A random sample of 61 statistics examinations was taken The

A random sample of 61 statistics examinations was taken. The average score, in the sample, was 75 with a sample variance of 8.25. Find the 90% confidence interval for the average examination score of the population of the examinations.

2. In a sample of 200 students, 166 indicated they never miss any classes during the week. Find the 95% confidence interval of students who never miss class.

Solution

Given a=1-0.9=0.1, Z(0.05) = 1.645 (from standard normal table)

So the lower bound is

xbar - Z*s/vn = 75 -1.645*sqrt(8.25/61) =74.39504

So the upper bound is

xbar + Z*s/vn = 75 +1.645*sqrt(8.25/61) =75.60496

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p=166/200 =0.83

Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

So the lower bound is

p - Z*sqrt(p*(1-p)/n) = 0.83 -1.96*sqrt(0.83*(1-0.83)/200) =0.77794

So the upper bound is

p + Z*sqrt(p*(1-p)/n) = 0.83 +1.96*sqrt(0.83*(1-0.83)/200) =0.88206