Question

1. Insomnia has become an epidemic in the United States. Much research has been done in the development of new pharmaceuticals to aide those who suffer from insomnia. Alternatives to the pharmaceuticals are being sought by sufferers. A new relaxation technique has been tested to see if it is effective in treating the disorder. Sixty insomnia sufferers between the ages of 18 to 40 with no underlying health conditions volunteered to participate in a clinical trial. They were randomly assigned to either receive the relaxation treatment or a proven pharmaceutical treatment. Thirty were assigned to each group. The amount of time it took each of them to fall asleep was measured and recorded. The data is shown below. Run an independent samples t-test to determine if the relaxation treatment is more effective than the pharmaceutical treatment at a level of significance of 0.05. Report the test statistic using correct APA formatting and interpret the results.

Relaxation	Pharmaceutical
98	20
117	35
51	130
28	83
65	157
107	138
88	49
90	142
105	157
73	39
44	46
53	194
20	94
50	95
92	161
112	154
71	75
96	57
86	34
92	118
75	41
41	145
102	148
24	117
96	177
108	119
102	186
35	22
46	61
74	75

Answer 1

T-test for two Means – Unknown Population Standard Deviations

Relaxation(x1)	(x1-x̅1)	(x1-x̅1)2	Pharmaceutical (x2)	x2-x̅2	(x2-x̅2)2
98	23.3	542.89	20	-82.3	6773.29
117	42.3	1789.29	35	-67.3	4529.29
51	-23.7	561.69	130	27.7	767.29
28	-46.7	2180.89	83	-19.3	372.49
65	-9.7	94.09	157	54.7	2992.09
107	32.3	1043.29	138	35.7	1274.49
88	13.3	176.89	49	-53.3	2840.89
90	15.3	234.09	142	39.7	1576.09
105	30.3	918.09	157	54.7	2992.09
73	-1.7	2.89	39	-63.3	4006.89
44	-30.7	942.49	46	-56.3	3169.69
53	-21.7	470.89	194	91.7	8408.89
20	-54.7	2992.09	94	-8.3	68.89
50	-24.7	610.09	95	-7.3	53.29
92	17.3	299.29	161	58.7	3445.69
112	37.3	1391.29	154	51.7	2672.89
71	-3.7	13.69	75	-27.3	745.29
96	21.3	453.69	57	-45.3	2052.09
86	11.3	127.69	34	-68.3	4664.89
92	17.3	299.29	118	15.7	246.49
75	0.3	0.09	41	-61.3	3757.69
41	-33.7	1135.69	145	42.7	1823.29
102	27.3	745.29	148	45.7	2088.49
24	-50.7	2570.49	117	14.7	216.09
96	21.3	453.69	177	74.7	5580.09
108	33.3	1108.89	119	16.7	278.89
102	27.3	745.29	186	83.7	7005.69
35	-39.7	1576.09	22	-80.3	6448.09
46	-28.7	823.69	61	-41.3	1705.69
74	-0.7	0.49	75	-27.3	745.29
Total 2241	Total 0	Total 24304.3	Total 3069	Total 0	Total 83302.3
x̅1	74.70	x̅2	102.30
s1	28.94959949	s1	53.59564444

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ =μ2​ ; the relaxation treatment is as effective as the pharmaceutical treatment

Ha: μ1​ < μ2 ; the relaxation treatment is more effective than the pharmaceutical treatment ​ i.e mean sleeping time for relaxation treatment is less than pharmaceutical treatment.

This corresponds to a left-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=58. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this left-tailed test is tc​=−1.672, for α=0.05 and df=58. (t value is calculated using t distribution table)

The rejection region for this left-tailed test is R={t:t<−1.672}.

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that t=−2.482 < tc​=−1.672, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.008, and since p=0.008<0.05, it is concluded that the null hypothesis is rejected.

(Here p value is calculated using t distribution with df =58,

p = P[t<-2.482] or due to symmetry

p=P[t>2.482] )

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean \μ1​ is less than μ2​, at the 0.05 significance level.

The relaxation treatment is more effective than the pharmaceutical treatment ​ i.e mean sleeping time for relaxation treatment is less than pharmaceutical treatment.

Graphically

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