When performing a rank correlation test, one alternative to using the Critical Values of Spearman's Rank Correlation Coefficient table to find critical values is to compute them using this approximation: $$r _ { s } = \pm \sqrt { \frac { t ^ { 2 } } { t ^ { 2 } + n - 2 } }$$
where $t$ is the $t$-score from the $t$ Distribution table corresponding to $n - 2$ degrees of freedom. Use this approximation to find critical values of $\mathrm { r } _ { \mathrm { S } }$ for the case where $\mathrm { n } = 40$ and $\alpha = 0.10$.
A) $\pm 0.202$
B) $\pm 0.264$
C) $\pm 0.312$
D) $\pm 0.304$

Question

Quizplus · Accepted Answer

Using the given formula, we can find the critical values of $\mathrm { r } _ { \mathrm { S } }$:

$$t = t_{0.10, 38} = 1.684$$

$$r _ { s } = \pm \sqrt { \frac { t ^ { 2 } } { t ^ { 2 } + n - 2 } } = \pm \sqrt {\frac { (1.684) ^ { 2 } } { (1.684) ^ { 2 } + 40 - 2 } } = \pm 0.264$$

Therefore, the answer is (B).

When Performing a Rank Correlation Test, One Alternative to Using rs=±t2t2+n−2r _ { s } = \pm \sqrt { \frac { t ^ { 2 } } { t ^ { 2 } + n - 2 } }rs​=±t2+n−2t2​​

When Performing a Rank Correlation Test, One Alternative to Using $r _ { s } = \pm \sqrt { \frac { t ^ { 2 } } { t ^ { 2 } + n - 2 } }$