Question

PROBLEM #3: Tardive dyskinesia denotes a syndrome comprising a variety of abnormal involuntary movements assumed to follow long-term use of antipsychotic drugs. In an experiment to see whether the drug deanol produced a greater improvement from base-line scores than a placebo treatment (“Double Blien Evaluation of Deanol in Tardive Dyskinesia," Amer. Med. Assn., 1978, pp. 1997-1998), the two treatments were administered for four weeks each in random order to 14 patients, resulting in the following total severity index (TSI) scores: Patient: Deanol: Placebo: 1 12.4 9.2 2 6.8' 10.2 3 12.6 12.2 4 13.2 12.7 5 12.4 12.1 6 76 9.0 7 12.1 12.4 Patient: Deanol: Placebo:: 8 5.9 5. 9 9 12.0 8.5 10 1.1 4.8 11 11.5 7.8 12 13.0 9.1 13 5.1 3.5 14 9.6 6.4, (The investigator attributed placebo improvement to frequent attention and the nature of the experiment.) Does the data indicate than on average deanol yields a higher TSI than the placebo treatment? Test using a = 10. (25 pts)

Answer 1

Here we have paired data. Let d = Deanol - Placebo

Following table shows the calculations:

Deanol	Placebo	d=test1-test2	(d-mean)^2
12.4	9.2	3.2	5.6644
6.8	10.2	-3.4	17.8084
12.6	12.2	0.4	0.1764
13.2	12.7	0.5	0.1024
12.4	12.1	0.3	0.2704
7.6	9	-1.4	4.9284
12.1	12.4	-0.3	1.2544
5.9	5.9	0	0.6724
12	8.5	3.5	7.1824
1.1	4.8	-3.7	20.4304
11.5	7.8	3.7	8.2944
13	9.1	3.9	9.4864
5.1	3.5	1.6	0.6084
9.6	6.4	3.2	5.6644
Total		11.5	82.5436

Sample size n 14 Sample sum of difference: d 11.5 Sample mean of differences: Σά 11.5/14 0.82 Sample standard deviation of differences: Σ(d-) =sqrt(82.5436/(14-1)) n-1 -2.5198

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Hypotheses are Level of significance: aa 0.1 Test is one tailed (right tailed) df =n-1-13 Degree of freedom: Test Statistics: d-Ha (0.82-0)/(2.5198/sqrt(14)) = 1.218 Critical value: 1.35 If t 1.35, Reject HO Rejection Region: Decision: Since test statistics does not lie in rejection region so we fail to reject the null hypothesis P-value 0.1224 P-value: Decision: Since p-value is not less than level of significance so we fail to reject the null hypothesis =TINV(0.2,13) Excel function for critical value: =TDIST (1.218,13,1) Excel function for p-value: