Question

Finding Critical Values An alternative to using Table A–9 to find critical values for rank correlation is to compute them using this approximation:

\({r_s} = \pm \sqrt {\frac{{{t^2}}}{{{t^2} + n - 2}}} \)

Here, t is the critical t value from Table A-3 corresponding to the desired significance level and n - 2 degrees of freedom. Use this approximation to find critical values of\({r_s}\)for Exercise 15 “Blood Pressure.” How do the resulting critical values compare to the critical values that would be found by using Formula 13-1 on page 633?

Short answer

The critical values of\({r_s}\)using the given formula are equal to -0.159 and 0.159.

The critical values of\({r_s}\)using Formula 13-1 are equal to -0.159 and 0.159.

The two critical values are the same.

Step-by-step solution

Given information

Data are given onthe blood pressures of males.

Critical values using the given formula

Consider\(\alpha = 0.05\).

The number of observations (n) = 153.

The degrees of freedom are computed as follows:

\(\begin{array}{c}df = n - 1\\ = 153 - 1\\ = 152\end{array}\)

The value oft corresponding to the degrees of freedom equal to 152 and\(\alpha = 0.05\)is 1.975.

Thus, the critical values using the formula are obtained as:

\(\begin{array}{c}{r_s} = \pm \sqrt {\frac{{{t^2}}}{{{t^2} + n - 2}}} \\ = \pm \sqrt {\frac{{{{1.975}^2}}}{{{{1.975}^2} + 153 - 2}}} \\ = \left( { - 0.159,0.159} \right)\end{array}\)

Thus, the critical values using the given formula are -0.159 and 0.159.

Critical values using the original formula

The value of z corresponding to\(\alpha = 0.05\)for a two-tailed test is equal to 1.96.

The critical values of\({r_s}\)using the following formula are:

\(\begin{array}{c}{r_s} = \pm \frac{z}{{\sqrt {n - 1} }}\\ = \pm \frac{{1.96}}{{\sqrt {153 - 1} }}\\ = \left( { - 0.159,0.159} \right)\end{array}\)

Thus, the critical values using Formula 13-1 are -0.159 and 0.159.

Comparison

The critical values calculated using the two formulas are the same.