
Devi P. answered 12/03/14
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mean = 12hr
std dev = 3hr
margin of error = 15min = 0.25hr
CI = 95%
Based on the information given in the problem, you are not given the population standard deviation, so you will use t-values. The critical t value associated with a 95% confidence interval is 1.96
Margin of Error = (critical t value) x [standard deviation/sqrt(sample size)]
0.25 = 1.96 x [3/sqrt(n)]
0.25sqrt(n) = 5.88
sqrt(n) = 23.52
n = (23.52)^2
n = 553.19
n = 554 <-- you cannot have a partial person, so always round up