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Statistics 12 Online
OpenStudy (anonymous):

A random sample of 139 forty-year-old men contains 26%smokers. Find the P-value for a test of the claim that thepercentage of forty year-old men that smoke is 22%

OpenStudy (anonymous):

I can't find the right answer

OpenStudy (amistre64):

isnt this:\[\dfrac{\hat p-p}{\sqrt{\large \frac{pq}{n}}}\]??

OpenStudy (anonymous):

Yes buti don't know how to solve it

OpenStudy (amistre64):

\[\dfrac{.26-.22}{\sqrt{\large \frac{.22~.78}{139}}}\] gives us a zscore to play with

OpenStudy (amistre64):

hopefully i used the right phat and p

OpenStudy (amistre64):

do you recall if we subtract the claim proportion, or the sample proportion?

OpenStudy (amistre64):

i think we subtract the claim, in which case -.22 is the way to go

OpenStudy (amistre64):

that gives us a test statistic of 1.14

OpenStudy (amistre64):

the claim is: 22% of 40yr smoke therefore the counter claim is that: 22% do not smoke this is a two tailed test. so we double the results of a tail to get the P-value

OpenStudy (amistre64):

|dw:1371130697510:dw|

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