Question

Employers often use standardized measures to gauge how likely it is that a new employee with little to no experience will succeed in their company. One such factor is intelligence, measured using the Intelligence Quotient (IQ). To show that this factor is related to job success, an organizational psychologist measures the IQ score and job performance (in units sold per day) in a sample of 10 new employees Job Performance IQ 100 17 115 38 108 33 98 18 120 46 147 51 132 47 85 53 105 24 110 35 (a) Convert the following data to ranks and then compute a Spearman correlation coefficient. (Round your answer to three decimal places.) 430 X (b) Using a two-tailed test at a 0.05 level of significance, state the decision to retain or reject the null hypothesis. O Retain the null hypothesis. OReject the null hypothesis.

Answer 1

YRa YRa-My XRa XRa- Mx Sum Diffs X Values Y Values 100 11.25 17 3.00 -2.50 1.00 -4.50 0.75 115 38 7.00 1.50 6.00 0.50 0.75 108 33 -0.50 5.00 4.00 -1.50 12.25 98 18 2.00 -3.50 2.00 -3.50 120 46 3.75 7.00 8.00 2.50 1.50 15.75 147 51 10.00 4.50 3.50 9.00 132 47 53 8.75 3.50 9.00 8.00 2.50 -20.25 85 1.00 -4.50 10.00 4.50 3.75 24 105 -2.50 4.00 -1.50 3.00 110 -0.25 35 6.00 0.50 5.00 -0.50

Calculation R CoVariance / (XRa St. Dev. * YRa St. Dev.) Кey XRa Ranks of x Values; YRa = Ranks of Y Values XRa - Mx X rank minus mean of X ranks YRa My Y rank minus mean of Y ranks Sum Diffs (XRa - Mx) * (YRa - My) -

X Ranks Mean: 5.5 Standard Dev: 3.03 Y Ranks Mean: 5.5 Standard Dev: 3.03 Combined Covariance 36.5 /9 4.06 R 4.06/(3.03* 3.03) 0.442