Question

Blanking Out on Tests In a study of students blanking out on tests, the arrangement of test items was studied for its effect on anxiety. The following scores are measures of “debilitating test anxiety” (based on data from “Item Arrangement, Cognitive Entry Characteristics, Sex and Test Anxiety as Predictors of Achievement in Examination Performance,” by Klimko, Journal of Experimental Education, Vol. 52, No. 4.) Is there sufficient evidence to support the claim that the two samples are from populations with different medians? Is there sufficient evidence to support the claim that the arrangement of the test items has an effect on the score? Use a 0.01 significance level.

Questions Arranged From Easy To Difficult
24.64	39.29	16.32	32.83	28.02
33.31	20.6	21.13	26.69	28.9
26.43	24.23	7.1	32.86	21.06
28.89	28.71	31.73	30.02	21.96
25.49	38.81	27.85	30.29	30.72

Questions Arranged From Difficult To Easy
33.62	34.02	26.63	30.26
35.91	26.68	29.49	35.32
27.24	32.34	29.34	33.53
27.62	42.91	30.2	32.54

Short answer

There is not enough evidence to conclude thatthere is a difference in the median values of the populations from which the two samples are taken.

Since the test result is insignificant, there is not enough evidence to conclude that there is an effect of the arrangement of the test items on anxiety scores.

Step-by-step solution

Given information

Two samples are given to measure the anxiety scores of students.

It is claimed that the two samples are from populations with different medians, and hence the arrangement of the test items does not affect the anxiety score.

Frame the statistical hypotheses

The Wilcoxon rank-sum test is performed to test the difference in the medians of the given two samples.

The null hypothesis is as follows:

There is no difference in the median values of thepopulations from which the two samples are taken.

The alternative hypothesis is as follows:

There is a difference in the median values of the populations from which the two samples are taken.

It is a two-tailed test.

Assigning ranks  

• Combine the two samples and label each observation with the sample name/number it comes from.
• The smallest observation is assigned rank 1; the next smallest observation is assigned rank 2, and so on until the largest value.
• If two observations have the same value, the mean of the ranks is assigned to them.

The following table shows the ranks:

Anxiety scores	Sample number	Ranks
24.64	Sample 1	8
39.29	Sample 1	40
16.32	Sample 1	2
32.83	Sample 1	31
28.02	Sample 1	17
33.31	Sample 1	33
20.6	Sample 1	3
21.13	Sample 1	5
26.69	Sample 1	13
28.9	Sample 1	20
26.43	Sample 1	10
24.23	Sample 1	7
7.1	Sample 1	1
32.86	Sample 1	32
21.06	Sample 1	4
28.89	Sample 1	19
28.71	Sample 1	18
31.73	Sample 1	28
30.02	Sample 1	23
21.96	Sample 1	6
25.49	Sample 1	9
38.81	Sample 1	39
27.85	Sample 1	16
30.29	Sample 1	26
30.72	Sample 1	27
33.62	Sample 2	35
34.02	Sample 2	36
26.63	Sample 2	11
30.26	Sample 2	25
35.91	Sample 2	38
26.68	Sample 2	12
29.49	Sample 2	22
35.32	Sample 2	37
27.24	Sample 2	14
32.34	Sample 2	29
29.34	Sample 2	21
33.53	Sample 2	34
27.62	Sample 2	15
42.91	Sample 2	41
30.2	Sample 2	24
32.54	Sample 2	30

The sum of the ranks

Compute the sum of the ranks for sample 1 as follows:

\(\begin{array}{l}R = 8 + 40 + 2 + .... + 27 = 437\\\end{array}\)

Thus, the sum of the ranks for sample 1 is equal to 437 and is denoted by R.

Calculate the mean and the standard deviation

Let\({n_1}\)be the sample size of sample 1.

Let\({n_2}\)be the sample size of sample 2.

Here,

\(\begin{array}{l}{n_1} = 25\\{n_2} = 16\end{array}\)

Compute the mean value\(\left( {{\mu _R}} \right)\)as follows:

\(\begin{array}{c}{\mu _R} = \frac{{{n_1}\left( {{n_1} + {n_2} + 1} \right)}}{2}\\ = \frac{{25\left( {25 + 16 + 1} \right)}}{2}\\ = 525\end{array}\)

Compute the standard deviation\(\left( {{\sigma _R}} \right)\)as follows:

\(\begin{array}{c}{\sigma _R} = \sqrt {\frac{{{n_1}{n_2}\left( {{n_1} + {n_2} + 1} \right)}}{{12}}} \\ = \sqrt {\frac{{25 \times 16\left( {25 + 16 + 1} \right)}}{{12}}} \\ = 37.42\end{array}\)

Calculate the test statistic

The test statistic is:

\(\begin{array}{c}z = \frac{{R - {\mu _R}}}{{{\sigma _R}}}\; \sim N\left( {0,1} \right)\\ = \frac{{437 - 525}}{{37.42}}\\ = - 2.35\end{array}\)

The absolute value of the z score is equal to 2.35.

The critical value of z from the standard normal table for a two-tailed test with\(\alpha = 0.01\)is equal to 2.5758.

As the absolute value of the test statistic is less than the critical value, there is a failure to reject the null hypothesis.

Conclusion

There is not enough evidence to conclude thatthere is no difference in the median values of the populations from which the two samples are taken.

Since there is a failure to reject the null hypothesis, it can be inferred that there is no difference in the anxiety score of the two types of arrangement.

Thus, it can be concluded that the arrangement of the test items does not affect the anxiety score.