Question

You are a state inspector for the Division of Weights and Measures. Your responsibility is to be sure the net weight found on all containers is correctly reflected on the label (note: the label is a claim). You are inspecting ABC Company who makes dry dog food in boxes and bags. The boxes you are checking indicate a net mean weight of 32 ounces. You check a sample of 200 boxes and found the average content to be 31.7 ounces. The standard deviation allowed for this type of product is 2.2 ounces. Can you conclude at a .02 level of significance that the boxes of dog food are being under filled

Answer 1

Answer:

No. There is not enough evidence to support the claim that the boxes of dog food are being under filled (P-vaue=0.027).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the boxes of dog food are being under filled.

Then, the null and alternative hypothesis are:

$H_0: \mu=32\\\\H_a:\mu< 32$

The significance level is 0.02.

The sample has a size n=200.

The sample mean is M=31.7.

The standard deviation of the population is known and has a value of σ=2.2.

We can calculate the standard error as:

$\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{2.2}{\sqrt{200}}=0.156$

Then, we can calculate the z-statistic as:

$z=\dfrac{M-\mu}{\sigma_M}=\dfrac{31.7-32}{0.156}=\dfrac{-0.3}{0.156}=-1.928$

This test is a left-tailed test, so the P-value for this test is calculated as:

$\text{P-value}=P(z<-1.928)=0.027$

As the P-value (0.027) is bigger than the significance level (0.02), the effect is  not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the boxes of dog food are being under filled.