Question

A computer manufacturer is testing a batch of processors. They place a simple random sample of processors from the batch under a stress test and record the number of failures, their guidelines specify the percentage of failures should be under 4%. Of the 300 processors tested, there were 54 failures.

a. State the hypotheses for the test. Assume the manufacturer wants to assume there is a problem with a batch, that is, they will only accept that there are fewer than 4% failures in the population if they have evidence for it.

b. Calculate the test statistic and p-value. Your test statistic should be either a z-value or a t-value, whichever is appropriate for the problem

Answer 1

Answer:

a)We need to conduct a hypothesis in order to test the claim that the true proportion is lower than 0.04 or no.:  

Null hypothesis:$p \leq 0.04$  

Alternative hypothesis:$p > 0.04$  

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

b) We need to use a z statistic

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

$z=\frac{0.135 -0.04}{\sqrt{\frac{0.04(1-0.04)}{300}}}=8.396$  

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

$p_v =P(z>8.396) \approx 0$  

Step-by-step explanation:

Data given and notation

n=400 represent the random sample taken

X=54 represent the number of failures

$\hat p=\frac{54}{400}=0.135$ estimated proportion of adults that said that it is morally wrong to not report all income on tax returns

$p_o=0.04$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

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

Part a: Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is lower than 0.04 or no.:  

Null hypothesis:$p \leq 0.04$  

Alternative hypothesis:$p > 0.04$  

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

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

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

Part b: Calculate the statistic  

We need to use a z statistic

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

$z=\frac{0.135 -0.04}{\sqrt{\frac{0.04(1-0.04)}{300}}}=8.396$  

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 right tailed test the p value would be:  

$p_v =P(z>8.396) \approx 0$  

So the p value obtained was a very low value and using the significance level for example $\alpha=0.05$ 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 defectives is significantly higher than 0.04 or 4%