SOLUTION: 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

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.
I am stuck on this simple question and not sure why so any help would be great.
Thanks.
SS
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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.
------
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
========================
Cheers,
Stan H.

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