The Bureau of Labor Statistics shows that the average insurance cost to a compan
ID: 3182432 • Letter: T
Question
The Bureau of Labor Statistics shows that the average insurance cost to a company per employee per hour is $1.84 for managers and $1.99 for professional specialty workers. Suppose these figures were obtained from 14 managers and 15 professional specialty workers and that their respective population standard deviation are $0.38 and $0.51. Assume that such insurance costs are normally distributed in the population.
a. Calculate a 98% 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?
b. Test to determine whether there is a significant difference in the hourly rates employers pay for insurance between managers and professional specialty workers. Use =0.02.
Explanation / Answer
(a)
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 at df =(n1+n2-2) = 27 is 2.473
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) ± (2.473 * 0.4521 * Sqrt( 1/14 + 1/15)) ]
= [-0.5654, 0.2654]
(b)
The test hypothesis is
Ho:1=2
Ha:1 not equal to 2
The test statistic is
Z=(xbar1-xbar2)/[s1^2/n1+s2^2/n2]
=(1.84-1.99)/sqrt(0.38^2/14+0.51^2/15)
= -0.902
Given a=0.02, the crtical value is Z(0.01)=-2.33 (check standard normal table)
Since -0.902 is greater than -2.33, we fail to reject Ho.
So we can conclude that there is no significant difference.
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