Question

A random sample of 25 boxes of cereal have a mean of 372.5grams and a standard deviation of 15 grams. Does an average box of cereal contain more than 368 grams of cereal? Test hypothesis at the 5% significance level, outline all steps involved. (9mks)​

Answer 1

Using the t-distribution, it is found that since the test statistic is less than the critical value for the right-tailed test, 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 tested if the average box of cereal does not contain more than 368 grams, that is:

$H_0: \mu \leq 368$

At the alternative hypothesis, it is tested if it contains, that is:

$H_1: \mu > 368$

We have the standard deviation for the sample, hence, the t-distribution is used to solve this question.

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

• $\overline{x}$ is the sample mean.

• $\mu$ is the value tested at the null hypothesis.

• s is the standard deviation of the sample.

• n is the sample size.

In this problem, the values of the parameters are: $\overline{x} = 372.5, \mu = 368, s = 15, n = 25$.

Hence, the value of the test statistic is:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{372.5 - 368}{\frac{15}{\sqrt{25}}}$

$t = 1.5$

The critical value for a right-tailed test, as we are testing if the mean is greater than a value, with a significance level of 0.05, with 25 -1 = 24 df, is of $t^{\ast} = 1.71$.

Since the test statistic is less than the critical value for the right-tailed test, 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