sample mean in hypothesis

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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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8.4.3 Hypothesis Testing for the Mean

$\quad$ $H_0$: $\mu=\mu_0$, $\quad$ $H_1$: $\mu \neq \mu_0$.

$\quad$ $H_0$: $\mu \leq \mu_0$, $\quad$ $H_1$: $\mu > \mu_0$.

$\quad$ $H_0$: $\mu \geq \mu_0$, $\quad$ $H_1$: $\mu \lt \mu_0$.

Two-sided Tests for the Mean:

Therefore, we can suggest the following test. Choose a threshold, and call it $c$. If $|W| \leq c$, accept $H_0$, and if $|W|>c$, accept $H_1$. How do we choose $c$? If $\alpha$ is the required significance level, we must have

• As discussed above, we let \begin{align}%\label{} W(X_1,X_2, \cdots,X_n)=\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}}. \end{align} Note that, assuming $H_0$, $W \sim N(0,1)$. We will choose a threshold, $c$. If $|W| \leq c$, we accept $H_0$, and if $|W|>c$, accept $H_1$. To choose $c$, we let \begin{align} P(|W| > c \; | \; H_0) =\alpha. \end{align} Since the standard normal PDF is symmetric around $0$, we have \begin{align} P(|W| > c \; | \; H_0) = 2 P(W>c | \; H_0). \end{align} Thus, we conclude $P(W>c | \; H_0)=\frac{\alpha}{2}$. Therefore, \begin{align} c=z_{\frac{\alpha}{2}}. \end{align} Therefore, we accept $H_0$ if \begin{align} \left|\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} \right| \leq z_{\frac{\alpha}{2}}, \end{align} and reject it otherwise.
• We have \begin{align} \beta (\mu) &=P(\textrm{type II error}) = P(\textrm{accept }H_0 \; | \; \mu) \\ &= P\left(\left|\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} \right| \lt z_{\frac{\alpha}{2}}\; | \; \mu \right). \end{align} If $X_i \sim N(\mu,\sigma^2)$, then $\overline{X} \sim N(\mu, \frac{\sigma^2}{n})$. Thus, \begin{align} \beta (\mu)&=P\left(\left|\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} \right| \lt z_{\frac{\alpha}{2}}\; | \; \mu \right)\\ &=P\left(\mu_0- z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}} \leq \overline{X} \leq \mu_0+ z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}}\right)\\ &=\Phi\left(z_{\frac{\alpha}{2}}+\frac{\mu_0-\mu}{\sigma / \sqrt{n}}\right)-\Phi\left(-z_{\frac{\alpha}{2}}+\frac{\mu_0-\mu}{\sigma / \sqrt{n}}\right). \end{align}
• Let $S^2$ be the sample variance for this random sample. Then, the random variable $W$ defined as \begin{equation} W(X_1,X_2, \cdots, X_n)=\frac{\overline{X}-\mu_0}{S / \sqrt{n}} \end{equation} has a $t$-distribution with $n-1$ degrees of freedom, i.e., $W \sim T(n-1)$. Thus, we can repeat the analysis of Example 8.24 here. The only difference is that we need to replace $\sigma$ by $S$ and $z_{\frac{\alpha}{2}}$ by $t_{\frac{\alpha}{2},n-1}$. Therefore, we accept $H_0$ if \begin{align} |W| \leq t_{\frac{\alpha}{2},n-1}, \end{align} and reject it otherwise. Let us look at a numerical example of this case.

$\quad$ $H_0$: $\mu=170$, $\quad$ $H_1$: $\mu \neq 170$.

• Let's first compute the sample mean and the sample standard deviation. The sample mean is \begin{align}%\label{} \overline{X}&=\frac{X_1+X_2+X_3+X_4+X_5+X_6+X_7+X_8+X_9}{9}\\ &=165.8 \end{align} The sample variance is given by \begin{align}%\label{} {S}^2=\frac{1}{9-1} \sum_{k=1}^9 (X_k-\overline{X})^2&=68.01 \end{align} The sample standard deviation is given by \begin{align}%\label{} S&= \sqrt{S^2}=8.25 \end{align} The following MATLAB code can be used to obtain these values: x=[176.2,157.9,160.1,180.9,165.1,167.2,162.9,155.7,166.2]; m=mean(x); v=var(x); s=std(x); Now, our test statistic is \begin{align} W(X_1,X_2, \cdots, X_9)&=\frac{\overline{X}-\mu_0}{S / \sqrt{n}}\\ &=\frac{165.8-170}{8.25 / 3}=-1.52 \end{align} Thus, $|W|=1.52$. Also, we have \begin{align} t_{\frac{\alpha}{2},n-1} = t_{0.025,8} \approx 2.31 \end{align} The above value can be obtained in MATLAB using the command $\mathtt{tinv(0.975,8)}$. Thus, we conclude \begin{align} |W| \leq t_{\frac{\alpha}{2},n-1}. \end{align} Therefore, we accept $H_0$. In other words, we do not have enough evidence to conclude that the average height in the city is different from the average height in the country.

Let us summarize what we have obtained for the two-sided test for the mean.

One-sided Tests for the Mean:

• As before, we define the test statistic as \begin{align}%\label{} W(X_1,X_2, \cdots,X_n)=\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}}. \end{align} If $H_0$ is true (i.e., $\mu \leq \mu_0$), we expect $\overline{X}$ (and thus $W$) to be relatively small, while if $H_1$ is true, we expect $\overline{X}$ (and thus $W$) to be larger. This suggests the following test: Choose a threshold, and call it $c$. If $W \leq c$, accept $H_0$, and if $W>c$, accept $H_1$. How do we choose $c$? If $\alpha$ is the required significance level, we must have \begin{align} P(\textrm{type I error}) &= P(\textrm{Reject }H_0 \; | \; H_0) \\ &= P(W > c \; | \; \mu \leq \mu_0) \leq \alpha. \end{align} Here, the probability of type I error depends on $\mu$. More specifically, for any $\mu \leq \mu_0$, we can write \begin{align} P(\textrm{type I error} \; | \; \mu) &= P(\textrm{Reject }H_0 \; | \; \mu) \\ &= P(W > c \; | \; \mu)\\ &=P \left(\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}}> c \; | \; \mu\right)\\ &=P \left(\frac{\overline{X}-\mu}{\sigma / \sqrt{n}}+\frac{\mu-\mu_0}{\sigma / \sqrt{n}}> c \; | \; \mu\right)\\ &=P \left(\frac{\overline{X}-\mu}{\sigma / \sqrt{n}}> c+\frac{\mu_0-\mu}{\sigma / \sqrt{n}} \; | \; \mu\right)\\ &\leq P \left(\frac{\overline{X}-\mu}{\sigma / \sqrt{n}}> c \; | \; \mu\right) \quad (\textrm{ since }\mu \leq \mu_0)\\ &=1-\Phi(c) \quad \big(\textrm{ since given }\mu, \frac{\overline{X}-\mu}{\sigma / \sqrt{n}} \sim N(0,1) \big). \end{align} Thus, we can choose $\alpha=1-\Phi(c)$, which results in \begin{align} c=z_{\alpha}. \end{align} Therefore, we accept $H_0$ if \begin{align} \frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} \leq z_{\alpha}, \end{align} and reject it otherwise.

$\quad$ $H_0$: $\mu \geq \mu_0$, $\quad$ $H_1$: $\mu \lt \mu_0$,

Statistics Tutorial

Descriptive statistics, inferential statistics, stat reference, statistics - hypothesis testing a mean.

A population mean is an average of value a population.

Hypothesis tests are used to check a claim about the size of that population mean.

Hypothesis Testing a Mean

The following steps are used for a hypothesis test:

• Check the conditions
• Define the claims
• Decide the significance level
• Calculate the test statistic

For example:

• Population : Nobel Prize winners
• Category : Age when they received the prize.

And we want to check the claim:

"The average age of Nobel Prize winners when they received the prize is more than 55"

By taking a sample of 30 randomly selected Nobel Prize winners we could find that:

The mean age in the sample (\(\bar{x}\)) is 62.1

The standard deviation of age in the sample (\(s\)) is 13.46

From this sample data we check the claim with the steps below.

1. Checking the Conditions

The conditions for calculating a confidence interval for a proportion are:

• The sample is randomly selected
• The population data is normally distributed
• Sample size is large enough

A moderately large sample size, like 30, is typically large enough.

In the example, the sample size was 30 and it was randomly selected, so the conditions are fulfilled.

Note: Checking if the data is normally distributed can be done with specialized statistical tests.

2. Defining the Claims

We need to define a null hypothesis (\(H_{0}\)) and an alternative hypothesis (\(H_{1}\)) based on the claim we are checking.

The claim was:

In this case, the parameter is the mean age of Nobel Prize winners when they received the prize (\(\mu\)).

The null and alternative hypothesis are then:

Null hypothesis : The average age was 55.

Alternative hypothesis : The average age was more than 55.

Which can be expressed with symbols as:

\(H_{0}\): \(\mu = 55 \)

\(H_{1}\): \(\mu > 55 \)

This is a ' right tailed' test, because the alternative hypothesis claims that the proportion is more than in the null hypothesis.

If the data supports the alternative hypothesis, we reject the null hypothesis and accept the alternative hypothesis.

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3. Deciding the Significance Level

The significance level (\(\alpha\)) is the uncertainty we accept when rejecting the null hypothesis in a hypothesis test.

The significance level is a percentage probability of accidentally making the wrong conclusion.

Typical significance levels are:

• \(\alpha = 0.1\) (10%)
• \(\alpha = 0.05\) (5%)
• \(\alpha = 0.01\) (1%)

A lower significance level means that the evidence in the data needs to be stronger to reject the null hypothesis.

There is no "correct" significance level - it only states the uncertainty of the conclusion.

Note: A 5% significance level means that when we reject a null hypothesis:

We expect to reject a true null hypothesis 5 out of 100 times.

4. Calculating the Test Statistic

The test statistic is used to decide the outcome of the hypothesis test.

The test statistic is a standardized value calculated from the sample.

The formula for the test statistic (TS) of a population mean is:

\(\displaystyle \frac{\bar{x} - \mu}{s} \cdot \sqrt{n} \)

\(\bar{x}-\mu\) is the difference between the sample mean (\(\bar{x}\)) and the claimed population mean (\(\mu\)).

\(s\) is the sample standard deviation .

\(n\) is the sample size.

In our example:

The claimed (\(H_{0}\)) population mean (\(\mu\)) was \( 55 \)

The sample mean (\(\bar{x}\)) was \(62.1\)

The sample standard deviation (\(s\)) was \(13.46\)

The sample size (\(n\)) was \(30\)

So the test statistic (TS) is then:

\(\displaystyle \frac{62.1-55}{13.46} \cdot \sqrt{30} = \frac{7.1}{13.46} \cdot \sqrt{30} \approx 0.528 \cdot 5.477 = \underline{2.889}\)

You can also calculate the test statistic using programming language functions:

With Python use the scipy and math libraries to calculate the test statistic.

With R use built-in math and statistics functions to calculate the test statistic.

5. Concluding

There are two main approaches for making the conclusion of a hypothesis test:

• The critical value approach compares the test statistic with the critical value of the significance level.
• The P-value approach compares the P-value of the test statistic and with the significance level.

Note: The two approaches are only different in how they present the conclusion.

The Critical Value Approach

For the critical value approach we need to find the critical value (CV) of the significance level (\(\alpha\)).

For a population mean test, the critical value (CV) is a T-value from a student's t-distribution .

This critical T-value (CV) defines the rejection region for the test.

The rejection region is an area of probability in the tails of the standard normal distribution.

Because the claim is that the population mean is more than 55, the rejection region is in the right tail:

The student's t-distribution is adjusted for the uncertainty from smaller samples.

This adjustment is called degrees of freedom (df), which is the sample size \((n) - 1\)

In this case the degrees of freedom (df) is: \(30 - 1 = \underline{29} \)

Choosing a significance level (\(\alpha\)) of 0.01, or 1%, we can find the critical T-value from a T-table , or with a programming language function:

With Python use the Scipy Stats library t.ppf() function find the T-Value for an \(\alpha\) = 0.01 at 29 degrees of freedom (df).

With R use the built-in qt() function to find the t-value for an \(\alpha\) = 0.01 at 29 degrees of freedom (df).

Using either method we can find that the critical T-Value is \(\approx \underline{2.462}\)

For a right tailed test we need to check if the test statistic (TS) is bigger than the critical value (CV).

If the test statistic is bigger than the critical value, the test statistic is in the rejection region .

When the test statistic is in the rejection region, we reject the null hypothesis (\(H_{0}\)).

Here, the test statistic (TS) was \(\approx \underline{2.889}\) and the critical value was \(\approx \underline{2.462}\)

Here is an illustration of this test in a graph:

Since the test statistic was bigger than the critical value we reject the null hypothesis.

This means that the sample data supports the alternative hypothesis.

And we can summarize the conclusion stating:

The sample data supports the claim that "The average age of Nobel Prize winners when they received the prize is more than 55" at a 1% significance level .

The P-Value Approach

For the P-value approach we need to find the P-value of the test statistic (TS).

If the P-value is smaller than the significance level (\(\alpha\)), we reject the null hypothesis (\(H_{0}\)).

The test statistic was found to be \( \approx \underline{2.889} \)

For a population proportion test, the test statistic is a T-Value from a student's t-distribution .

Because this is a right tailed test, we need to find the P-value of a t-value bigger than 2.889.

The student's t-distribution is adjusted according to degrees of freedom (df), which is the sample size \((30) - 1 = \underline{29}\)

We can find the P-value using a T-table , or with a programming language function:

With Python use the Scipy Stats library t.cdf() function find the P-value of a T-value bigger than 2.889 at 29 degrees of freedom (df):

With R use the built-in pt() function find the P-value of a T-Value bigger than 2.889 at 29 degrees of freedom (df):

Using either method we can find that the P-value is \(\approx \underline{0.0036}\)

This tells us that the significance level (\(\alpha\)) would need to be bigger than 0.0036, or 0.36%, to reject the null hypothesis.

This P-value is smaller than any of the common significance levels (10%, 5%, 1%).

So the null hypothesis is rejected at all of these significance levels.

The sample data supports the claim that "The average age of Nobel Prize winners when they received the prize is more than 55" at a 10%, 5%, or 1% significance level .

Note: An outcome of an hypothesis test that rejects the null hypothesis with a p-value of 0.36% means:

For this p-value, we only expect to reject a true null hypothesis 36 out of 10000 times.

Calculating a P-Value for a Hypothesis Test with Programming

Many programming languages can calculate the P-value to decide outcome of a hypothesis test.

Using software and programming to calculate statistics is more common for bigger sets of data, as calculating manually becomes difficult.

The P-value calculated here will tell us the lowest possible significance level where the null-hypothesis can be rejected.

With Python use the scipy and math libraries to calculate the P-value for a right tailed hypothesis test for a mean.

Here, the sample size is 30, the sample mean is 62.1, the sample standard deviation is 13.46, and the test is for a mean bigger than 55.

With R use built-in math and statistics functions find the P-value for a right tailed hypothesis test for a mean.

Left-Tailed and Two-Tailed Tests

This was an example of a right tailed test, where the alternative hypothesis claimed that parameter is bigger than the null hypothesis claim.

You can check out an equivalent step-by-step guide for other types here:

• Left-Tailed Test
• Two-Tailed Test

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Chapter 10: Inference for Means

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

• 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.

Learn By Doing

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.

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

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8.1: The Elements of Hypothesis Testing

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Learning Objectives

• To understand the logical framework of tests of hypotheses.
• To learn basic terminology connected with hypothesis testing.
• To learn fundamental facts about hypothesis testing.

Types of Hypotheses

A hypothesis about the value of a population parameter is an assertion about its value. As in the introductory example we will be concerned with testing the truth of two competing hypotheses, only one of which can be true.

Definition: null hypothesis and alternative hypothesis

• The null hypothesis , denoted \(H_0\), is the statement about the population parameter that is assumed to be true unless there is convincing evidence to the contrary.
• The alternative hypothesis , denoted \(H_a\), is a statement about the population parameter that is contradictory to the null hypothesis, and is accepted as true only if there is convincing evidence in favor of it.

Definition: statistical procedure

Hypothesis testing is a statistical procedure in which a choice is made between a null hypothesis and an alternative hypothesis based on information in a sample.

The end result of a hypotheses testing procedure is a choice of one of the following two possible conclusions:

• Reject \(H_0\) (and therefore accept \(H_a\)), or
• Fail to reject \(H_0\) (and therefore fail to accept \(H_a\)).

The null hypothesis typically represents the status quo, or what has historically been true. In the example of the respirators, we would believe the claim of the manufacturer unless there is reason not to do so, so the null hypotheses is \(H_0:\mu =75\). The alternative hypothesis in the example is the contradictory statement \(H_a:\mu <75\). The null hypothesis will always be an assertion containing an equals sign, but depending on the situation the alternative hypothesis can have any one of three forms: with the symbol \(<\), as in the example just discussed, with the symbol \(>\), or with the symbol \(\neq\). The following two examples illustrate the latter two cases.

Example \(\PageIndex{1}\)

A publisher of college textbooks claims that the average price of all hardbound college textbooks is \(\$127.50\). A student group believes that the actual mean is higher and wishes to test their belief. State the relevant null and alternative hypotheses.

The default option is to accept the publisher’s claim unless there is compelling evidence to the contrary. Thus the null hypothesis is \(H_0:\mu =127.50\). Since the student group thinks that the average textbook price is greater than the publisher’s figure, the alternative hypothesis in this situation is \(H_a:\mu >127.50\).

Example \(\PageIndex{2}\)

The recipe for a bakery item is designed to result in a product that contains \(8\) grams of fat per serving. The quality control department samples the product periodically to insure that the production process is working as designed. State the relevant null and alternative hypotheses.

The default option is to assume that the product contains the amount of fat it was formulated to contain unless there is compelling evidence to the contrary. Thus the null hypothesis is \(H_0:\mu =8.0\). Since to contain either more fat than desired or to contain less fat than desired are both an indication of a faulty production process, the alternative hypothesis in this situation is that the mean is different from \(8.0\), so \(H_a:\mu \neq 8.0\).

In Example \(\PageIndex{1}\), the textbook example, it might seem more natural that the publisher’s claim be that the average price is at most \(\$127.50\), not exactly \(\$127.50\). If the claim were made this way, then the null hypothesis would be \(H_0:\mu \leq 127.50\), and the value \(\$127.50\) given in the example would be the one that is least favorable to the publisher’s claim, the null hypothesis. It is always true that if the null hypothesis is retained for its least favorable value, then it is retained for every other value.

Thus in order to make the null and alternative hypotheses easy for the student to distinguish, in every example and problem in this text we will always present one of the two competing claims about the value of a parameter with an equality. The claim expressed with an equality is the null hypothesis. This is the same as always stating the null hypothesis in the least favorable light. So in the introductory example about the respirators, we stated the manufacturer’s claim as “the average is \(75\) minutes” instead of the perhaps more natural “the average is at least \(75\) minutes,” essentially reducing the presentation of the null hypothesis to its worst case.

The first step in hypothesis testing is to identify the null and alternative hypotheses.

The Logic of Hypothesis Testing

Although we will study hypothesis testing in situations other than for a single population mean (for example, for a population proportion instead of a mean or in comparing the means of two different populations), in this section the discussion will always be given in terms of a single population mean \(\mu\).

The null hypothesis always has the form \(H_0:\mu =\mu _0\) for a specific number \(\mu _0\) (in the respirator example \(\mu _0=75\), in the textbook example \(\mu _0=127.50\), and in the baked goods example \(\mu _0=8.0\)). Since the null hypothesis is accepted unless there is strong evidence to the contrary, the test procedure is based on the initial assumption that \(H_0\) is true. This point is so important that we will repeat it in a display:

The test procedure is based on the initial assumption that \(H_0\) is true.

The criterion for judging between \(H_0\) and \(H_a\) based on the sample data is: if the value of \(\overline{X}\) would be highly unlikely to occur if \(H_0\) were true, but favors the truth of \(H_a\), then we reject \(H_0\) in favor of \(H_a\). Otherwise we do not reject \(H_0\).

Supposing for now that \(\overline{X}\) follows a normal distribution, when the null hypothesis is true the density function for the sample mean \(\overline{X}\) must be as in Figure \(\PageIndex{1}\): a bell curve centered at \(\mu _0\). Thus if \(H_0\) is true then \(\overline{X}\) is likely to take a value near \(\mu _0\) and is unlikely to take values far away. Our decision procedure therefore reduces simply to:

• if \(H_a\) has the form \(H_a:\mu <\mu _0\) then reject \(H_0\) if \(\bar{x}\) is far to the left of \(\mu _0\);
• if \(H_a\) has the form \(H_a:\mu >\mu _0\) then reject \(H_0\) if \(\bar{x}\) is far to the right of \(\mu _0\);
• if \(H_a\) has the form \(H_a:\mu \neq \mu _0\) then reject \(H_0\) if \(\bar{x}\) is far away from \(\mu _0\) in either direction.

Think of the respirator example, for which the null hypothesis is \(H_0:\mu =75\), the claim that the average time air is delivered for all respirators is \(75\) minutes. If the sample mean is \(75\) or greater then we certainly would not reject \(H_0\) (since there is no issue with an emergency respirator delivering air even longer than claimed).

If the sample mean is slightly less than \(75\) then we would logically attribute the difference to sampling error and also not reject \(H_0\) either.

Values of the sample mean that are smaller and smaller are less and less likely to come from a population for which the population mean is \(75\). Thus if the sample mean is far less than \(75\), say around \(60\) minutes or less, then we would certainly reject \(H_0\), because we know that it is highly unlikely that the average of a sample would be so low if the population mean were \(75\). This is the rare event criterion for rejection: what we actually observed \((\overline{X}<60)\) would be so rare an event if \(\mu =75\) were true that we regard it as much more likely that the alternative hypothesis \(\mu <75\) holds.

In summary, to decide between \(H_0\) and \(H_a\) in this example we would select a “rejection region” of values sufficiently far to the left of \(75\), based on the rare event criterion, and reject \(H_0\) if the sample mean \(\overline{X}\) lies in the rejection region, but not reject \(H_0\) if it does not.

The Rejection Region

Each different form of the alternative hypothesis Ha has its own kind of rejection region:

• if (as in the respirator example) \(H_a\) has the form \(H_a:\mu <\mu _0\), we reject \(H_0\) if \(\bar{x}\) is far to the left of \(\mu _0\), that is, to the left of some number \(C\), so the rejection region has the form of an interval \((-\infty ,C]\);
• if (as in the textbook example) \(H_a\) has the form \(H_a:\mu >\mu _0\), we reject \(H_0\) if \(\bar{x}\) is far to the right of \(\mu _0\), that is, to the right of some number \(C\), so the rejection region has the form of an interval \([C,\infty )\);
• if (as in the baked good example) \(H_a\) has the form \(H_a:\mu \neq \mu _0\), we reject \(H_0\) if \(\bar{x}\) is far away from \(\mu _0\) in either direction, that is, either to the left of some number \(C\) or to the right of some other number \(C′\), so the rejection region has the form of the union of two intervals \((-\infty ,C]\cup [C',\infty )\).

The key issue in our line of reasoning is the question of how to determine the number \(C\) or numbers \(C\) and \(C′\), called the critical value or critical values of the statistic, that determine the rejection region.

Definition: critical values

The critical value or critical values of a test of hypotheses are the number or numbers that determine the rejection region.

Suppose the rejection region is a single interval, so we need to select a single number \(C\). Here is the procedure for doing so. We select a small probability, denoted \(\alpha\), say \(1\%\), which we take as our definition of “rare event:” an event is “rare” if its probability of occurrence is less than \(\alpha\). (In all the examples and problems in this text the value of \(\alpha\) will be given already.) The probability that \(\overline{X}\) takes a value in an interval is the area under its density curve and above that interval, so as shown in Figure \(\PageIndex{2}\) (drawn under the assumption that \(H_0\) is true, so that the curve centers at \(\mu _0\)) the critical value \(C\) is the value of \(\overline{X}\) that cuts off a tail area \(\alpha\) in the probability density curve of \(\overline{X}\). When the rejection region is in two pieces, that is, composed of two intervals, the total area above both of them must be \(\alpha\), so the area above each one is \(\alpha /2\), as also shown in Figure \(\PageIndex{2}\).

The number \(\alpha\) is the total area of a tail or a pair of tails.

Example \(\PageIndex{3}\)

In the context of Example \(\PageIndex{2}\), suppose that it is known that the population is normally distributed with standard deviation \(\alpha =0.15\) gram, and suppose that the test of hypotheses \(H_0:\mu =8.0\) versus \(H_a:\mu \neq 8.0\) will be performed with a sample of size \(5\). Construct the rejection region for the test for the choice \(\alpha =0.10\). Explain the decision procedure and interpret it.

If \(H_0\) is true then the sample mean \(\overline{X}\) is normally distributed with mean and standard deviation

\[\begin{align} \mu _{\overline{X}} &=\mu \nonumber \\[5pt] &=8.0 \nonumber \end{align} \nonumber \]

\[\begin{align} \sigma _{\overline{X}}&=\dfrac{\sigma}{\sqrt{n}} \nonumber \\[5pt] &= \dfrac{0.15}{\sqrt{5}} \nonumber\\[5pt] &=0.067 \nonumber \end{align} \nonumber \]

Since \(H_a\) contains the \(\neq\) symbol the rejection region will be in two pieces, each one corresponding to a tail of area \(\alpha /2=0.10/2=0.05\). From Figure 7.1.6, \(z_{0.05}=1.645\), so \(C\) and \(C′\) are \(1.645\) standard deviations of \(\overline{X}\) to the right and left of its mean \(8.0\):

\[C=8.0-(1.645)(0.067) = 7.89 \; \; \text{and}\; \; C'=8.0 + (1.645)(0.067) = 8.11 \nonumber \]

The result is shown in Figure \(\PageIndex{3}\). α = 0.1

The decision procedure is: take a sample of size \(5\) and compute the sample mean \(\bar{x}\). If \(\bar{x}\) is either \(7.89\) grams or less or \(8.11\) grams or more then reject the hypothesis that the average amount of fat in all servings of the product is \(8.0\) grams in favor of the alternative that it is different from \(8.0\) grams. Otherwise do not reject the hypothesis that the average amount is \(8.0\) grams.

The reasoning is that if the true average amount of fat per serving were \(8.0\) grams then there would be less than a \(10\%\) chance that a sample of size \(5\) would produce a mean of either \(7.89\) grams or less or \(8.11\) grams or more. Hence if that happened it would be more likely that the value \(8.0\) is incorrect (always assuming that the population standard deviation is \(0.15\) gram).

Because the rejection regions are computed based on areas in tails of distributions, as shown in Figure \(\PageIndex{2}\), hypothesis tests are classified according to the form of the alternative hypothesis in the following way.

Definitions: Test classifications

• If \(H_a\) has the form \(\mu \neq \mu _0\) the test is called a two-tailed test .
• If \(H_a\) has the form \(\mu < \mu _0\) the test is called a left-tailed test .
• If \(H_a\) has the form \(\mu > \mu _0\)the test is called a right-tailed test .

Each of the last two forms is also called a one-tailed test .

Two Types of Errors

The format of the testing procedure in general terms is to take a sample and use the information it contains to come to a decision about the two hypotheses. As stated before our decision will always be either

• reject the null hypothesis \(H_0\) in favor of the alternative \(H_a\) presented, or
• do not reject the null hypothesis \(H_0\) in favor of the alternative \(H_0\) presented.

There are four possible outcomes of hypothesis testing procedure, as shown in the following table:

As the table shows, there are two ways to be right and two ways to be wrong. Typically to reject \(H_0\) when it is actually true is a more serious error than to fail to reject it when it is false, so the former error is labeled “ Type I ” and the latter error “ Type II ”.

Definition: Type I and Type II errors

In a test of hypotheses:

• A Type I error is the decision to reject \(H_0\) when it is in fact true.
• A Type II error is the decision not to reject \(H_0\) when it is in fact not true.

Unless we perform a census we do not have certain knowledge, so we do not know whether our decision matches the true state of nature or if we have made an error. We reject \(H_0\) if what we observe would be a “rare” event if \(H_0\) were true. But rare events are not impossible: they occur with probability \(\alpha\). Thus when \(H_0\) is true, a rare event will be observed in the proportion \(\alpha\) of repeated similar tests, and \(H_0\) will be erroneously rejected in those tests. Thus \(\alpha\) is the probability that in following the testing procedure to decide between \(H_0\) and \(H_a\) we will make a Type I error.

Definition: level of significance

The number \(\alpha\) that is used to determine the rejection region is called the level of significance of the test. It is the probability that the test procedure will result in a Type I error .

The probability of making a Type II error is too complicated to discuss in a beginning text, so we will say no more about it than this: for a fixed sample size, choosing \(alpha\) smaller in order to reduce the chance of making a Type I error has the effect of increasing the chance of making a Type II error . The only way to simultaneously reduce the chances of making either kind of error is to increase the sample size.

Standardizing the Test Statistic

Hypotheses testing will be considered in a number of contexts, and great unification as well as simplification results when the relevant sample statistic is standardized by subtracting its mean from it and then dividing by its standard deviation. The resulting statistic is called a standardized test statistic . In every situation treated in this and the following two chapters the standardized test statistic will have either the standard normal distribution or Student’s \(t\)-distribution.

Definition: hypothesis test

A standardized test statistic for a hypothesis test is the statistic that is formed by subtracting from the statistic of interest its mean and dividing by its standard deviation.

For example, reviewing Example \(\PageIndex{3}\), if instead of working with the sample mean \(\overline{X}\) we instead work with the test statistic

\[\frac{\overline{X}-8.0}{0.067} \nonumber \]

then the distribution involved is standard normal and the critical values are just \(\pm z_{0.05}\). The extra work that was done to find that \(C=7.89\) and \(C′=8.11\) is eliminated. In every hypothesis test in this book the standardized test statistic will be governed by either the standard normal distribution or Student’s \(t\)-distribution. Information about rejection regions is summarized in the following tables:

Every instance of hypothesis testing discussed in this and the following two chapters will have a rejection region like one of the six forms tabulated in the tables above.

No matter what the context a test of hypotheses can always be performed by applying the following systematic procedure, which will be illustrated in the examples in the succeeding sections.

Systematic Hypothesis Testing Procedure: Critical Value Approach

• Identify the null and alternative hypotheses.
• Identify the relevant test statistic and its distribution.
• Compute from the data the value of the test statistic.
• Construct the rejection region.
• Compare the value computed in Step 3 to the rejection region constructed in Step 4 and make a decision. Formulate the decision in the context of the problem, if applicable.

The procedure that we have outlined in this section is called the “Critical Value Approach” to hypothesis testing to distinguish it from an alternative but equivalent approach that will be introduced at the end of Section 8.3.

Key Takeaway

• A test of hypotheses is a statistical process for deciding between two competing assertions about a population parameter.
• The testing procedure is formalized in a five-step procedure.

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8.3: Hypothesis Test Examples for Means with Unknown Standard Deviation

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Full Hypothesis Test Examples

Example \(\PageIndex{6}\)

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 scores 65 65 70 67 66 63 63 68 72 71. He performs a hypothesis test using a 5% level of significance. The data are assumed to be from a normal distribution.

Set up the hypothesis test:

A 5% level of significance means that \(\alpha = 0.05\). This is a test of a single population mean .

\(H_{0}: \mu = 65  H_{a}: \mu > 65\)

Since the instructor thinks the average score is higher, use a "\(>\)". The "\(>\)" means the test is right-tailed.

Determine the distribution needed:

Random variable: \(\bar{X} =\) average score on the first statistics test.

Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given . You are only given \(n = 10\) sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a student's \(t\).

Use \(t_{df}\). Therefore, the distribution for the test is \(t_{9}\) where \(n = 10\) and \(df = 10 - 1 = 9\).

Calculate the \(p\)-value using the Student's \(t\)-distribution:

\(p\text{-value} = P(\bar{x} > 67) = 0.0396\) where the sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data.

Interpretation of the p -value: If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 65 or more.

Compare \(\alpha\) and the \(p-\text{value})\:

Since \( α = 0.05\) and \(p\text{-value} = 0.0396\). \(\alpha > p\text{-value}\).

Make a decision: Since \(\alpha > p\text{-value}\), reject \(H_{0}\).

This means you reject \(\mu = 65\). In other words, you believe the average test score is more than 65.

Conclusion: At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.

The \(p\text{-value}\) can easily be calculated.

Put the data into a list. Press STAT and arrow over to TESTS . Press 2:T-Test . Arrow over to Data and press ENTER . Arrow down and enter 65 for \(\mu_{0}\), the name of the list where you put the data, and 1 for Freq: . Arrow down to \(\mu\): and arrow over to \(> \mu_{0}\). Press ENTER . Arrow down to Calculate and press ENTER . The calculator not only calculates the \(p\text{-value}\) (p = 0.0396) but it also calculates the test statistic ( t -score) for the sample mean, the sample mean, and the sample standard deviation. \(\mu > 65\) is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with \( t = 1.9781\) (test statistic) and \(p = 0.0396\) (\(p\text{-value}\)). Make sure when you use Draw that no other equations are highlighted in \(Y =\) and the plots are turned off.

Exercise \(\PageIndex{6}\)

It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, find the p -value, state your conclusion, and identify the Type I and Type II errors.

• \(H_{0}: \mu = 5\)
• \(H_{a}: \mu < 5\)
• \(p = 0.0082\)

Because \(p < \alpha\), we reject the null hypothesis. There is sufficient evidence to suggest that the stock price of the company grows at a rate less than $5 a week.

• Type I Error: To conclude that the stock price is growing slower than $5 a week when, in fact, the stock price is growing at $5 a week (reject the null hypothesis when the null hypothesis is true).
• Type II Error: To conclude that the stock price is growing at a rate of $5 a week when, in fact, the stock price is growing slower than $5 a week (do not reject the null hypothesis when the null hypothesis is false).

Example \(\PageIndex{10}\)

The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.

1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98 1.02; .95; .95

Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05. Assume the population is normal.

Let’s follow a four-step process to answer this statistical question.

• \(H_{0}: \mu \leq 1\)
• \(H_{a}: \mu > 1\)
• Plan : We are testing a sample mean without a known population standard deviation. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal.
• Do the calculations : We will input the sample data into the TI-83 as follows.

4. State the Conclusions : Since the \(p\text{-value} (p = 0.036)\) is less than our alpha value, we will reject the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.

The hypothesis test itself has an established process. This can be summarized as follows:

• Determine \(H_{0}\) and \(H_{a}\). Remember, they are contradictory.
• Determine the random variable.
• Determine the distribution for the test.
• Draw a graph, calculate the test statistic, and use the test statistic to calculate the \(p\text{-value}\). (A z -score and a t -score are examples of test statistics.)
• Compare the preconceived α with the p -value, make a decision (reject or do not reject H 0 ), and write a clear conclusion using English sentences.

Notice that in performing the hypothesis test, you use \(\alpha\) and not \(\beta\). \(\beta\) is needed to help determine the sample size of the data that is used in calculating the \(p\text{-value}\). Remember that the quantity \(1 – \beta\) is called the Power of the Test . A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same.If the power is low, the null hypothesis might not be rejected when it should be.

• Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC.
• Data from Bloomberg Businessweek . Available online at http://www.businessweek.com/news/2011- 09-15/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html.
• Data from energy.gov. Available online at http://energy.gov (accessed June 27. 2013).
• Data from Gallup®. Available online at www.gallup.com (accessed June 27, 2013).
• Data from Growing by Degrees by Allen and Seaman.
• Data from La Leche League International. Available online at www.lalecheleague.org/Law/BAFeb01.html.
• Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013).
• Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013).
• Data from the Bureau of Labor Statistics. Available online at http://www.bls.gov/oes/current/oes291111.htm .
• Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013)
• Data from the U.S. Census Bureau, available online at quickfacts.census.gov/qfd/states/00000.html (accessed June 27, 2013).
• Data from the United States Census Bureau. Available online at www.census.gov/hhes/socdemo/language/.
• Data from Toastmasters International. Available online at http://toastmasters.org/artisan/deta...eID=429&Page=1 .
• Data from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013).
• Federal Bureau of Investigations. “Uniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.” Available online at http://www.disastercenter.com/kentucky/crime/3868.htm (accessed June 27, 2013).
• “Foothill-De Anza Community College District.” De Anza College, Winter 2006. Available online at research.fhda.edu/factbook/DA...t_da_2006w.pdf.
• Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. “Cellular Telephones and Cancer—a Nationwide Cohort Study in Denmark.” Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http://www.ncbi.nlm.nih.gov/pubmed/11158188 (accessed June 27, 2013).
• Rape, Abuse & Incest National Network. “How often does sexual assault occur?” RAINN, 2009. Available online at www.rainn.org/get-information...sexual-assault (accessed June 27, 2013).

Contributors and Attributions

Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] .

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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• How to Write a Strong Hypothesis | Steps & Examples

How to Write a Strong Hypothesis | Steps & Examples

Published on May 6, 2022 by Shona McCombes . Revised on November 20, 2023.

A hypothesis is a statement that can be tested by scientific research. If you want to test a relationship between two or more variables, you need to write hypotheses before you start your experiment or data collection .

Example: Hypothesis

Daily apple consumption leads to fewer doctor’s visits.

Table of contents

What is a hypothesis, developing a hypothesis (with example), hypothesis examples, other interesting articles, frequently asked questions about writing hypotheses.

A hypothesis states your predictions about what your research will find. It is a tentative answer to your research question that has not yet been tested. For some research projects, you might have to write several hypotheses that address different aspects of your research question.

A hypothesis is not just a guess – it should be based on existing theories and knowledge. It also has to be testable, which means you can support or refute it through scientific research methods (such as experiments, observations and statistical analysis of data).

Variables in hypotheses

Hypotheses propose a relationship between two or more types of variables .

• An independent variable is something the researcher changes or controls.
• A dependent variable is something the researcher observes and measures.

If there are any control variables , extraneous variables , or confounding variables , be sure to jot those down as you go to minimize the chances that research bias  will affect your results.

In this example, the independent variable is exposure to the sun – the assumed cause . The dependent variable is the level of happiness – the assumed effect .

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Step 1. Ask a question

Writing a hypothesis begins with a research question that you want to answer. The question should be focused, specific, and researchable within the constraints of your project.

Step 2. Do some preliminary research

Your initial answer to the question should be based on what is already known about the topic. Look for theories and previous studies to help you form educated assumptions about what your research will find.

At this stage, you might construct a conceptual framework to ensure that you’re embarking on a relevant topic . This can also help you identify which variables you will study and what you think the relationships are between them. Sometimes, you’ll have to operationalize more complex constructs.

Step 3. Formulate your hypothesis

Now you should have some idea of what you expect to find. Write your initial answer to the question in a clear, concise sentence.

4. Refine your hypothesis

You need to make sure your hypothesis is specific and testable. There are various ways of phrasing a hypothesis, but all the terms you use should have clear definitions, and the hypothesis should contain:

• The relevant variables
• The specific group being studied
• The predicted outcome of the experiment or analysis

5. Phrase your hypothesis in three ways

To identify the variables, you can write a simple prediction in  if…then form. The first part of the sentence states the independent variable and the second part states the dependent variable.

In academic research, hypotheses are more commonly phrased in terms of correlations or effects, where you directly state the predicted relationship between variables.

If you are comparing two groups, the hypothesis can state what difference you expect to find between them.

6. Write a null hypothesis

If your research involves statistical hypothesis testing , you will also have to write a null hypothesis . The null hypothesis is the default position that there is no association between the variables. The null hypothesis is written as H 0 , while the alternative hypothesis is H 1 or H a .

• H 0 : The number of lectures attended by first-year students has no effect on their final exam scores.
• H 1 : The number of lectures attended by first-year students has a positive effect on their final exam scores.

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

• Sampling methods
• Simple random sampling
• Stratified sampling
• Cluster sampling
• Likert scales
• Reproducibility

 Statistics

• Null hypothesis
• Statistical power
• Probability distribution
• Effect size
• Poisson distribution

Research bias

• Optimism bias
• Cognitive bias
• Implicit bias
• Hawthorne effect
• Anchoring bias
• Explicit bias

A hypothesis is not just a guess — it should be based on existing theories and knowledge. It also has to be testable, which means you can support or refute it through scientific research methods (such as experiments, observations and statistical analysis of data).

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

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

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Statistics and probability

Course: statistics and probability   >   unit 13.

• Statistical significance of experiment
• Statistical significance on bus speeds
• Hypothesis testing in experiments
• Difference of sample means distribution
• Confidence interval of difference of means
• Clarification of confidence interval of difference of means

Hypothesis test for difference of means

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Video transcript

8.2.3.1 - One Sample Mean t Test, Formulas

Five step hypothesis testing procedure.

Data must be quantitative. In order to use the t distribution to approximate the sampling distribution either the sample size must be large (\(\ge\ 30\)) or the population must be known to be normally distributed. The possible combinations of null and alternative hypotheses are:

where \( \mu_{0} \) is the hypothesized population mean.

For the test of one group mean we will be using a \(t\) test statistic:

\(t=\dfrac{\overline{x}-\mu_0}{\frac{s}{\sqrt{n}}}\)

\(\overline{x}\) = sample mean \(\mu_{0}\) = hypothesized population mean \(s\) = sample standard deviation \(n\) = sample size

Note that structure of this formula is similar to the general formula for a test statistic:

\(\dfrac{sample\;statistic-null\;value}{standard\;error}\)

When testing hypotheses about a mean or mean difference, a \(t\) distribution is used to find the \(p\)-value. These \(t\) distributions are indexed by a quantity called degrees of freedom, calculated as \(df = n – 1\) for the situation involving a test of one mean or test of mean difference. The \(p\)-value can be found using Minitab.

If \(p \leq \alpha\) reject the null hypothesis.

If \(p>\alpha\) fail to reject the null hypothesis.

Based on your decision in Step 4, write a conclusion in terms of the original research question.

The new few pages will walk you through examples before giving you the opportunity to do two on your own.

8.2.3.1.1 - Video Example: Book Costs

Research question: Does the average Penn State student spend more than \$300 each semester on textbooks?

In a sample of 226 Penn State students, the mean cost of a student’s textbooks was \$344 with a standard deviation of \$106.

8.2.3.1.2 : Example: Pulse Rate

A research study measured the pulse rates of 57 college men and found a mean pulse rate of 70.4211 beats per minute with a standard deviation of 9.9480 beats per minute. Researchers want to know if the mean pulse rate for all college men is different from the current standard of 72 beats per minute.

Pulse rates are quantitative. The sampling distribution will be approximately normally distributed because \(n \ge 30\).

This is a two-tailed test because we want to know if the mean pulse rate is  different  from 72.

\(H_{0}:\mu=72 \) \(H_{a}: \mu\neq 72 \)

\(t=\dfrac{\overline{x}-\mu_0}{\dfrac{s}{\sqrt{n}}}\)

\(t=\dfrac{\overline{x}-\mu_0}{\dfrac{s}{\sqrt{n}}}=\dfrac{70.4211-72}{\dfrac{9.9480}{\sqrt{57}}}=-1.198\)

Our \(t\) test statistic is -1.198

\(df=n-1=57-1=56\)

\(p=0.117981+0.117981=0.235962\)

Given that the null hypothesis is true and \(\mu=72\), the probability of taking a random sample of \(n=57\) and finding a sample mean this or more extremely different is 0.235962. This is our p-value. 

\(p>.05\), therefore we fail to reject the null hypothesis.

There is not sufficient evidence to state that the mean pulse of college men is different from 72.

8.2.3.1.3 - Example: Coffee

In the population of Americans who drink coffee, the average daily consumption is 3 cups per day. A university wants to know if their students tend to drink more coffee than the national average. They ask a random sample of 50 students how many cups of coffee they drink each day and found \(\overline{x}=3.8\) and \(s=1.5\). Do they have evidence that their students drink more than the national average?

Amount of coffee consumed is a quantitative variable. We are given that random sampling methods were employed. Because \(n \ge 30\), we can approximate the sampling distribution using a t distribution. 

This is a right-tailed test because we want to know if the mean in the sample is  greater than  the national average.

\(H_{0}:\mu= 3\) \(H_{a}:\mu>3\)

\(t=\dfrac{\overline{x}-\mu_0}{\dfrac{s}{\sqrt{n}}}=\dfrac{3.8-3}{\dfrac{1.5}{\sqrt{50}}}=3.771\)

Our \(t\) test statistic is 3.771

\(df=n-1=50-1=49\)

Using Minitab, we can find that \(P(t > 3.771) =0.0002191\)

p-value = 0.0002191

If \(\mu=3\), then the probability of taking a random sample of \(n=50\) and finding \(\overline{x} \geq 3\) is 0.0002191

\(p\leq.05\), therefore we reject the null hypothesis.

There is evidence to state the mean number of coffees consumed in the population of all students at this university is greater than 3.

8.2.3.1.4 - Example: Transportation Costs

According to  CNN , in 2011, the average American spent \$16,803 on housing. A suburban community wants to know if their residents spent less than this national average. In a survey of 30 randomly selected residents, they found that they spent an annual average of \$15,800 with a standard deviation of \$2,600.

Housing costs are quantitative. Because \(n \ge 30\), the sampling distribution can be approximated using the \(t\) distribution.  

This is a left-tailed test because we want to know if residents of this community spent  less than  the national average.

\(H_{0}:\mu=16803\) \(H_{a}:\mu<16803\)

\(t=\dfrac{\overline{x}-\mu_0}{\dfrac{s}{\sqrt{n}}}=\dfrac{15800-16803}{\dfrac{2600}{\sqrt{30}}}=-2.113\)

Our t test statistics is -2.113

\(df=n-1=30-1=29\)

This is a left-tailed test so we want to know the probability of \(t < -2.113\)

Using Minitab we can find that \(p=0.0216634\)

\(p\leq .05\), therefore we reject the null hypothesis.

There is evidence to state that on average residents of this community spent less than the national average on housing in 2011.

Example of performing a basic hypothesis test

You can follow six basic steps to correctly set up and perform a hypothesis test. For example, the manager of a pipe manufacturing facility wants to determine whether the average diameter of their pipes is different from 5cm. The manager follows the basic steps for doing a hypothesis test.

You should determine the criteria for the test and the required sample size before you collect the data.

First, the manager formulates the hypotheses. The null hypothesis is: The population mean of all the pipes is equal to 5 cm. Formally, this is written as: H 0 : μ = 5

Because they need to ensure that the pipes are not larger or smaller than 5 cm, the manager chooses the two-sided alternative hypothesis, which states that the population mean of all the pipes is not equal to 5 cm. Formally, this is written as H 1 : μ ≠ 5

• Choose a significance level (also called alpha or α). The manager selects a significance level 0.05, which is the most commonly used significance level.
• Determine the power and sample size for the test. The manager uses a power and sample size calculation to determine how many pipes they need to measure to have a good chance of detecting a difference of 0.1 cm or more from the target diameter.
• Collect the data. They collect a sample of pipes and measure their diameters.
• Compare the p-value from the test to the significance level. After they perform the hypothesis test, the manager obtains a p-value of 0.004. The p-value is less than the significance level of 0.05.
• Decide whether to reject or fail to reject the null hypothesis. The manager rejects the null hypothesis and concludes that the mean pipe diameter of all pipes is not equal to 5cm.
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