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.

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:

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

The test statistic is:

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

23.6 is tested at the null hypothesis:

This means that

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

Value of the test statistic:

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.

here is the

mark brainliest pl

maybe i can