Question

The average weight of a package of rolled oats is supposed to be at most 18 ounces. A sample of 18 packages shows a mean of 18.20 ounces with a sample standard deviation of 0.50 ounces. (a) at the 5 percent level of significance, is the true mean larger than the specification

Answer 1

Answer:

Pvalue of 0.0446 < 0.05, which means that we reject the null hypothesis and accept the alternative hypothesis, that the true mean is larger than the specification.

Step-by-step explanation:

The average weight of a package of rolled oats is supposed to be at most 18 ounces.

This means that the null hypothesis is:

$H_{0}: \mu \leq 18$

Is the true mean larger than the specification?

Due to the question asked, the alternate hypothesis is:

$H_{a}: \mu > 18$

The test statistic is:

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

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

18 is tested at the null hypothesis:

This means that $\mu = 18$

A sample of 18 packages shows a mean of 18.20 ounces with a sample standard deviation of 0.50 ounces.

This means that $n = 18, X = 18.2, \sigma = 0.5$

Value of the test statistic:

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

$z = \frac{18.2 - 18}{\frac{0.5}{\sqrt{18}}}$

$z = 1.7$

Pvalue of the test:

Probability of finding a sample mean above 18.2, which is 1 subtracted by the pvalue of z = 1.7.

Looking at the z-table, z = 1.7 has a pvalue of 0.9554.

1 - 0.9554 = 0.0446

0.0446 < 0.05, which means that we reject the null hypothesis and accept the alternative hypothesis, that the true mean is larger than the specification.