The human resources department of a consulting firm gives a standard creativity
ID: 3275039 • Letter: T
Question
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, 55 new hires look the test and scored a mean of 113.1 points with a standard deviation of 17.1 Last year, 100 new hires took the test and scored a mean of 116.9 points with a standard deviation of 17.5. Assume that the population standard deviations of 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 90% confluence interval for mu_1 - mu_2, the difference between the mean test score mu_2 of new hires from the current year and the mean test score mu_2 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 (it necessary, consult a list of formulas). What is the lower limit of the 90% confidence interval? What is the upper limit of the 90% confidence interval?Explanation / Answer
x1(bar) 113.10 x2(bar) 116.90 s1 17.10 s2 17.50 n1 55 n2 100 SE = sqrt[ (s12/n1) + (s22/n2) ] (s12/n1) 5.3165 (s22/n2) 3.0625 SE 2.8947 DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] } [ (s12 / n1)2 / (n1 - 1) ] 0.523 [ (s22 / n2)2 / (n2 - 1) ] 0.09 (s12/n1 + s22/n2)2 70.21 DF = 114 x1bar - x2bar -3.80 SE 2.8947 CI 90% DF 114 t-value 1.6583 ME = t*SE 4.8003 Confidence Interval (x1bar - x2bar) +/- ME Lower bound -8.60030 Upper bound 1.00030 Confidence Interval (-8.6003 , 1.0003 )
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