Question

Do certain drugs slow down the reaction? If so, the government will try to test the differences taking this drug. Thirty random samples were examined before and after drug administration. And their reaction time (in seconds) (see table below). Verify that there is a difference at the significance level of 5%.

person	before	after
1	1.42	1.48
2	1.87	1.75
3	1.34	1.31
4	0.98	1.22
5	1.51	1.58
6	1.43	1.57
7	1.52	1.48
8	1.61	1.55
9	1.37	1.54
10	1.49	1.37
11	0.95	1.07
12	1.32	1.35
13	1.68	1.77
14	1.44	1.44
15	1.17	1.27
16	1.83	1.75
17	1.4	1.51
18	1.75	1.67
19	1.56	1.72
20	1.56	1.63
21	2.03	1.81
22	1.38	1.48
23	1.42	1.35
24	1.69	1.75
25	1.5	1.39
26	1.12	1.24
27	1.38	1.4
28	1.71	1.65
29	0.91	1.11
30	1.59	1.85

Answer 1

Hypothesis :

H0 : Mean 1 = Mean 2

Ha : Mean 1 ≠ Mean 2

Before

N₁: 30
df₁ = N - 1 = 30 - 1 = 29
M₁: 1.46
SS₁: 2
s²₁ = SS₁/(N - 1) = 2/(30-1) = 0.07


After

N₂: 30
df₂ = N - 1 = 30 - 1 = 29
M₂: 1.5
SS₂: 1.27
s²₂ = SS₂/(N - 1) = 1.27/(30-1) = 0.04

person	before	after	(X1 - M1)	(X1 - M1)^2	(X2 - M2)	(X2 - M2)^2
1	1.42	1.48	-0.04	0	-0.02	0
2	1.87	1.75	0.41	0.16	0.25	0.06
3	1.34	1.31	-0.12	0.02	-0.19	0.04
4	0.98	1.22	-0.48	0.23	-0.28	0.08
5	1.51	1.58	0.05	0	0.08	0.01
6	1.43	1.57	-0.03	0	0.07	0
7	1.52	1.48	0.06	0	-0.02	0
8	1.61	1.55	0.15	0.02	0.05	0
9	1.37	1.54	-0.09	0.01	0.04	0
10	1.49	1.37	0.03	0	-0.13	0.02
11	0.95	1.07	-0.51	0.26	-0.43	0.19
12	1.32	1.35	-0.14	0.02	-0.15	0.02
13	1.68	1.77	0.22	0.05	0.27	0.07
14	1.44	1.44	-0.02	0	-0.06	0
15	1.17	1.27	-0.29	0.09	-0.23	0.05
16	1.83	1.75	0.37	0.13	0.25	0.06
17	1.4	1.51	-0.06	0	0.01	0
18	1.75	1.67	0.29	0.08	0.17	0.03
19	1.56	1.72	0.1	0.01	0.22	0.05
20	1.56	1.63	0.1	0.01	0.13	0.02
21	2.03	1.81	0.57	0.32	0.31	0.09
22	1.38	1.48	-0.08	0.01	-0.02	0
23	1.42	1.35	-0.04	0	-0.15	0.02
24	1.69	1.75	0.23	0.05	0.25	0.06
25	1.5	1.39	0.04	0	-0.11	0.01
26	1.12	1.24	-0.34	0.12	-0.26	0.07
27	1.38	1.4	-0.08	0.01	-0.1	0.01
28	1.71	1.65	0.25	0.06	0.15	0.02
29	0.91	1.11	-0.55	0.31	-0.39	0.15
30	1.59	1.85	0.13	0.02	0.35	0.12
			M1: 1.46	SS1: 2.00	M2: 1.50	SS2: 1.27

T-value Calculation

s²_p = ((df₁/(df₁ + df₂)) * s²₁) + ((df₂/(df₂ + df₂)) * s²₂) = ((29/58) * 0.07) + ((29/58) * 0.04) = 0.06

s²_M1 = s²_p/N₁ = 0.06/30 = 0
s²_M2 = s²_p/N₂ = 0.06/30 = 0

t = (M₁ - M₂)/√(s²_M1 + s²_M2) = -0.04/√0 = -0.61

The t-value is -0.61412. The p-value is .541535. The result is not significant at p < .05.

We do not reject the null hypothesis

Therefore there is no difference before and after taking drugs at 5% significance level