Question 1129146: On a survey on entertainment activities, 322 of 2,023 adult Americans surveyed report reported that they have watched a movie in cinema at least once in the past year. Assume that this sample is representative of the population of adult Americans.
A. Find the sample proportion of adult Americans who have seen a movie in cinema in the last year at least once.
C.calculate the margin of error.
D. Construct 90% confidence level
E. Construct 95% confidence level
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! A. sample proportion(P) = 322/2023 = 0.1592
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Standard Error(SE) of P = square root(0.1592 * (1-0.1592) / 2023) = 0.0001
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Margin of error(ME) = Critical value(CV) x SE of P
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Since the sample size is sufficiently large, that is, sample size is greater than 30, we can use the Normal Distribution to determine CV
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To determine the CV for a confidence level use the following steps
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alpha = 1 - (confidence level/100)
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critical probability(p*) = 1 - (a/2)
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CV is the z-score associated with p*
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CV for a 90% confidence level is 1.645
CV for a 95% confidence level is 1.960
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C.
ME for 90% = 1.645 * 0.0001 = 0.0001645 is approximately 0.0002
ME for 95% = 1.960 * 0.0001 = 0.0001960 is approximately 0.0002
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D. 0.1592 + or - 0.0002 is (0.1590, 0.1594)
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E. 0.1592 + or - 0.0002 is (0.1590, 0.1594)
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Note the problem does not state the accuracy level so rounding to 4 significant digits yields the same confidence interval for D and E
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