A sample of 12 apartments on the east side of town found tha

A sample of 12 apartments on the east side of town found that rents had increased by a mean of $50 per month since last year with a standard deviation of $18 and a sample of 15 apartments on the west side found a mean increase of $35 per month with a standard deviation of $12. Assuming the increases in rents on both the east and west side are normally distributed, find a 99% confidence interval for difference between the mean increase in monthly rent on the west and east sides

Solution

  
Calculating the means of each group,              
              
X1 =    50          
X2 =    35          
              
Calculating the standard deviations of each group,              
              
s1 =    18          
s2 =    12          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    12          
n2 = sample size of group 2 =    15          
Thus, df = n1 + n2 - 2 =    25          
Also, sD =    6.049793385          
              
      
              
For the   0.99   confidence level, then      
              
alpha/2 = (1 - confidence level)/2 =    0.005          
t(alpha/2) =    2.787435814          
              
lower bound = [X1 - X2] - t(alpha/2) * sD =    -1.863410746          
upper bound = [X1 - X2] + t(alpha/2) * sD =    31.86341075          
              
Thus, the confidence interval is              
              
(   -1.863410746   ,   31.86341075   ) [ANSWER]

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