Question

In a sample of 37, the mean is  x bar= 95.30 with a standard deviation of s = 7.30. Assuming the results follow a normal distribution, construct the 90% confidence interval for the population mean. The t Table = 1.688.

• A. 90% confidence interval is 95.30 ± 2.33

• B. 90% confidence interval is 95.20 ± 1.688

• C.90% confidence interval is between 93.27 and 97.33

• D. 90% confidence interval is 97.33

Answer 1

Sol;

n=37

df=n-1=37-1=36

alpha=0.10

alpha/2=0.10/2=0.05

t critical in excel

==T.INV(0.05,36)

=1.6883

90% confidence interval for mean is

xbar-t*s/sqrtt(n),xbar+t*s/sqrtt(n)

95.30+_1.6883*7.30/sqrt(37)

95.30+_2.02615

95.30-2.02615,95.30+2.02615

93.27385,97.32615

93.27,97.33

ANSWER

C.90% confidence interval is between 93.27 and 97.33