Question

A sample of 1600 computer chips revealed that 60% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that above 57% do not fail in the first 1000 hours of their use. Is there sufficient evidence at the 0.01 level to support the company's claim? State the null and alternative hypotheses for the above scenario.

Answer 1

If we compare the p value obtained and the significance level given
`\alpha=0.01` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of chips that do not fail in the first 1000 hours of their use is significantly higher than 0.57 or 57%.

Explanation:

1) Data given and notation

n=1600 represent the random sample taken

X=0.6*1600=960 represent the number of chips do not fail in the first 1000 hours of their use

`\hat p=(960)/(1600)=0.6` estimated proportion of chips do not fail in the first 1000 hours of their use

`p_o=0.57` is the value that we want to test

`\alpha=0.01` represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that above 57% of chips do not fail in the first 1000 hours of their use :

Null hypothesis:
`p\leq 0.57`

Alternative hypothesis:
`p > 0.57`

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`.

3) Calculate the statistic

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

`z=\frac{0.6 -0.57}{\sqrt{(0.57(1-0.57))/(1600)}}=2.42`

4) 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=0.01`. The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

`p_v =P(z>2.42)=0.0078`

If we compare the p value obtained and the significance level given
`\alpha=0.01` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of chips that do not fail in the first 1000 hours of their use is significantly higher than 0.57 or 57%.