Question

A company distributes candies in bags labeled 23.6 ounces. The local bureau of weights and Measures randomly selects 60 bags of candies and obtain a sample mean of 24 ounces . Assuming that the standard deviation is 3.2. At 0.05 level of significance , test the claim that the bags contain more than 23.6 ounces . what is your conclusion about the claim.

Answer 1

Answer:

The p-value of the test is 0.166 > 0.05, which means that there is not sufficient evidence at the 0.05 significance level to conclude that the bags contain more than 23.6 ounces.

Step-by-step explanation:

A company distributes candies in bags labeled 23.6 ounces. Test if the mean is more than this:

At the null hypothesis, we test if the mean is of 23.6, that is:

$H_0: \mu = 23.6$

At the alternative hypothesis, we test if the mean is of more than 23.6, that is:

$H_1: \mu > 23.6$

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.

23.6 is tested at the null hypothesis:

This means that $\mu = 23.6$

The local bureau of weights and Measures randomly selects 60 bags of candies and obtain a sample mean of 24 ounces. Assuming that the standard deviation is 3.2.

This means that $n = 60, X = 24, \sigma = 3.2$

Value of the test statistic:

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

$z = \frac{24 - 23.6}{\frac{3.2}{\sqrt{60}}}$

$z = 0.97$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 24, which is 1 subtracted by the p-value of z = 0.97.

Looking at the z-table, z = 0.97 has a p-value of 0.834.

1 - 0.834 = 0.166

The p-value of the test is 0.166 > 0.05, which means that there is not sufficient evidence at the 0.05 significance level to conclude that the bags contain more than 23.6 ounces.