Question

A sample of 900900 computer chips revealed that 25%25% of the chips do not fail in the first 10001000 hours of their use. The company's promotional literature states that 28%28% of the chips do not fail in the first 10001000 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. Is there enough evidence at the 0.020.02 level to support the manager's claim?

Answer 1

`z=\frac{0.25 -0.28}{\sqrt{(0.28(1-0.28))/(900)}}=-2.004`

`p_v =2*P(z<-2.004)=0.045`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.02` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 2% of significance the proportion of chips that do not fail in the first 100 hours is not significantly different from 0.28, and the claim from the quality control manager is correct

Explanation:

Data given and notation

n=900 represent the random sample taken

`\hat p=0.25` estimated proportion of chips that do not fail in the first 100 hours

`p_o=0.28` is the value that we want to test

`\alpha=0.02` represent the significance level

Confidence=98% or 0.98

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 true proportion is different from 0.28 or no:

Null hypothesis:
`p=0.28`

Alternative hypothesis:
`p \\eq 0.28`

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

`z=\frac{\hat p -p_o}{\sqrt{(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{0.25 -0.28}{\sqrt{(0.28(1-0.28))/(900)}}=-2.004`

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided
`\alpha=0.02`. The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

`p_v =2*P(z<-2.004)=0.045`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.02` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 2% of significance the proportion of chips that do not fail in the first 100 hours is not significantly different from 0.28, and the claim from the quality control manager is correct