SOLUTION: A sample of 64 was taken from a population, with the mean of the sample being 119 and the standard deviation of the sample being 16. What is the 95% confidence interval for the mea
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-> SOLUTION: A sample of 64 was taken from a population, with the mean of the sample being 119 and the standard deviation of the sample being 16. What is the 95% confidence interval for the mea
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Question 1071281: A sample of 64 was taken from a population, with the mean of the sample being 119 and the standard deviation of the sample being 16. What is the 95% confidence interval for the mean of the population? Found 2 solutions by Boreal, rothauserc:Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! CI half-interval is t(df=63, 0.975)s/sqrt(n)
I will use 2 for t
=2(16)/8=4
So the 95% CI is 119+/-4
(115, 123)
You can put this solution on YOUR website! The standard error(SE) of the mean is 16 / square root(64) = 2
:
alpha(a) = 1 - (95/100) = 0.05
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The critical probability (p*): p* = 1 - a/2 = 1 - 0.05/2 = 0.975
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Use z-table(sample > 40) to find crical value(CV) associated with p*
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CV = 1.96
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margin of error(ME) = CV * SE = 1.96 * 2 = 3.92
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95% confidence interval(CI) = sample statistic + or - ME
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95% CI = 119 + or - 3.92 = (115.08, 122.92)
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