Question

Please help me answer this question for my Statistical Methods subject. PLEASE answer it seriously and completely, or better not answer it if not then, really appreciate it, thank you.

A consumer agency wanted to find out if the mean time taken for each of three brands of medicines to provide headache relief is the same. The first drug was administered to six patients, the second to four patients and the third to five patients. All were randomly administered. The following table gives the summary statistics of time taken (in minutes) to get headache relief after the patients had taken the medicine. 2 Drug 1 44.833 Drug 3 = 53.6 Drug 2 -20.0 182.167 =17.333 175.3 Xa=41.133 - Test for homogeneity of variances. Use a 0.05 (a) (30 marks) (b) At a 0.025, can the consumer agency conclude that the mean time taken to provide headache relief is the same for each of the three drugs (30 marks) (c) If the null hypothesis is rejected, test all possible pair wise comparisons for the mean time taken to provide headache relief of the three drugs (40 marks)

Answer 1

1. (a) Null hypothesis, Ho oo0 (i.e. homogeneity of variance) vs. Alternative hypothesis, H1 Ho is true. 3 i-1 MSE of square due to error mean sum 3 175.3)/12 138.6695 = 5, n n1 + n2 +n3 _ = (5 * 182.167 3 17.333 +4 where, n 4, n3 6, n2 15 1 n 3 12, a 3, C1+ 1 - 1.1167 3(a 1) 1 ni -1 3 (n 1) log s log(MSE) Test statistic U 3.5256 _ i-1 p - value Hence we fail to reject Ho at 5% level of significance and conclude that variances are equal P(U 3.5256|U 0.1716 0.05 _ across groups

(b) Let i Null hypothesis, Ho vs. Alternative hypothesis, H1 of ith Drug; i 1,2,3 true mean H2H3 At least one i is different from other s. 3 XGM26 (44.833 SSB of square between group sum 77 =1 41.133)24 (20 41.133)5* (53.6 41.133)2 2645.685 SSB MSB 2645.685/2 = 1322.842 3 1 MSB/MSE 1322.842/138.6695 P(F 9.5395F reject Ho at 5% level of significance and conclude that F ratio 9.5395 _ p- value Hence F2,12) 0.0033<0.05 we for all durg the mean time taken to provide headache relief is not same

(c) Since we so we need to per form post hoc test reject Ho Tukey HSD (honest significant difference Here we use HSD(i,jHSD for (i, j)th pair MSE 1 2 ni ni HSD(1, 2) (44.833 20)/V(138.6695/2) HSD(1, 3) (44.833 53.6)/V(138.6695/2) (1/6 1/5) HSD(2, 3) (20 53.6)/V(138.6695/2) Now, (1/6 1/4) 4.62 1.74 (1/4 1/5)-6.02 of ith drug and mean of jth drug significantly di fferent if mean are |HSD(i, j)>o.05,3,12 3.77 Hence Drug 1 and Drug 2 are and Drug 2 and Drug 3 are significantly different significantly different.