Using the z-distribution, it is found that since the p-value is less than 0.05, there is evidence that the mean amount of cereal in each box is different from 16 ounces at 0.05 significance.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if the mean is not different to 16 ounces, that is:

$H_0: \mu = 16$

At the alternative hypothesis, it is tested if the mean is different, hence:

$H_1: \mu \neq 16$

<h3>What is the test statistic?</h3>

The test statistic is:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which:

$\overline{x}$ is the sample mean.

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

$\sigma$ is the standard deviation of the population.

n is the sample size.

The parameters for this problem are:

$\overline{x} = 15.75, \mu = 16, \sigma = 1.46, n = 150$ .

Hence:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{15.75 - 16}{\frac{1.46}{\sqrt{150}}}$

z = -2.1

<h3>What is the p-value and the conclusion?</h3>

Using a z-distribution calculator, for a two-tailed test, as we are testing if the mean is different of a value, with z = -2.1, the p-value is of 0.0357.

Since the p-value is less than 0.05, there is evidence that the mean amount of cereal in each box is different from 16 ounces at 0.05 significance.

More can be learned about the z-distribution at brainly.com/question/16313918

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