A random sample of 87 observations produced a mean x=25.8 and a standard deviati

Question

A random sample of 87 observations produced a mean x=25.8 and a standard deviation equals=2.7.

a. Find a 95% confidence interval for .

b. Find a 90% confidence interval for .

c. Find a 99% confidence interval for .

a. The 95% confidence interval is ( _,_ )

(Use integers or decimals for any numbers in the expression. Round to two decimal places as needed.)

b. The 90% confidence interval is (_,_)

(Use integers or decimals for any numbers in the expression. Round to two decimal places as needed.)

c. The 99% confidence interval is (_,_)

(Use integers or decimals for any numbers in the expression. Round to two decimal places as needed.)

Explanation / Answer

mean x=25.8

standard deviation =2.7

sample size n = 87

standard deviation for = standard deviation /n = 2.7/87 = 0.2895

a)

95% confidence interval is (mean-1.96*sd,mean+1.96*sd)

95% confidence interval for is (25.8-1.96*0.2895, 25.8+1.96*0.2895) i.e (25.23, 26.37)

b)

90% confidence interval is (mean-1.645*sd,mean+1.645*sd)

90% confidence interval for is (25.8-1.645*0.2895, 25.8+1.645*0.2895) i.e (25.32, 26.28)

c)

99% confidence interval is (mean-2.575*sd,mean+2.575*sd)

99% confidence interval for is (25.8-2.575*0.2895, 25.8+2.575*0.2895) i.e (25.05, 26.55)

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