Question

Khalid, a statistics lecturer at King Faisal University, wanted to find the effects of special tutoring on scores achieved by 12 students. The lecturer administered the same exame before and after the tutoring period. The scores are displayed as below.

Before: 75, 64, 58, 82, 90, 55, 72, 57, 64, 88, 63, 79, 64, 65, 64, 71, 85, 74, 76, 78

After: 78, 62, 64, 82, 92, 54, 70, 68, 75, 92, 65,75, 62, 63, 70, 72, 80, 71, 78, 80

..

Use the sign test at 1% level of significance to test the null hypothesis that on the average the effects of special tutoring on scores before and after are equal against the alternative hypothesis that the effect after the special tutoring is better than before. [6 marks]

Answer 1

The null hypothesis, H₀: No difference in median of the signed differences (After minus Before)
The alternate hypothesis, H₁: Median of the signed differences is greater than zero.

Check the data for the 20 samples given.

Write a '+' symbol if the score after the special tutoring is better than before
Write an 'O' symbol if the score after the special tutoring is worse than before
Write a '-' symbol if the score after the special tutoring is equal to before

Before	75	64	58	82	90	55	72	57	64	88	63	79	64	65	64	71	85	74	76	78
After	78	62	64	82	92	54	70	68	75	92	65	75	62	63	70	72	80	71	78	80
Sign	+	-	+	O	+	-	-	+	+	+	+	-	-	-	+	+	-	-	+	+

So,

No. of positives = 11
No. of negatives = 8
No. of 'O' = 1

Subtract the number of 'O' from the original sample size. So, the effective sample size (n) = 20 - 1 = 19

The p-value of the test = 1 - BINOM.DIST(11,19,0.5,1) = 0.180

The p-value of the test is more than the type-I error which is 0.01. So, the null hypothesis cannot be rejected. So, the effect of special tutoring is insignificant in improving the score.