Question 132572: Suppose that 78% of all Americans support a proposed law to require a police permit before someone is allowed to buy a gun. If a simple random sample of 45 Americans is to be taken, and each person is asked if they support the proposed law, and the sample proportion who do favor such a law is calculated:
a.Specify the sampling distribution of the sample proportion.
b.What is the probability that the sample proportion exceeds 80%?
Thanks For any help with this
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Suppose that 78% of all Americans support a proposed law to require a police permit before someone is allowed to buy a gun. If a simple random sample of 45 Americans is to be taken, and each person is asked if they support the proposed law, and the sample proportion who do favor such a law is calculated:
a.Specify the sampling distribution of the sample proportion.
The mean of the sample-proportions is 0.78
The standard deviation of the sample-proportions is sqrt[0.78*0.22/45]
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b.What is the probability that the sample proportion exceeds 80%?
Find the z-score of 0.80
z(0.80) = (0.80-0.78)/sqrt[0.78*0.22/45] = 0.323875...
P(p-hat >0.80) = P(z>0.323875...) = 0.3730
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Cheers,
Stan H.
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