Chapter 16, Problem 40CE

Professor Bert Forman believes the students who complete his examinations in the shortest time receive the highest grades and those who take the longest to complete them receive the lowest grades. To verify his suspicion, he assigns a rank to the order of finish and then grades the examinations. The results are shown below:

Convert the test scores to a rank and find the coefficient of rank correlation between the order of completion and the rank of the test score. At the .05 significance level, can Professor Forman conclude there is a positive association between the order of finish and the test scores?

To determine

Convert the test scores into ranks:

Obtain the correlation between the rankings of order of completion and test score.

State whether it can be concluded that there is a positive association between the rankings of order of completion and test score.

Answer to Problem 40CE

The correlation between the rankings of the order of completion and test score is 0.788.

The conclusion is that there is evidence that there is a positive association between the rankings of order of completion and test score.

Explanation of Solution

Here, the ranking of the test is done in context with descending order.

The rank of test scores is given below:

48	48	43	49	50	47	39	30	37	35	36	33
3.5	3.5	6	2	1	5	7	12	8	10	9	11

Spearman’s coefficient of rank correlation:

$r_s=1-\frac{6{\displaystyle \sum d^2}}{n\left(n^2-1\right)}$

Where d is the difference between ranks of each pair.

n is the number of paired observations.

The table represents the difference between ranks of each pair:

(1) Student	(2) Rank by Order of completion	(3) Rank by Score	(4) Difference, $d=\left\{\left(2\right)-\left(3\right)\right\}$	(5) $d^2$
1	1	3.5	$-2.5\left(=1-3.5\right)$	6.25
2	2	3.5	$-1.5\left(=2-3.5\right)$	2.25
3	3	6	$-3\left(=3-6\right)$	9
4	4	2	$2\left(=4-2\right)$	4
5	5	1	$4\left(=5-1\right)$	16
6	6	5	$1\left(=6-5\right)$	1
7	7	7	$0\left(=7-7\right)$	0
8	8	12	$-4\left(=8-12\right)$	16
9	9	8	$1\left(=9-8\right)$	1
10	10	10	$0\left(=10-10\right)$	0
11	11	9	$2\left(=11-9\right)$	4
12	12	11	$1\left(=12-11\right)$	1
				${\displaystyle \sum d^2=}60.50$

In this context, the number of paired observation, n is 12.

The Spearman’s coefficient of rank correlation is obtained as given below:

Substitute the corresponding values to get the rank correlation.

$\begin{array}{c}{r}_s=1-\frac{6\left(60.50\right)}{12\left({12}^2-1\right)}\\ =1-\frac{363}{1,716}\\ =1-0.212\\ =0.788\end{array}$

Thus, the Spearman’s coefficient of rank correlation is 0.788.

The rank correlation value of 0.788 reveals that there is a strong positive correlation between the order of completion and test score.

The test hypotheses are given as follows:

Null hypothesis:

$H_0:$ The rank correlation in the population is zero.

Alternative hypothesis:

$H_1:$ There is a positive association between the rankings of order of completion and test score.

If the sample size is greater than 10, then the sampling distribution of $r_s$ follows the t distribution with n–2 df.

Hypothesis test for rank correlation:

$t=r_s\sqrt{\frac{n-2}{1-r_s^2}}$

Degrees of freedom:

$\begin{array}{c}n-2=12-2\\ =10\end{array}$

Decision rule:

• If $t>t_{0.05}$, reject the null hypothesis.
• Otherwise fail to reject the null hypothesis.

In this context, the critical value $t_{0.05}(t_{\alpha })$ for the right-tailed test is obtained as 1.812 using the EXCEL formula, “=T.INV (0.95,10)”.

From Part (a), the rank correlation, $r_s$ is 0.788.

The test statistic will be obtained as given below:

Substitute $r_s$ as 0.788 and n as 12.

$\begin{array}{c}t=0.788\sqrt{\frac{12-2}{1-{\left(0.788\right)}^2}}\\ =0.788\sqrt{\frac{10}{0.379}}\\ =0.788\left(5.14\right)\\ =4.05\end{array}$

Conclusion:

Here, the test statistic is greater than the critical value.

Therefore, by the decision rule, reject the null hypothesis.

Therefore, there is evidence to conclude that there is a positive association between the rankings of order of completion and test score.