A psychologist investigates the hypothesis that birth order affects assertivenes

Question

A psychologist investigates the hypothesis that birth order affects assertiveness. Her subjects are 20 young adults between 20 and 25 years of age. There are seven first-born, six second-born, and seven third-born subjects. Each subject is given an assertiveness test, with the following results. High scores indicate greater assertiveness. Assume the data are so far from normally distributed that the F test can't be used, but the data are at least of ordinal scaling. Use = 0.01 to evaluate the data. What is your conclusion?

Condition 1
First-Born Condition 2
Second-Born Condition 3
Third-Born 16 19 5 10 13 20 2 2 1 19 25 30 27 22 20 32 2 7 8 12

Explanation / Answer

State the hypotheses:

H0:mu1=mu2=mu3 (mean assertive scores are same for three conditions)

H1:not all the means are equal.

From information given, the samples are drawn randomly from population, that is three conditions are independent, the level of measurement is ordinal, and the distribution of scores for three conditions are far away from normal. Therefore, ANOVA test is inappropriate but Kruskal-Wallis test is suitable one.

To obtain the test statistic,H do as follows:

Condition1   Rank1   Condition2   Rank2   Condition3   Rank3
16 11 19 12.5 5 5.0
10 8 13 10 20 14.5
2 3 2 3 1 1.0
19 12.5 25 17.0 30 19.0
27 18 22 16 20 14.5
32 20 2 3.0 7 6.0
8 7 12 9.0

The sum of ranks for three conditions are as follows:

R1=79.50, R2=61.50, and R3=69

Now substitute the obtained values in the following formula.

H=12/n(n+1)sigma j=1k R^2j/nj-3(n+1), where, n=(7+6+7)=20

=12/20(20+1)(79.50^2/7+61.50^2/6+69^2/7)-3(20+1)

=0.24

The critical H at 2 df (df=k-1=3-1) and alpha=0.01 is 9.21.

Conclusion: The observed H is less than critical H, therefore, fail to reject H0. There is insufficient sample evidence to conclude that not all the means are equal.

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