Question

Bags of a certain brand of tortilla chips claim to have a net weight of 14 ounces. The net weights actually vary slightly from bag to bag and are normally distributed with mean μ. A representative of a consumer advocacy group wishes to see if there is any evidence that the mean net weight is less than advertised.
For this, the representative randomly selects 16 bags of this brand and determines the net weight of each. He finds the sample mean to be 13.82 and the sample standard deviation to be 0.24.
Use these data to perform an appropriate test of hypothesis at 5% significance level.

Answer 1

Answer:

The p-value of the test is of 0.0045 < 0.05, which means that there is enough evidence to reject the null hypothesis and accept the alternate hypothesis that the mean net weight is less than advertised.

Step-by-step explanation:

Bags of a certain brand of tortilla chips claim to have a net weight of 14 ounces. A representative of a consumer advocacy group wishes to see if there is any evidence that the mean net weight is less than advertised.

At the null hypothesis, we test that the mean is of 14, that is:

$H_0: \mu = 14$

At the alternate hypothesis, we test tha tthe mean is less than advertised, that is, less than 14. So:

$H_a: \mu < 14$

The test statistic is:

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

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

14 is tested at the null hypothesis:

This means that $\mu = 14$

For this, the representative randomly selects 16 bags of this brand and determines the net weight of each. He finds the sample mean to be 13.82 and the sample standard deviation to be 0.24.

This means that $n = 16, X = 13.82, s = 0.24$

Test statistic:

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

$t = \frac{13.82 - 14}{\frac{0.24}{\sqrt{16}}}$

$t = -3$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean below 13.82, which is the p-value of t = -3, using a left-tailed test with 16 - 1 = 15 degrees of freedom.

With the help of a calculator, the p-value is of 0.0045.

The p-value of the test is of 0.0045 < 0.05, which means that there is enough evidence to reject the null hypothesis and accept the alternate hypothesis that the mean net weight is less than advertised.