A sample of 40 taken from mound shaped population 1 has Xbar

A sample of 40, taken from mound shaped population #1, has Xbar₁= 98 and s₁=6.5

A sample of 20, taken from a normal population #2, has xbar₂ = 90, and s₂ = 7.3

Approximate a 90% confidence level for ₁ - _2: Show work

Can we conclude which is larger with 90% confidence:

Solution

Calculating the standard deviations of each group,              
              
s1 =    6.5          
s2 =    7.3          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    40          
n2 = sample size of group 2 =    20          
Thus, df = n1 + n2 - 2 =    58          
Also, sD =    1.928924571          
              
For the   0.9   confidence level, then      
              
alpha/2 = (1 - confidence level)/2 =    0.05          
t(alpha/2) =    1.671552762          
              
lower bound = [X1 - X2] - t(alpha/2) * sD =    4.775700805          
upper bound = [X1 - X2] + t(alpha/2) * sD =    11.22429919          
              
Thus, the confidence interval is              
              
(   4.775700805   ,   11.22429919   ) [ANSWER]

Notice that this interval is purely positive.

Thus, at 90% confidence, we conclude that the mean of population 1 is higher. [ANSWER]