In 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 One-Sample Mean
Section* *

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 t-distribution with \(n-1 \) 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 5-1
Section* *

**very different**from 8.5?

This depends on the standard deviation of \(\bar{x} \) .

\begin{align} t^*&=\dfrac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\\&=\dfrac{8.3-8.5}{\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.