A random sample of size 36 from a normal population yields m

A random sample of size 36 from a normal population yields mean X-bar = 32.8 and standard deviation s = 4.51. Construct a 95 percent confidence interval for ?.

Solution

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

So the lower bound is

xbar - Z*s/vn =32.8 -1.96*4.51/sqrt(36) =31.32673

So the upper bound is

xbar + Z*s/vn =32.8 +1.96*4.51/sqrt(36) =34.27327