5.3  Hypothesis Testing for OneSample Mean
5.3  Hypothesis Testing for OneSample MeanIn the previous section, we learned how to perform a hypothesis test for one proportion. The concepts of hypothesis testing remain constant for any hypothesis test. In these next few sections, we will present the hypothesis test for one mean. We start with our knowledge of the sampling distribution of the sample mean.
Hypothesis Test for OneSample Mean
Recall that under certain conditions, the sampling distribution of the sample mean, \(\bar{x} \), is approximately normal with mean, \(\mu \), standard error \(\dfrac{\sigma}{\sqrt{n}} \), and estimated standard error \(\dfrac{s}{\sqrt{n}} \).
Null:
\(H_0\colon \mu=\mu_0\)
Conditions:
 The distribution of the population is Normal
 The sample size is large \( n>30 \).
Test Statistic:
If at least one of conditions are satisfied, then...
\( t=\dfrac{\bar{x}\mu_0}{\frac{s}{\sqrt{n}}} \)
will follow a tdistribution with \(n1 \) degrees of freedom.
Notice when working with continuous data we are going to use a t statistic as opposed to the z statistic. This is due to the fact that the sample size impacts the sampling distribution and needs to be taken into account. We do this by recognizing “degrees of freedom”. We will not go into too much detail about degrees of freedom in this course.
Let’s look at an example.
Example 51
This depends on the standard deviation of \(\bar{x} \) .
\begin{align} t^*&=\dfrac{\bar{x}\mu}{\frac{s}{\sqrt{n}}}\\&=\dfrac{8.38.5}{\frac{1.2}{\sqrt{61}}}\\&=1.3 \end{align}
Thus, we are asking if \(1.3\) is very far away from zero, since that corresponds to the case when \(\bar{x}\) is equal to \(\mu_0 \). If it is far away, then it is unlikely that the null hypothesis is true and one rejects it. Otherwise, one cannot reject the null hypothesis.
5.3.1 Steps in Conducting a Hypothesis Test for \(\mu\)
5.3.1 Steps in Conducting a Hypothesis Test for \(\mu\) Step 1: Set up the hypotheses and check conditions
 One Mean ttest Hypotheses

\( H_0\colon \mu=\mu_0 \)
\( H_a\colon \mu\ne \mu_0 \)
Conditions: The data comes from an approximately normal distribution or the sample size is at least 30
 Step 2: Decide on the significance level, \(\alpha \)
Typically, 5%. If \(\alpha\) is not specified, use 5%
 Step 3: Calculate the test statistic
One Mean ttest: \( t^*=\dfrac{\bar{x}\mu_0}{\frac{s}{\sqrt{n}}} \)
 Step 4: Compute the appropriate pvalue based on our alternative hypothesis
Typically we will let Minitab handle this for us. But if you are really interested, you can look p values up in probability tables found in the appendix of your textbook!
 Step 5: Make a decision about the null hypothesis
If the pvalue is less than the significance level, \(\alpha\), then reject \(H_0\) (and conclude \(H_a \)). If it is greater than the significance level, then do not reject \(H_0 \).
 Step 6
State an overall conclusion.
Minitab^{®}
Conduct a OneSample Mean tTest
Note that these steps are very similar to those for onemean confidence interval. The differences occur in steps 4 through 8.
To conduct the one sample mean ttest in Minitab...
 Choose Stat > Basic Stat > 1 Sample t.
 In the dropdown box use "One or more samples, each in a column" if you have the raw data, otherwise select "Summarized data" if you only have the sample statistics.
 If using the raw data, enter the column of interest into the blank variable window below the drop down selection. If using summarized data, enter the sample size, sample mean, and sample standard deviation in their respective fields.
 Choose the check box for "Perform hypothesis test" and enter the null hypothesis value.
 Choose Options.
 Enter the confidence level associated with alpha (e.g. 95% for alpha of 5%).
 From the drop down list for "Alternative hypothesis" select the correct alternative.
 Click OK and OK.