Question

A sample of 1100 computer chips revealed that 77% of the chips fail in the first 1000 hours of their use. The company's promotional literature states that 76% of the chips fail in the first 1000 hours of their use. The quality control manager wants to test the claim that the actual percentage that fail is different from the stated percentage. Find the value of the test statistic. Round your answer to two decimal places.

Answer 1

Answer:

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

Replacing the info given we got:

$z=\frac{0.77 -0.76}{\sqrt{\frac{0.76(1-0.76)}{1100}}}=0.778$  

Step-by-step explanation:

Information given

n=1100 represent the random sample taken

$\hat p=0.77$ estimated proportion of chips that fall in the first 1000 hours of their use

$p_o=0.76$ is the value that we want to test

z would represent the statistic

$p_v$ represent the p value

Solution

We need to conduct a hypothesis in order to check if the true proportion is equal to 0.76.:  

Null hypothesis:$p=0.76$  

Alternative hypothesis:$p \neq 0.76$  

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

Replacing the info given we got:

$z=\frac{0.77 -0.76}{\sqrt{\frac{0.76(1-0.76)}{1100}}}=0.778$