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In a sample of 60 adults selected randomly from one town, it is found that 12 of them have been exposed to a particular strain of the flu. The test statistic is z=1.08. Find the P-value for a test of the claim that the proportion of all adults in the town that have been exposed to this strain of the flu is different to 15% (H1: p ≠0.15 ).

User Corion
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Answer:


z=\frac{0.2 -0.15}{\sqrt{(0.15(1-0.15))/(60)}}=1.085


p_v =2*P(z>1.085)=0.2779

Explanation:

Data given and notation

n=60 represent the random sample taken

X=12 represent the adults exposed to a particular strain of the flu


\hat p=(12)/(60)=0.2 estimated proportion of adults exposed to a particular strain of the flu


p_o=0.15 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)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is 0.15 or no.:

Null hypothesis:
p=0.15

Alternative hypothesis:
p \\eq 0.15

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{(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.

Calculate the statistic

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


z=\frac{0.2 -0.15}{\sqrt{(0.15(1-0.15))/(60)}}=1.085

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 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>1.085)=0.2779

User IVerzin
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