Question 460938: Suppose that a company CEO claims that a majority of his employees carry secondary health insurance. You decide to test his claim using a significance level of 1 = 0.05. A sample of 150 employees finds that 87 of them carry secondary health insurance. 1st, you set up your hypotheses as follows:
Ho:ρ ≤ 0.50
H1:ρ > 0.50 (claim)
z = 87/150 - 0.50/√0.50 x 0.50/150
= 1.9596
Compute the probability of getting a sample statistic at least as extreme as z = 1.9596 and interpret this probability value.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Suppose that a company CEO claims that a majority of his employees carry secondary health insurance.
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You decide to test his claim using a significance level of alpha = 0.05.
A sample of 150 employees finds that 87 of them carry secondary health insurance. 1st, you set up your hypotheses as follows:
Ho:ρ ≤ 0.50
H1:ρ > 0.50 (claim)
z = [(87/150) - 0.50]/sqrt[0.50 x 0.50/150}
= 1.9596
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Compute the probability of getting a sample statistic at least as extreme as z = 1.9596 and interpret this probability value.
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P(z > 1.9596) = normalcdf(1.9696,100) = 0.0250
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Cheers,
Stan H.
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