Using the t-distribution, as we have the standard deviation for the sample, it is found that since the test statistic is less than the critical value for the right-tailed test, there is evidence to support the belief that the mean is less than 15. 25 ounces.
<h3>What are the hypothesis tested?</h3>
At the null hypothesis, it is tested if the mean is of 15.25 ounces, that is:
![H_0: \mu = 15.25](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu%20%3D%2015.25)
At the alternative hypothesis, it is tested if the mean is of less than 15.25 pounds, that is:
.
<h3>What is the test statistic?</h3>
The test statistic is given by:
![t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5Coverline%7Bx%7D%20-%20%5Cmu%7D%7B%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%7D)
The parameters are:
is the sample mean.
is the value tested at the null hypothesis.
- s is the standard deviation of the sample.
In this problem, the values of the parameters are:
.
Hence, the value of the test statistic is:
![t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5Coverline%7Bx%7D%20-%20%5Cmu%7D%7B%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%7D)
![t = \frac{15.18 - 15.25}{\frac{0.12}{\sqrt{36}}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B15.18%20-%2015.25%7D%7B%5Cfrac%7B0.12%7D%7B%5Csqrt%7B36%7D%7D%7D)
![t = -3.5](https://tex.z-dn.net/?f=t%20%3D%20-3.5)
<h3>What is the decision?</h3>
Considering a <em>left-tailed test</em>, as we are testing if the mean is less than a value, with 36 - 1 = <em>35 df and the standard significance level of 0.05</em>, it is found that the critical value is
.
Since the test statistic is less than the critical value for the right-tailed test, there is evidence to support the belief that the mean is less than 15. 25 ounces.
More can be learned about the t-distribution at brainly.com/question/16313918