SOLUTION: The human resources department of a consulting firm gives a standard creativity test to a randomly selected group of new hires every year. This year, 65 new hires took the test a

Question 1136460: The human resources department of a consulting firm gives a standard creativity test to a randomly selected group of
new hires every year. This year, 65 new hires took the test and scored a mean of 112.8 points with a standard deviation of 15.6. Last year,
90 new hires took the test and scored a mean of 117.2 points with a standard deviation of. Assume that the population standard deviations of 17.2 the test scores of all new hires in the current year
and the test scores of all new hires last year can be estimated by the sample standard deviations, as the
samples used were quite large. Construct a 95% confidence interval for u1-u2, the difference between the mean test
score u1 of new hires from the current year and the mean test score u2 of new hires from last year. Then
complete the table below.
Carry your intermediate computations to at least three decimal places.
Round your answers to at least two decimal places. (If necessary, consult a list of formulas.)
What is the lower limit of the 95% confidence interval?
What is the upper limit of the 95% confidence interval?

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
There needs to be clarification of
1. What is the sd of the 90 hires?
2. This should be a 2-sample t-test given that s is being used to estimate sigma. Where does the "population standard deviation" come from? Is s of the prior year 17.2?
half-interval is t*(sp*sqrt((1/n1)+(1/n2)))
If I can assume the sd of the prior year for 17.2, and that was a sample,
then sp^2=(n1-1)s1^2+(n2-2)s2^2/(n1+n2-2)
sp^2=64*15.6^2+89*17.2^2/(153)
sp^2=273.888; sp=16.550
t df=153,0.975=1.976
difference in means is -4.40
denominator of (sqrt ((1/n1)+(1/n2)))=0.02650; sp*sqrt((1/n1)+(1/n2))=2.694.
Multiply that by t of 1.976 to get 5.32
interval is (-9.72, 0.82)
(0 is in the interval so it would not be significant)

RELATED QUESTIONS
The human resources division in a large government department has shown that illness is... (answered by Boreal)

The human resources department of a manufacturing firm found the average number of... (answered by ewatrrr)

In 2009, a consulting firm surveyed a random sample... (answered by richwmiller)

My problem is finding the proper way to find probabilities; Of 104 individuals,what is... (answered by oscargut)

A human resource consulting firm designed an experiment to test the effect of the... (answered by stanbon)

The most common human blood group is O which is estimated to be 45% in US population. If... (answered by Boreal)

a human resources director for a large corporation claims the probability a randomly... (answered by Theo)

a human resources director for a large corporation claims the probability a randomly... (answered by Boreal)

a human resources director for a large corporation claims the probability a randomly... (answered by ikleyn)