Using the t-distribution, it is found that since the <u>test statistic is less than the critical value for the right-tailed test</u>, there is not enough evidence to conclude that an average box of cereal contain more than 368 grams of cereal.
At the null hypothesis, it is <u>tested if the average box of cereal does not contain more than 368 grams</u>, that is:
![H_0: \mu \leq 368](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu%20%5Cleq%20368)
At the alternative hypothesis, it is <u>tested if it contains</u>, that is:
![H_1: \mu > 368](https://tex.z-dn.net/?f=H_1%3A%20%5Cmu%20%3E%20368)
We have the <em>standard deviation for the sample</em>, hence, the t-distribution is used to solve this question.
The test statistic is given by:
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 <em>parameters </em>are:
.
Hence, the value of the <em>test statistic</em> 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{372.5 - 368}{\frac{15}{\sqrt{25}}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B372.5%20-%20368%7D%7B%5Cfrac%7B15%7D%7B%5Csqrt%7B25%7D%7D%7D)
![t = 1.5](https://tex.z-dn.net/?f=t%20%3D%201.5)
The critical value for a <u>right-tailed test</u>, as we are testing if the mean is greater than a value, with a <u>significance level of 0.05,</u> with 25 -1 = <u>24 df</u>, is of
.
Since the <u>test statistic is less than the critical value for the right-tailed test</u>, there is not enough evidence to conclude that an average box of cereal contain more than 368 grams of cereal.
You can learn more about the use of the t-distribution to test an hypothesis at brainly.com/question/13873630