The Bureau of Labor Statistics shows that the average insura

The Bureau of Labor Statistics shows that the average insurance cost a company pays per employee per hour is $1.84 for managers with a standard deviation of $0.38 from 14 managers and $1.99 for 15 professional specialty workers with a standard deviation of $0.51. Suppose these populations are normally distributed with equal variances, calculate a 90% confidence interval to estimate the difference in the mean hourly company expenditures for insurance between these two groups of employees. Calculate a 90% confidence interval to estimate the difference in the mean hourly company expenditures for insurance for these two groups. What is the value of the point estimate?

Solution

C.I. WHEN SD ARE EQUAL              
              
CI = (x1 - x2) ± t a/2 * S^2 * Sqrt ( 1 / n1 + 1 / n2 )              
Where,               
x1 = Mean of Sample 1, x2 = Mean of sample2              
sd1 = SD of Sample 1, sd2 = SD of sample2              
a = 1 - (Confidence Level/100)              
ta/2 = t-table value              
CI = Confidence Interval               
Value Pooled variance S^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
S^2 = (13*0.1444 + 14*0.2601) / (29- 2 )
S^2 = 0.2044  
S = 0.4521      
t a/2 =(n1+n2-2) i.e 27 d.f is 2.052  
CI = [ ( 1.84-1.99) ± t a/2 * S * Sqrt( 1/14 + 1/15)]              
= [ (-0.15) ± t a/2 * 0.4521 * Sqrt( 1/14 + 1/15) ]              
= [ (-0.15) ± (2.052 * 0.4521 * Sqrt( 1/14 + 1/15)) ]              
= [-0.4947, 0.1947]