Pineapple Corporation (PC) maintains that their cans have always contained an average of 12 ounces of fruit. The production group believes that the mean weight has changed. They take a sample of 15 cans and find a sample mean of 12.05 ounces and a sample standard deviation of .08 ounces. What conclusion can we make from the appropriate hypothesis test at the .01 level of significance

Respuesta :

Answer:

We accept the null hypothesis, that is, that the mean weight of the cans is still of 12 ounces of fruit.

Step-by-step explanation:

Pineapple Corporation (PC) maintains that their cans have always contained an average of 12 ounces of fruit.

This means that the null hypothesis is: [tex]H_0: \mu = 12[/tex]

The production group believes that the mean weight has changed.

This means that the alternate hypothesis is:

[tex]H_a: \mu \neq 12[/tex]

The test statistic is:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

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

12 is tested at the null hypothesis:

This means that [tex]\mu = 12[/tex]

They take a sample of 15 cans and find a sample mean of 12.05 ounces and a sample standard deviation of .08 ounces.

This means, respectibely, that [tex]n = 15, X = 12.05, \sigma = 0.08[/tex]

Value of the test statistic:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{12.05 - 12}{\frac{0.08}{\sqrt{15}}}[/tex]

[tex]z = 2.42[/tex]

Pvalue of the test:

We are testing if the mean is different from a value, which means that the pvalue is 2 multiplied by 1 subtracted by the pvalue of z = 2.42.

Looking at the z-table, z = 2.42 has a pvalue of 0.9922

1 - 0.9922 = 0.0078

2*0.0078 = 0.0156

What conclusion can we make from the appropriate hypothesis test at the .01 level of significance?

0.0156 > 0.01. This means that at the 0.01 level, we accept the null hypothesis, that is, that the mean weight of the cans is still of 12 ounces of fruit.