The Bureau of Labor Statistics shows that the average insurance cost a company p

Question

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?

Explanation / Answer

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]  

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