Example 15

EXAMPLE 15 Performing the Wilcoxon rank sum test using technology

A study investigated whether there was a difference in physical activity levels between female adolescents with anorexia nervosa (AN) and those without AN.⁹ In this study, the amount of physical activity (in minutes for the year) of $n_{1} = 314$ randomly selected female adolescents with AN (patients) and $n_{2} = 340$ randomly selected female adolescents without AN (controls) was estimated by interviewing their mothers. The samples were drawn independently. Use Minitab and SPSS to test whether the population median minutes of physical activity for the patients differs from that for the controls, using level of significance $α = 0.01$ .

Solution

Because both the patients and the controls were randomly selected, because $n_{1} > 10$ and $n_{2} > 10$ , and because we assume that both population shapes are the same, we may proceed with the hypothesis test.

Step 1 State the hypotheses.

$\begin{matrix} H_{0} : M_{1} = M_{2} & versus & H_{a} : M_{1} \neq M_{2} \end{matrix}$

where $M_{1}$ and $M_{2}$ represent the population median activity level (in minutes) of the patients and the controls, respectively.
Step 2 Find the critical value and state the rejection rule. The level of significance is $α = 0.01$ , so our critical value is $Z_{crit} = 2.58$ . We will reject $H_{0}$ if $Z_{data} ≤ − 2.58$ or if $Z_{data} ≥ 2.58$ .
Step 3 Find the value of the test statistic. We use the instructions provided in the Step-by-Step Technology Guide at the end of this section. Figure 16 shows the Minitab results from the Mann-Whitney test, which is equivalent to the Wilcoxon rank sum test for independent samples. Figure 17 shows the SPSS results from the Mann-Whitney test.

FIGURE 16 Minitab results.

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FIGURE 17 SPSS results.

The highlighted $" W = 109336.5 "$ represents $R_{1}$ in our notation. We have

$\begin{matrix} μ_{R} = \frac{n_{1} (n_{1} + n_{2} + 1)}{2} = \frac{314 (314 + 340 + 1)}{2} = 102, 835 \\ σ_{R} = \sqrt{\frac{n_{1} n_{2} (n_{1} + n_{2} + 1)}{12}} = \sqrt{\frac{314 (340) (314 + 340 + 1)}{12}} \approx 2413.9836 \end{matrix}$

So that

$Z_{data} = \frac{R_{1} - μ_{R}}{σ_{R}} \approx \frac{109, 336.5 - 102, 835}{2413.9836} \approx 2.6933$
Step 4 State the conclusion and the interpretation. We said we will reject $H_{0}$ if $Z_{data} ≤ − 2.58$ or if $Z_{data} ≥ 2.58$ . We have $Z_{data} ≈ 2.6933$ , which is greater than 2.58. Therefore, we reject $H_{0}$ . There is evidence that the population median amount of physical activity for female adolescents with AN differs from the population median amount of physical activity for female adolescents without AN.