Mathematics, 10.12.2019 04:31 lilquongohard

Arandom 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.

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Answer from: alexsandra120505

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

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

Answer from: Quest

p = 6s

step-by-step explanation:

Answer from: Quest

divide the numbers

step-by-step explanation:

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Arandom sample of 300 circuits generated 13 defectives .use the data to test the hypothesish0: p =...

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