Arandom sample of 300 circuits generated 13 defectives.
use the data to test the hypothesis
h0: p = 0.05 againsth1: p ≠ 0.05. useα = 0.08. find the p-value for thetest.
round your answer to four decimal places (e. g. 12.3456).

(reject or not reject) null hypothesis

the p-value is ?

If we compare the p value and the significance level given we have so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 8% of significance the true proportion of defectives is different from 0.05.  

Step-by-step explanation:

1) Data given and notation n  

n=300 represent the random sample taken

X=13 represent the defectives

estimated proportion of defectives

is the value that we want to test

represent the significance level

Confidence=92% or 0.92

z would represent the statistic (variable of interest)

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 proportion of defective is different from 0.05.:  

Null hypothesis:  

Alternative hypothesis:  

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

(1)  

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

3) Calculate the statistic  

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

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

Since is a bilateral test the p value would be:  

If we compare the p value and the significance level given we have so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 8% of significance the true proportion of defectives is different from 0.05.