Question 765249: Approximately 19.4% of the U.S population age 5 years and above speaks a language other than English at home. In a large metropolitan area, it was found that 94 residents (age 5 years and above) out of 400 randomly selected residents spoke a language other than English at home. Use the z-test for for a proportion, and test the claim that the proportion for this metropolitan area is higher than the national proportion. Use alpha=0.01 and report the p-value.
Any help with this will be deeply appreciated.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Approximately 19.4% of the U.S population age 5 years and above speaks a language other than English at home. In a large metropolitan area, it was found that 94 residents (age 5 years and above) out of 400 randomly selected residents spoke a language other than English at home. Use the z-test for for a proportion, and test the claim that the proportion for this metropolitan area is higher than the national proportion. Use alpha=0.01 and report the p-value.
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Ho: p = 0.194
Ha: p > 0.194 (claim)
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sample proportion = p-hat = 94/400 = 0.235
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test stat:: z(0.235) = (0.235-0.194)/sqrt[0.194*0.806/400] = 2.0737
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p-value = P(z > 2.0737) = invNorm(2.0737,100) = 0.0191
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Conclusion: Since the p-value is greater tha 1% fail to reject Ho.
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The test results do not support the claim.
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Cheers,
Stan H.
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