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.

IQ	Job Performance
100	17
115	34
108	24
98	18
120	42
147	51
132	43
85	55
105	29
110	38

(a) Convert the following data to ranks and then compute a Spearman correlation 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 retain or reject the null hypothesis.

Retain the null hypothesis.Reject the null hypothesis.   

Answer 1

Answer Date: 27/03/2018 a) The Spearman correlation coefficient is The table for ranking groupl and group 2 ascending order smallest to largest and equal two ranks are tied is, Rank IQ _Rank Job Performance |Difference |Square 4 4 0 0 10 10 4 0 0 0 81 0 0 96 Total

6x96 10x(102-1) = 0.418 The Spearman correlation coefficient is 0.418 b) To test the hypothesis is that IQ score is related to job performance at 5% level of significance The null and alternative hypothesis is, The null hypothesis is that IQ score is not related to job performance The alternative hypothesis is that IQ score is related to job performance. The t-test statistics is,

0.418 x V10-2 v1-(0.418)-2 The t-test statistics is 1.30 The t-critical value is, The sample size is small and two-tailed test. Look in the column headed α = 0.05 and the row headed in the t distribution table by using degree of freedom is =10-2 The t-critical value is 2.306 Decision

The statistical decision is that the t-test statistics which is 1.30 is less than t-critical value which is 2.306, so the null hypothesis is not rejected at 5% level of significance. There is insufficient evidence to indicate that IQ score is related to job performance. The result is not statistically significant