Question

A sample of 1500 computer chips revealed that 37% of the chips fail in the first 1000 hours of their use. The company's promotional literature states that 39% 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:

$z=\frac{0.37 -0.39}{\sqrt{\frac{0.39(1-0.39)}{1500}}}=-1.59$  

Step-by-step explanation:

Data given and notation

n=1500 represent the random sample taken

$\hat p=0.37$ estimated proportion of of chips fail in the first 100 hours

$p_o=0.39$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that  the actual percentage that fail is different from the stated percentage. Find the value of the test statistic.:  

Null hypothesis:$p=0.39$  

Alternative hypothesis:$p \neq 0.39$  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

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

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

$z=\frac{0.37 -0.39}{\sqrt{\frac{0.39(1-0.39)}{1500}}}=-1.59$