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

Answer:

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

$p_v =2*P(z$  

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 .  

Step-by-step explanation:

1) Data given and notation  

n=300 represent the random sample taken

X=13 represent the number of defectives

$\hat p=\frac{13}{300}=0.0433$ 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:$p \neq 0.05$  

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

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

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

3) Calculate the statistic  

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

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

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:  

$p_v =2*P(z$  

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 .