Question

A sample of 1300 computer chips revealed that 58% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature states that 61% of the chips do not fail in the first 1000 hours of their use. The quality control manager wants to test the claim that the actual percentage that do not fail is different from the stated percentage. Find the value of the test statistic. Round your answer to two decimal places.

Answer 1

Answer:

There is enough evidence to support the claim that that the actual percentage that do not fail is different from the stated percentage (61%).

Test statistic z = -2.19.

P-value = 0.03.

Step-by-step explanation:

This is a hypothesis test for a proportion.

The claim is that that the actual percentage that do not fail is different from the stated percentage (61%).

Then, the null and alternative hypothesis are:

$H_0: \pi=0.61\\\\H_a:\pi\neq 0.61$

The significance level is assumed to be 0.05.

The sample has a size n=1300.

The sample proportion is p=0.58.

 

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.61*0.39}{1300}}\\\\\\ \sigma_p=\sqrt{0.000183}=0.014$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.58-0.61+0.5/1300}{0.014}=\dfrac{-0.03}{0.014}=-2.189$

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

$\text{P-value}=2\cdot P(z<-2.189)=0.03$

As the P-value (0.03) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that that the actual percentage that do not fail is different from the stated percentage (61%).