Hi,

re: Your reply (Idea being..with a 7% SD ... the 3% difference in the test sample

would be expected to be consistent with the average of 20%) 

I.  Yes, .07/sqrt(100)  

II.  With Excel function: P(z <4.29)) = NORMDIST(4.29)  = .999



Ho: p = 0.2

Ha: p > 0.2 (claim)

sample proportion = .23

z(.23) = (.23-.2)/.07/sqrt(100) = .03/.007  = 4.29

p-value = P(z < 4.29) = .999 OR 99.9% (10% significance level)



p-value greater than 10%, accept Ho

The attrition rate is same as 20%

 

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