hypothesis test of the sample mean

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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.

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

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.

Hypothesis tests about the mean

by Marco Taboga , PhD

This lecture explains how to conduct hypothesis tests about the mean of a normal distribution.

We tackle two different cases:

when we know the variance of the distribution, then we use a z-statistic to conduct the test;

when the variance is unknown, then we use the t-statistic.

In each case we derive the power and the size of the test.

We conclude with two solved exercises on size and power.

Table of contents

Known variance: the z-test

The null hypothesis, the test statistic, the critical region, the decision, the power function, the size of the test, how to choose the critical value, unknown variance: the t-test, how to choose the critical values, solved exercises.

The assumptions are the same we made in the lecture on confidence intervals for the mean .

A test of hypothesis based on it is called z-test .

Otherwise, it is not rejected.

We explain how to do this in the page on critical values .

This case is similar to the previous one. The only difference is that we now relax the assumption that the variance of the distribution is known.

The test of hypothesis based on it is called t-test .

Otherwise, we do not reject it.

The page on critical values explains how this equation is solved.

Below you can find some exercises with explained solutions.

Suppose that a statistician observes 100 independent realizations of a normal random variable.

The mean and the variance of the random variable, which the statistician does not know, are equal to 1 and 4 respectively.

Find the probability that the statistician will reject the null hypothesis that the mean is equal to zero if:

she runs a t-test based on the 100 observed realizations;

A statistician observes 100 independent realizations of a normal random variable.

She performs a t-test of the null hypothesis that the mean of the variable is equal to zero.

How to cite

Please cite as:

Taboga, Marco (2021). "Hypothesis tests about the mean", Lectures on probability theory and mathematical statistics. Kindle Direct Publishing. Online appendix. https://www.statlect.com/fundamentals-of-statistics/hypothesis-testing-mean.

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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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