Question 1056922: A group of 463 first-year college students were asked, “About how many hours do you study during a typical week?” The mean response is 15.3 hours. Assume that the study time is normally distributed with a population standard deviation of 8.5 hours. Construct a 99% confidence level interval for the mean study time of all first-year students.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! Since we are given the sample standard deviation, we can use the following formula for the Margin of Error(ME)
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ME = critical value(CV) * standard error(SE)
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We need to compute the CV first, since this is a normal distribution we can use the tables of z-values
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alpha(a) = 1 - (confidence level / 100) = 1 - 0.99 = 0.01
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critical [probability(p*) = 1 - (a / 2) = 1 - (0.01 / 2) = 0.995
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CV is the z-score associated with p*, that is, CV = 2.58
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SE = standard dev / square root(n) = 8.5 / square root(463) = 0.395
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ME = 2.58 * 0.395 = 1.0191 approximately 1.02
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The 99% confidence interval is 15.3 + or - 1.02
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