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a.
Obtain the degree of association between the citizens ranked with sports and citizens ranked with world
a.
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Answer to Problem 38CE
The degree of association between the citizens ranked with sports and citizens ranked with world events is 0.486.
Explanation of Solution
rs=1−6∑d2n(n2−1)
Here, 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) Citizen |
(2) Sports |
(3) Rank 1 |
(4) World Events |
(5) Rank 2 |
(6) Difference ={(3)−(5)} |
(7) d2 |
1 | 47 | 5 | 49 | 4 | 1(=5−4) | 1 |
2 | 12 | 1 | 10 | 1 | 0(=1−1) | 0 |
3 | 62 | 10 | 76 | 11 | −1(=10−11) | 1 |
4 | 81 | 11 | 92 | 14 | −3(=11−14) | 9 |
5 | 90 | 14 | 86 | 12.5 | 1.5(=14−12.5) | 2.25 |
6 | 35 | 3 | 42 | 3 | 0(=3−3) | 0 |
7 | 61 | 9 | 61 | 7 | 2(=9−7) | 4 |
8 | 87 | 12.5 | 75 | 9.5 | 3(=12.5−9.5) | 9 |
9 | 59 | 7 | 86 | 12.5 | −5.5(=7−12.5) | 30.25 |
10 | 40 | 4 | 61 | 7 | −3(=4−7) | 9 |
11 | 87 | 12.5 | 18 | 2 | 10.5(=12.5−2) | 110.25 |
12 | 16 | 2 | 75 | 9.5 | −7.5(=2−9.5) | 56.25 |
13 | 50 | 6 | 51 | 5 | 1(=6−5) | 1 |
14 | 60 | 8 | 61 | 7 | 1(=8−7) | 1 |
∑d2=234 |
In this context, the number of paired observation, n is 14.
The Spearman’s coefficient of rank correlation obtained is given below:
Substitute the corresponding values to get the rank correlation.
rs=1−6(234)14(142−1)=1−1,4042,730=1−0.5143=0.486
Thus, the Spearman’s coefficient of rank correlation is 0.486.
b.
State whether the rank
b.
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Answer to Problem 38CE
The rank correlation between the sports and world events “knowledge” scores is greater than zero.
Explanation of Solution
The test hypothesis is given as follows:
Null hypothesis:
H0: The rank correlation in the population is zero.
Alternative hypothesis:
H1: The rank correlation between the sports and world events “knowledge” scores is greater than zero.
In this context, the number of paired observations is 14.
If the
Hypothesis test for rank correlation:
t=rs√n−21−r2s
Degrees of freedom:
n−2=14−2=12
Decision rule:
- If t>t0.05, reject the null hypothesis.
- Otherwise fail to reject the null hypothesis.
In this context, the critical value t0.05(tα) for right-tailed test is obtained as 1.782 using the EXCEL formula, “=T.INV (0.95,12)”.
From Part (a), the rank correlation, rs is 0.486.
The test statistic will be obtained as follows:
Substitute rs as 0.486, n as 14.
t=0.486√14−21−(0.486)2=0.486√120.764=0.486(3.964)=1.926
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 support the claim that the rank correlation between the sports and world events “knowledge” scores is greater than zero.
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Chapter 16 Solutions
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