Employers often use standardized measures to gauge how likely it is that a new employee with little to no experience will suceed in thier company. One such factor is intelligence, measured using the intelligence quotient (IQ). To show that this factor is related to job success, an organizational pychologist measures the IQ score and job performance (i units sold per day) in a sample of 10 new employees
IQ Job performance
100 17
115 35
108 24
98 16
120 42
147 55
132 46
85 54
105 25
110 36
(a) convert the following data to ranks and then compute a spearman correlaion coefficient (round your answer to three decimal places)
(b) using a two-tailed test at a 0.05 level of significance, state the decision to reject or retain the null hypothesis
IQ | Job performance | x-y= | |||
x | y | x rank | y rank | d | d2 |
100 | 17 | 8 | 9 | -1 | 1 |
115 | 35 | 4 | 6 | -2 | 4 |
108 | 24 | 6 | 8 | -2 | 4 |
98 | 16 | 9 | 10 | -1 | 1 |
120 | 42 | 3 | 4 | -1 | 1 |
147 | 55 | 1 | 1 | 0 | 0 |
132 | 46 | 2 | 3 | -1 | 1 |
85 | 54 | 10 | 2 | 8 | 64 |
105 | 25 | 7 | 7 | 0 | 0 |
110 | 36 | 5 | 5 | 0 | 0 |
∑d2 | 76 |
Spearman correlation coefficient = | rs=1-6Σd2/(n(n2-1))= | 0.539 |
b)
for 0.05 level with two tailed test & n=10 , critical value = | 0.6480 |
since test statistic is not higher than critical value , retain the null hypothesis
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