Question 536908
Claim: The proportion of brown M&Ms more than 22%. A simple random sample of 150 M&Ms contains 26% brown. 
I have to identify the null and alternative hypothesis and identify the sampling distribution (normal, t, chi-square) of the test statistic or state that none of the sampling distributions would work.
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Ho: p <= 0.22
Ha: p > 0.22
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test statistic: z(0.26) = (0.26-0.22)/sqrt[0.22*0.78/150] = 1.1826
Distribution: Normal
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Cheers,
Stan H.