Question

A random sample of 300 circuits generated 13 defectives .Use the data to test the hypothesisH0: p = 0.05 against H1: p ≠ 0.05. Use α = 0.08. Find the P-value for thetest.

Answer 1

`z=\frac{0.0433 -0.05}{\sqrt{(0.05(1-0.05))/(300)}}=-0.532`

`p_v =2*P(z<-0.532)=0.595`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.08` 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 5% of significance the proportion of defectives is not significantly different from 0.05 .

Explanation:

1) Data given and notation

n=300 represent the random sample taken

X=13 represent the number of defectives

`\hat p=(13)/(300)=0.0433` estimated proportion of defectives

`p_o=0.05` is the value that we want to test

`\alpha=0.08` represent the significance level

Confidence=92% or 0.92

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 the true porportion is 0.05.:

Null hypothesis:
`p=0.05`

Alternative hypothesis:
`p \\eq 0.05`

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{(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{0.0433 -0.05}{\sqrt{(0.05(1-0.05))/(300)}}=-0.532`

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=0.08`. 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<-0.532)=0.595`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.08` 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 5% of significance the proportion of defectives is not significantly different from 0.05 .