A hospital in a community wishes to estimate the difference

A hospital in a community wishes to estimate the difference in the mean amount of sick leave taken by workers on the day shift and the night shift. A random sample of n1 = 36 day workers had a mean of X1 = 10.6 sick day leave last year. Assume standard deviation is sigma 1 = 3.3 days, in a random sample of n2 = 64 night shift workers, the mean was x2 = 12.9 sick day leave. Assume the standard deviation is sigma 2 = 5.2 days. Estimate by a 98% confidence interval the difference (mu l - mu2) in the means of day shift and the night shift workers.

Solution

  
Calculating the means of each group,              
              
X1 =    10.6          
X2 =    12.9          
              
Calculating the standard deviations of each group,              
              
s1 =    3.3          
s2 =    5.2          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    36          
n2 = sample size of group 2 =    64          

Thus, df = n1 + n2 - 2 =    98          

Also, sD =    0.851469318          
              
              
For the   0.98   confidence level, then      
              
alpha/2 = (1 - confidence level)/2 =    0.01          
z(alpha/2) =    2.326347874          
              
lower bound = [X1 - X2] - z(alpha/2) * sD =    -4.280813838          
upper bound = [X1 - X2] + z(alpha/2) * sD =    -0.319186162          
              
Thus, the confidence interval is              
              
(   -4.280813838   ,   -0.319186162   ) [ANSWER]