Question

The effect of social competence training is tested. 12 Subjects are tested before and after the training for social competence on a seven point scale. Data are ordinal scaled and not normally distributed. The question is, if social competence is enhanced by the training. Use Sign Test. Raw data are listed in the following table: Before 5 3 4 2 1 6 7 3 2 3 5 1 After 6 2 4 4 3 6 7 5 3 5 5 3

(x+0.5)-(n/2)/Squrt N/2

Ti-84

Stat edit enter both lists then stat calc 2 find X

Sub in and you get (3.5+.5)-(12/2)/ Squrt (12/2) =.2721 Got to A2 positive table or Cumulative area from the left for z - score and find z=.6064.

Am I doing this correctly and have I gone far enough please show your work thank you .

Answer 1

Before	After	Difference
5	6	1
3	2	-1
4	4	0
2	4	2
1	3	2
6	6	0
7	7	0
3	5	2
2	3	1
3	5	2
5	5	0
1	3	2
	Positives	7
	Negatives	1

There's alternative method too:

The provided number of positive signs is and the number of negative signs is

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: Median of Differences = 0

Ha: Median of Differences > 0

The test statistic is

or

To perform the test, we count number of + sign and number of - sign.
Number of plus sign = 7
Number of minus sign = 1
n = 7 + 1 = 8
μ=np=8×0.5=4

σ=√npq=√8×0.5×0.5=√2=1.4142

Applying the z-test statistic, we get
z=ˉx-μσ=7-41.4142=2.1213

(2) Rejection Region

The critical value for the signficance level provided and the type of tail is

or

The z-value is 2.12132.

At α=0.05, the critical value of Z0.05=1.6449

As calculated z=2.1213>1.6449

So, H0 is rejected.

The p-value is .01695. The result is significant at p < 0.05.

(3) Decision about the null hypothesis

Since in this case X=1≤X∗=1, there is enough evidence to claim that the population median of differences is greater than 0, at the 0.05 significance level.

Please rate my answer and comment for doubt. I hope it helped.