Question

A direct mailing company sells computers and computer parts by mail. The company claims that at least 90% of all orders are mailed within 72 hours after they are received. The quality control department at the company often takes samples to check if this claim is valid. A recently taken random sample of 150 orders showed that 129 of them were mailed within 72 hours. Set up the null and alternative hypotheses to test whether the company’s claim is true?

Answer 1

Answer:

The null and alternative hypothesis are:

$H_0: \pi=0.9\\\\H_a:\pi$

There is not enough evidence to support the claim that the company is worng and less than 90% of all orders are mailed within 72 hours after they are received.

Step-by-step explanation:

This is a hypothesis test for a proportion.

The company claims that at least 90% of all orders are mailed within 72 hours after they are received. The company's claim is the null hypothesis, the claim that is going to be tested is if this proportion is significantly less than 90%.

Then, the null and alternative hypothesis are:

$H_0: \pi=0.9\\\\H_a:\pi$

The significance level is 0.05.

The sample has a size n=150.

The sample proportion is p=0.86.

$p=X/n=129/150=0.86$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.9*0.1}{150}}\\\\\\ \sigma_p=\sqrt{0.0006}=0.024$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.86-0.9+0.5/150}{0.024}=\dfrac{-0.037}{0.024}=-1.497$

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

\text{P-value}=P(z<-1.497)=0.0672

As the P-value (0.0672) is greater than the significance level (0.05), the effect is  not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that less than 90% of all orders are mailed within 72 hours after they are received.