Question

Assume that you have a sample of n1=8​, with the sample mean X1=48​, and a sample standard deviation of S1=5​, and you have an independent sample of n2=14 from another population with a sample mean of X2=33​, and the sample standard deviation S2=8. Construct a 90​% confidence interval estimate of the population mean difference between mu 1 μ1 and mu 2 μ2. Assume that the two population variances are equal.

Answer 1

sp = sqrt((((n1 - 1)*s1^2 + (n2 - 1)*s2^2)/(n1 + n2 - 2))*(1/n1 + 1/n2))
sp = sqrt((((8 - 1)*5^2 + (14 - 1)*8^2)/(8 + 14 - 2))*(1/8 + 1/14))
sp = 3.1449

Given CI level is 0.9, hence α = 1 - 0.9 = 0.1                  
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.725                  
                  
Margin of Error                  
ME = tc * sp                  
ME = 1.725 * 3.1449                  
ME = 5.425                  
                  
CI = (x1bar - x2bar - tc * sp , x1bar - x2bar + tc * sp)                  
CI = (48 - 33 - 1.725 * 3.1449 , 48 - 33 - 1.725 * 3.1449                  
CI = (9.58 , 20.42)