Question

2. An experiment about a person’s ability to perform some task before and after taking one of the two drugs was conducted. The subjects performed the task that involved mental addition. The subjects were randomly divided into two groups. Each group drank a beverage containing one of two drugs, labeled A, B (placebo). After a period of time for the drugs to take effect, each subject repeated the mental addition test. We want to relate the after test score to the before test score and the drug that was taken. The data from the experiment are presented in Ex4a.xls.

b. Suppose there are four drugs labeled A, B (placebo), C, D and the subjects were randomly divided into four groups. We want to relate the after test score to the before test score and the drug that was taken without insignificant interaction. The data from the experiment are presented in Ex4b.xls. Which drug/drugs have an effect on the after mental task score that is different from the placebo at significance level .05?

Ex4B.xls

BEFORE	DRUG	AFTER
24	A	24
28	A	30
38	A	39
42	A	41
24	A	27
39	A	46
45	A	56
19	A	25
19	A	18
22	A	25
34	A	31
52	A	52
27	A	38
42	A	45
28	B	28
43	B	41
37	B	37
30	B	33
49	B	39
37	B	38
40	B	41
36	B	38
41	B	36
23	B	18
33	B	32
39	B	33
36	B	35
18	B	19
17	C	18
20	C	17
49	C	41
29	C	25
27	C	35
27	C	31
44	C	55
38	C	43
32	C	44
32	C	28
24	C	33
13	C	13
39	C	39
52	C	58
19	D	24
34	D	28
22	D	21
28	D	28
27	D	28
44	D	40
39	D	34
27	D	27
59	D	47
39	D	39
29	D	26
55	D	46
49	D	42
36	D	30

Answer 1

b)

We want to determine which drug resulted in improved. We will be doing pairwise t test for each Drug Separately

i)

Drug A

BEFORE	AFTER	Diff
24	24	0
28	30	-2
38	39	-1
42	41	1
24	27	-3
39	46	-7
45	56	-11
19	25	-6
19	18	1
22	25	-3
34	31	3
52	52	0
27	38	-11
42	45	-3
	Mean	-3
	Std Dev	4.33

Alpha = 0.05

Null and Alternate Hypothesis

H₀: µ_D = 0

H_a: µ_D < 0 (ie After Taking Drug, Scores Improved)

Test Statistic

t = -3 / (4.33/14^1/2) = -2.59

p-value = TDIST(2.59,14-1,1) = 0.011193

Result

Since the p-value is less than 0.05, we reject the null hypothesis.

Conclusion

Drug A resulted in better results.

ii)

Drug B

BEFORE	AFTER	Diff
28	28	0
43	41	2
37	37	0
30	33	-3
49	39	10
37	38	-1
40	41	-1
36	38	-2
41	36	5
23	18	5
33	32	1
39	33	6
36	35	1
18	19	-1
	Mean	1.571429
	Std Dev	3.65

Alpha = 0.05

Null and Alternate Hypothesis

H₀: µ_D = 0

H_a: µ_D < 0 (ie After Taking Drug, Scores Improved)

Test Statistic

t = 1.57 / (3.65/14^1/2) = 1.61

p-value = TDIST(1.61,14-1,1) = 0.066

Result

Since the p-value is greater than 0.05, we fail to reject the null hypothesis.

Conclusion

Drug B (Placebo) did not resulted in better results.

iii)

Drug C

BEFORE	AFTER	Diff
17	18	-1
20	17	3
49	41	8
29	25	4
27	35	-8
27	31	-4
44	55	-11
38	43	-5
32	44	-12
32	28	4
24	33	-9
13	13	0
39	39	0
52	58	-6
	Mean	-2.64286
	Std Dev	6.17

Alpha = 0.05

Null and Alternate Hypothesis

H₀: µ_D = 0

H_a: µ_D < 0 (ie After Taking Drug, Scores Improved)

Test Statistic

t = -2.64 / (6.17/14^1/2) = -1.60

p-value = TDIST(1.60,14-1,1) = 0.067

Result

Since the p-value is greater than 0.05, we fail to reject the null hypothesis.

Conclusion

Drug C did not result in better results.

iv)

Drug D

BEFORE	AFTER	Diff
19	24	-5
34	28	6
22	21	1
28	28	0
27	28	-1
44	40	4
39	34	5
27	27	0
59	47	12
39	39	0
29	26	3
55	46	9
49	42	7
36	30	6
	Mean	3.357143
	Std Dev	4.52

Alpha = 0.05

Null and Alternate Hypothesis

H₀: µ_D = 0

H_a: µ_D < 0 (ie After Taking Drug, Scores Improved)

Test Statistic

t = 3.357 / (4.52/14^1/2) = 2.78

p-value = TDIST(2.78,14-1,1) = 0.008

Result

Since the p-value is less than 0.05, we reject the null hypothesis.

Conclusion

Drug D resulted in better results.

Hence, only Drug A and D resulted in better results after taking the drug