Question

A sample of 1600 computer chips revealed that 54% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that more than 51% do not fail in the first 1000 hours of their use. Is there sufficient evidence at the 0.05 level to support the company's claim?

Answer 1

`p_v =P(z>2.40)=0.0082`

The p value obtained was a very low value and using the significance level given
`\alpha=0.05` 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 5% of significance the proportion of the chips that do not fail in the first 1000 hours of their use is higher than 0.51 or 51% (Company's claim).

Explanation:

1) Data given and notation n

n=1600 represent the random sample taken

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

`\hat p=0.54` estimated proportion of the chips do not fail in the first 1000 hours of their use

`p_o=0.51` is the value that we want to test

`\alpha=0.05` represent the significance level

Confidence=95% or 0.95

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 more than 51% do not fail in the first 1000 hours of their use, the system of hypothesis on this case would be:

Null hypothesis:
`p\leq 0.51`

Alternative hypothesis:
`p > 0.51`

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{0.54 -0.51}{\sqrt{(0.51(1-0.51))/(1600)}}=2.40`

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=0.05`. The next step would be calculate the p value for this test.

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

`p_v =P(z>2.40)=0.0082`

So the p value obtained was a very low value and using the significance level given
`\alpha=0.05` 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 5% of significance the proportion of the chips that do not fail in the first 1000 hours of their use is higher than 0.51 or 51%.