Question

5. The active ingredient of a particular kind of drug tablet has traditionally been measured by method A. A cheaper and quicker method B has been developed but its reliability has to be tested against method A. Each of 10 tablets is divided into two halves and the amount of active ingredient (in milligrams) is measured, one half-tablet by method A and the other by method B. The results are given below. Tablet Method A Method B Difference 97.2 97.2 0 1 ... : 10 Mean SD 95.2 97.46 4.086 94.8 95.4 4.085 0.4 2.06 3.086 Note that in each column the mean and standard deviation given in the table are of the 10 measurements in that column. (a) Assume the two samples are INDEPENDENT. i. Test whether there is any difference in the two means. Assume that the population standard deviations are equal i.e. OA = 0B. State clearly the hypotheses and the p-value. What is the conclusion at the 5% significance level? At the 10% level? ii. Find a 95% confidence interval for the difference in means. (b) Now consider the data as being obtained from a matched pairs study. i. Use the matched pairs t test to test if there is any significant difference between the measurements by the two methods. Use 5% significance level. State clearly the hypotheses, p-value and conclusion. ii. Find a 95% confidence interval for the difference in means.

Answer 1

b) ) =2.06 sa = 3.886 Hypother's, Ho: Md=0 vs Ha Ma Mato Test statistic a 2,06 - 2.1109 3.086/50 with ready p-value = 2(+2211109) =0.063 0.0640 The p-valus 0,0640 is greater than 5% significance ·level. Hence, fail to reject to and conclude that there is as not any significant difference between the measurements by the two methods. 11 95% confidence interval for the difference in means is Adrn = 2.06+2,26 x 3.086 a Į k 10,05 ng 10 = 2.06 1 2.215 The 95% of (-0.15, 4.27) includes the vahu zero Herrce, any difference & significant difference between the measurements by the most two methods. there is not