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diamong [38]
3 years ago
13

A random sample of 300 circuits generated 13 defectives .Use the data to test the hypothesisH0: p = 0.05 against H1: p ≠ 0.05. U

se α = 0.08. Find the P-value for thetest.
Mathematics
1 answer:
WINSTONCH [101]3 years ago
4 0

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 .  

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