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a) Since the p-value of the test is 0.15 > 0.05, there is not significant evidence to conclude that the mean weight is below 325-mg, that is, that it meets specifications.

b) Any larger sample size does not ensure that regulations are met, it has to be significantly large such that t will be less than the critical value for the test.

At the null hypothesis, we test if the mean sodium content is of at least 325-mg, that is:

$H_0: \mu \geq 325$

At the alternative hypothesis, we test if it is less than 325-mg, that is:

$H_1: \mu < 325$

Item a:

We have the standard deviation for the sample, thus, the t-distribution is used. 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.

For this problem, the values of the parameters are: $\overline{x} = 322, \mu = 325, s = 18, n = 40$

The value of the test statistic is:

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

$t = \frac{322 - 325}{\frac{18}{\sqrt{40}}}$

$t = -1.05$

The p-value of the test is found using a left-tailed test, as we test if the mean is less than a value, with t = -1.05 and 40 - 1 = 39 df.

Using a t-distribution calculator, this p-value is of 0.15.

Since the p-value of the test is 0.15 > 0.05, there is not significant evidence to conclude that the mean weight is below 325-mg, that is, that it meets specifications.

Item b:

Increasing the sample size decreases t which also decreases the p-value in a left-tailed test.

However, the specifications are only met is t is lower than the critical value of t, hence, there has to be a significant increase in sample size.

Hence:

Any larger sample size does not ensure that regulations are met, it has to be significantly large such that t will be less than the critical value for the test.

A similar problem is given at brainly.com/question/25454581

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2 years ago