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 16 115 38 108 23 98 20 120 48 147 56 132 47 85 57 105 28 110 35 (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

a)
`\tau = 1- (6\sum d^2)/(n^3 -n)=1-(6*96)/(10^3 -10)=0.418`

b)
`t =\sqrt{((10-2)0.418^2)/(1-0.418^2)}=1.301`

`P_v = 2*P(t_(8)>1.301) =0.229`

So using the significance level provided we see that
`p_v >\alpha` so we have enough evidence to FAIL to reject the null hypothesis that the Spearman Correlation coeffcint is equal to 0.

Explanation:

Dataset given

Number IQ Job performance

1 100 16

2 115 38

3 108 23

4 98 20

5 120 48

6 147 56

7 132 47

8 85 57

9 105 28

10 110 35

Previous concepts

Spearman's Rank correlation coefficient "is a value that measure the strength and direction (negative or positive) of a relationship between two variables. The result will always be between 1 and minus 1".

Solution to the problem

Part a

In order to calculate the sparman correlation coefficient we need to order the dataset like this:

Number IQ(x) Rank1 Job performance (y) Rank2 d d^2

1 85 10 57 1 9 81

2 98 9 20 9 0 0

3 100 8 16 10 -2 4

4 105 7 28 7 0 0

5 108 6 23 8 -2 4

6 110 5 35 6 -1 1

7 115 4 38 5 -1 1

8 120 3 48 3 0 0

9 132 2 47 4 -2 4

10 147 1 56 2 -1 1

The difference d is dfined as
`d= Rank_1 -Rank_2`

Then
`\sum d^2 = 96`

And now we can calculate the sparman correlation coeffcient like this:

`\tau = 1- (6\sum d^2)/(n^3 -n)=1-(6*96)/(10^3 -10)=0.418`

Part b

The system of hypothesis on this case are:

H0:
`\tau =0`

H1:
`\tau \\eq 0`

The statistic to check the hypothesis is given by:

`t =\sqrt{((n-2)\tau^2)/(1-\tau^2)}`

And replacing the value obtained we got:

`t =\sqrt{((10-2)0.418^2)/(1-0.418^2)}=1.301`

The degrees of freedom on this case are given by:

`df= n-2=10-2= 8`

And the p value since is a bilateral test is given by:

`P_v = 2*P(t_(8)>1.301) =0.229`

So using the significance level provided we see that
`p_v >\alpha` so we have enough evidence to FAIL to reject the null hypothesis that the Spearman Correlation coeffcint is equal to 0.