bar x_2 } = \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}} = \sqrt{\frac{2.3^2}{40}+\frac{1.9^2}{49}}= 0.453787912342 \]

STEP 6: Plot [latex]\hat p[/latex] and find the p-value. [latex]\bar x_1-\bar x_2[/latex] is plotted on the graph above. Since the test is right-tailed, the p-value is the area to the tail from [latex]\bar x_1-\bar x_2[/latex], shaded in purple on the graph above. To find the p-value, we can use one of the following methods:

lower: 2.64440715004 upper: 10^99 df: 75.5143574829 Paste

STEP 7: Make the decision by comparing p-value with α p-value is 0.00497452874312 which is less than [latex]\alpha[/latex] of 5%. Since p-value is less than [latex]alpha[/latex], we reject the null hypothesis ⇒ There’s enough evidence to support Ha.

STEP 8: Interpret your decision in the context of original claim The samples provide significant evidence at a significance level of 5% to conclude that students at One State drink more than students at State Two .

ONLINE CALCULATOR Approach

Go to Inference for the Mean @ rsubedi.com

Number of Samples

Independent or dependent samples?

Distribution Type for Test

For df use: Welch-Satterthwaite

Alternative Hypothesis

Confidence Interval? ← Optional

Sample Statistics Information:

CALCULATE Results show in a panel to the right. Test statistic t , p -value, and df are displayed.

Go to: Two Independent Samples With Statistics  from the list of online calculators

Enter the following values and press  Calculate .

CALCULATE Results displayed are:

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Null Hypothesis

Null Hypothesis , often denoted as H 0, is a foundational concept in statistical hypothesis testing. It represents an assumption that no significant difference, effect, or relationship exists between variables within a population. It serves as a baseline assumption, positing no observed change or effect occurring. The null is t he truth or falsity of an idea in analysis.

In this article, we will discuss the null hypothesis in detail, along with some solved examples and questions on the null hypothesis.

Table of Content

• What Is a Null Hypothesis?

Symbol of Null Hypothesis

Formula of null hypothesis, types of null hypothesis, principle of null hypothesis, how do you find null hypothesis, what is a null hypothesis.

Null Hypothesis in statistical analysis suggests the absence of statistical significance within a specific set of observed data. Hypothesis testing, using sample data, evaluates the validity of this hypothesis. Commonly denoted as H 0 or simply “null,” it plays an important role in quantitative analysis, examining theories related to markets, investment strategies, or economies to determine their validity.

Definition of Null Hypothesis

Null Hypothesis represent a default position, often suggesting no effect or difference, against which researchers compare their experimental results. The Null Hypothesis, often denoted as H 0 , asserts a default assumption in statistical analysis. It posits no significant difference or effect, serving as a baseline for comparison in hypothesis testing.

Null Hypothesis is represented as H 0 , the Null Hypothesis symbolizes the absence of a measurable effect or difference in the variables under examination.

Certainly, a simple example would be asserting that the mean score of a group is equal to a specified value like stating that the average IQ of a population is 100.

The Null Hypothesis is typically formulated as a statement of equality or absence of a specific parameter in the population being studied. It provides a clear and testable prediction for comparison with the alternative hypothesis. The formulation of the Null Hypothesis typically follows a concise structure, stating the equality or absence of a specific parameter in the population.

Mean Comparison (Two-sample t-test)

H 0 : μ 1 = μ 2

This asserts that there is no significant difference between the means of two populations or groups.

Proportion Comparison

H 0 : p 1 − p 2 = 0

This suggests no significant difference in proportions between two populations or conditions.

Equality in Variance (F-test in ANOVA)

H 0 : σ 1 = σ 2

This states that there’s no significant difference in variances between groups or populations.

Independence (Chi-square Test of Independence):

H 0 : Variables are independent

This asserts that there’s no association or relationship between categorical variables.

Null Hypotheses vary including simple and composite forms, each tailored to the complexity of the research question. Understanding these types is pivotal for effective hypothesis testing.

Equality Null Hypothesis (Simple Null Hypothesis)

The Equality Null Hypothesis, also known as the Simple Null Hypothesis, is a fundamental concept in statistical hypothesis testing that assumes no difference, effect or relationship between groups, conditions or populations being compared.

Non-Inferiority Null Hypothesis

In some studies, the focus might be on demonstrating that a new treatment or method is not significantly worse than the standard or existing one.

Superiority Null Hypothesis

The concept of a superiority null hypothesis comes into play when a study aims to demonstrate that a new treatment, method, or intervention is significantly better than an existing or standard one.

Independence Null Hypothesis

In certain statistical tests, such as chi-square tests for independence, the null hypothesis assumes no association or independence between categorical variables.

Homogeneity Null Hypothesis

In tests like ANOVA (Analysis of Variance), the null hypothesis suggests that there’s no difference in population means across different groups.

Examples of Null Hypothesis

• Medicine: Null Hypothesis: “No significant difference exists in blood pressure levels between patients given the experimental drug versus those given a placebo.”
• Education: Null Hypothesis: “There’s no significant variation in test scores between students using a new teaching method and those using traditional teaching.”
• Economics: Null Hypothesis: “There’s no significant change in consumer spending pre- and post-implementation of a new taxation policy.”
• Environmental Science: Null Hypothesis: “There’s no substantial difference in pollution levels before and after a water treatment plant’s establishment.”

The principle of the null hypothesis is a fundamental concept in statistical hypothesis testing. It involves making an assumption about the population parameter or the absence of an effect or relationship between variables.

In essence, the null hypothesis (H 0 ) proposes that there is no significant difference, effect, or relationship between variables. It serves as a starting point or a default assumption that there is no real change, no effect or no difference between groups or conditions.

Null Hypothesis Rejection

Rejecting the Null Hypothesis occurs when statistical evidence suggests a significant departure from the assumed baseline. It implies that there is enough evidence to support the alternative hypothesis, indicating a meaningful effect or difference. Null Hypothesis rejection occurs when statistical evidence suggests a deviation from the assumed baseline, prompting a reconsideration of the initial hypothesis.

Identifying the Null Hypothesis involves defining the status quotient, asserting no effect and formulating a statement suitable for statistical analysis.

When is Null Hypothesis Rejected?

The Null Hypothesis is rejected when statistical tests indicate a significant departure from the expected outcome, leading to the consideration of alternative hypotheses. It occurs when statistical evidence suggests a deviation from the assumed baseline, prompting a reconsideration of the initial hypothesis.

Null Hypothesis and Alternative Hypothesis

In the realm of hypothesis testing, the null hypothesis (H 0 ) and alternative hypothesis (H₁ or Ha) play critical roles. The null hypothesis generally assumes no difference, effect, or relationship between variables, suggesting that any observed change or effect is due to random chance. Its counterpart, the alternative hypothesis, asserts the presence of a significant difference, effect, or relationship between variables, challenging the null hypothesis. These hypotheses are formulated based on the research question and guide statistical analyses.

Null Hypothesis vs Alternative Hypothesis

The null hypothesis (H 0 ) serves as the baseline assumption in statistical testing, suggesting no significant effect, relationship, or difference within the data. It often proposes that any observed change or correlation is merely due to chance or random variation. Conversely, the alternative hypothesis (H 1 or Ha) contradicts the null hypothesis, positing the existence of a genuine effect, relationship or difference in the data. It represents the researcher’s intended focus, seeking to provide evidence against the null hypothesis and support for a specific outcome or theory. These hypotheses form the crux of hypothesis testing, guiding the assessment of data to draw conclusions about the population being studied.

Example of Alternative and Null Hypothesis

Let’s envision a scenario where a researcher aims to examine the impact of a new medication on reducing blood pressure among patients. In this context:

Null Hypothesis (H 0 ): “The new medication does not produce a significant effect in reducing blood pressure levels among patients.”

Alternative Hypothesis (H 1 or Ha): “The new medication yields a significant effect in reducing blood pressure levels among patients.”

The null hypothesis implies that any observed alterations in blood pressure subsequent to the medication’s administration are a result of random fluctuations rather than a consequence of the medication itself. Conversely, the alternative hypothesis contends that the medication does indeed generate a meaningful alteration in blood pressure levels, distinct from what might naturally occur or by random chance.

Also, Check

Solved Examples on Null Hypothesis

Example 1: A researcher claims that the average time students spend on homework is 2 hours per night.

Null Hypothesis (H 0 ): The average time students spend on homework is equal to 2 hours per night. Data: A random sample of 30 students has an average homework time of 1.8 hours with a standard deviation of 0.5 hours. Test Statistic and Decision: Using a t-test, if the calculated t-statistic falls within the acceptance region, we fail to reject the null hypothesis. If it falls in the rejection region, we reject the null hypothesis. Conclusion: Based on the statistical analysis, we fail to reject the null hypothesis, suggesting that there is not enough evidence to dispute the claim of the average homework time being 2 hours per night.

Example 2: A company asserts that the error rate in its production process is less than 1%.

Null Hypothesis (H 0 ): The error rate in the production process is 1% or higher. Data: A sample of 500 products shows an error rate of 0.8%. Test Statistic and Decision: Using a z-test, if the calculated z-statistic falls within the acceptance region, we fail to reject the null hypothesis. If it falls in the rejection region, we reject the null hypothesis. Conclusion: The statistical analysis supports rejecting the null hypothesis, indicating that there is enough evidence to dispute the company’s claim of an error rate of 1% or higher.

Null Hypothesis – Practice Problems

Q1. A researcher claims that the average time spent by students on homework is less than 2 hours per day. Formulate the null hypothesis for this claim?

Q2. A manufacturing company states that their new machine produces widgets with a defect rate of less than 5%. Write the null hypothesis to test this claim?

Q3. An educational institute believes that their online course completion rate is at least 60%. Develop the null hypothesis to validate this assertion?

Q4. A restaurant claims that the waiting time for customers during peak hours is not more than 15 minutes. Formulate the null hypothesis for this claim?

Q5. A study suggests that the mean weight loss after following a specific diet plan for a month is more than 8 pounds. Construct the null hypothesis to evaluate this statement?

Null Hypothesis – Frequently Asked Questions

How to form a null hypothesis.

A null hypothesis is formed based on the assumption that there is no significant difference or effect between the groups being compared or no association between variables being tested. It often involves stating that there is no relationship, no change, or no effect in the population being studied.

When Do we reject the Null Hypothesis?

In statistical hypothesis testing, if the p-value (the probability of obtaining the observed results) is lower than the chosen significance level (commonly 0.05), we reject the null hypothesis. This suggests that the data provides enough evidence to refute the assumption made in the null hypothesis.

What is a Null Hypothesis in Research?

In research, the null hypothesis represents the default assumption or position that there is no significant difference or effect. Researchers often try to test this hypothesis by collecting data and performing statistical analyses to see if the observed results contradict the assumption.

What Are Alternative and Null Hypotheses?

The null hypothesis (H0) is the default assumption that there is no significant difference or effect. The alternative hypothesis (H1 or Ha) is the opposite, suggesting there is a significant difference, effect or relationship.

What Does it Mean to Reject the Null Hypothesis?

Rejecting the null hypothesis implies that there is enough evidence in the data to support the alternative hypothesis. In simpler terms, it suggests that there might be a significant difference, effect or relationship between the groups or variables being studied.

How to Find Null Hypothesis?

Formulating a null hypothesis often involves considering the research question and assuming that no difference or effect exists. It should be a statement that can be tested through data collection and statistical analysis, typically stating no relationship or no change between variables or groups.

How is Null Hypothesis denoted?

The null hypothesis is commonly symbolized as H 0 in statistical notation.

What is the Purpose of the Null hypothesis in Statistical Analysis?

The null hypothesis serves as a starting point for hypothesis testing, enabling researchers to assess if there’s enough evidence to reject it in favor of an alternative hypothesis.

What happens if we Reject the Null hypothesis?

Rejecting the null hypothesis implies that there is sufficient evidence to support an alternative hypothesis, suggesting a significant effect or relationship between variables.

Is it Possible to Prove the Null Hypothesis?

No, statistical testing aims to either reject or fail to reject the null hypothesis based on evidence from sample data. It does not prove the null hypothesis to be true.

What are Test for Null Hypothesis?

Various statistical tests, such as t-tests or chi-square tests, are employed to evaluate the validity of the Null Hypothesis in different scenarios.

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