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Applied Statistics Using SPSS, STATISTICA, MATLAB and R

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214 5 Non-Parametric Tests of Hypotheses<br />

Example 5.22<br />

Q: Consider the Freshmen dataset <strong>and</strong> use the Kruskal-Wallis test in order to<br />

assess whether the freshmen performance (EXAMAVG) differs according to their<br />

attitude towards skipping the Initiation (Question 8).<br />

A: The mean ranks <strong>and</strong> results of the test are shown in Table 5.26. Based on the<br />

observed asymptotic significance, we reject the null hypothesis at a 5% level, i.e.,<br />

we have evidence that the freshmen answer Question 8 of the enquiry differently,<br />

depending on their average performance on the examinations.<br />

Table 5.26. Results, obtained with <strong>SPSS</strong>, for the Kruskal-Wallis test of average<br />

freshmen performance in 5 categories of answers to Question 8: a) ranks; b)<br />

significance.<br />

a<br />

Q8 N Mean Rank<br />

1 10 104.45<br />

2 22 75.16<br />

3 48 60.08<br />

4 39 59.04<br />

5 12 63.46<br />

Total 131<br />

EXAMAVG<br />

Chi-Square 14.081<br />

df 4<br />

Asymp. Sig. 0.007<br />

b<br />

Example 5.23<br />

Q: The variable ART of the Cork Stoppers’ dataset was analysed in section<br />

4.5.2.1 using the one-way ANOVA test. Perform the same analysis using the<br />

Kruskal-Wallis test <strong>and</strong> estimate its power for the alternative hypothesis<br />

corresponding to the sample means.<br />

A: We saw in 4.5.2.1 that a logarithmic transformation of ART was needed in<br />

order to be able to apply the ANOVA test. This transformation is not needed with<br />

the Kruskal-Wallist test, whose only assumption is the independency of the<br />

samples.<br />

Table 5.27 shows the results, from which we conclude that the null hypothesis<br />

of median equality of the three populations is rejected at a 5% significance level<br />

(or even at a smaller level).<br />

In order to estimate the power of this Kruskal-Wallis test, we notice that the<br />

sample size is large, <strong>and</strong> therefore, we expect the power to be the same as for the<br />

one-way ANOVA test using a number of cases equal to n = 50×0.955 ≈ 48. The<br />

power of the one-way ANOVA, for the alternative hypothesis corresponding to the<br />

sample means <strong>and</strong> with n = 48, is 1.

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