Question

a random sample of 10 college students are given an anxiety test before and after some intervention. Can we could that the intervention was effective to decrease the anxiety level of the students ? The > score means the > anxiety. 02.05 student Before 54 After 50 32 36 2 a + 5. 6 7 70 42 48 50 44 80 45 75 45 43 70 39 30 8 9 10 82 #A: If we measure the anxiety before and after intervention for male and female students what is the non parametric statistical test to analyze the differences? a. Wilcoxon sign ranked b. Spearman 6s Knustál-Wallis #16What parameter test would be used on a larger sample? a one sample test b. two proporhon test c. Two sample t-test

Answer 1

Let us denote the differeence between before and after for student i as

di = Beforej – After

For calculating the mean and variance of the differences, we need teh following table:

Student	Before	After	Diff	Diff^2
1	54	50	4	16
2	35	32	3	9
3	70	63	7	49
4	42	45	-3	9
5	73	75	-2	4
6	50	45	5	25
7	44	43	1	1
8	80	70	10	100
9	33	39	-6	36
10	82	70	12	144
		Total	31	393

The mean of the differences:  

d, d

  

31 d = 3.1 10

The variance of the difference :

1 2 * =Σ - (Σ)

1 2 2 Sa 393 1 10 (31) 10 – 1

$s_{d}^{2}=\frac{1}{9}\left [393-\frac{961}{10}\right ]$

2 d [393 - 96.1

$s_{d}^{2}=\frac{1}{9}\left [296.9\right ]$

  $s_{d}^{2}=32.9889$

$s_{d}=5.7436$

Here the claim is that the intervention reduces anxiety.

Null hypothesis:

$H_{0}:\bar{\delta }=0$

$H_{0}:\bar{\delta }>0$

Level of significance: $\alpha =0.05$

Test Statistic: We shall be using a paired t-test for the test:

$t=\frac{\bar{d}-0}{S_{d}/\sqrt{n}}$

  $t=\frac{3.1}{5.7436/\sqrt{10}}$

  $t=\frac{3.1}{5.7436/3.1623}$

$t=\frac{9.8031}{5.7436}$

$t=1.7068$. The critical value of t is 1.8331.

The p-value is 0.0610

Since the p-value >0.05, we fail to reject the null hypothesis. Hence we conclude that there are no sufficient evidence to claim that the intervention reduces anxiety.