Memorial University of Newfoundland

• STAT 3520STAT 3520: Experimental Design I, Fall 2016 Midterm Exam - Solution Key 1. (15 points) (a) (5 points) (Answer varies) This experiment was poorly designed because patients’ response may be due to the placebo effect. Therefore, this experiment was biased. It systematically favored gastric freezing because the placebo effect was confounded with the effect of the treatment. (b) (5 points) Control/Comparison: Comparing active treatment with control treatment or comparing two or more active treatment. Randomization: Use chance to assign subjects to treatments. Replication: Applying each treatment to two or more subjects. Blocking: Arranging of experimental units in groups (blocks) that are similar to one another. (c) (5 points) (Answer varies) Experiments should compare treatments rather than attempt to assess a single treatment in isolation. The subjects should be randomly assigned either to the treatment or placebo group. 2. (25 points) (a) (10 points) From the given data set, we have n1 = 9, n2, ¯yA = 10.63, y¯B = 11.04, s 2 A = 6.01, s 2 B = 6.34 and s 2 p = (n1 − 1)s 2 1 + (n2 − 1)s 2 2 n1 + n2 − 2 = 111.15 18 = 6.18, and, therefore, sp = √ 6.18 = 2.49. The test statistic is t0 = y¯1 − y¯2 sp q 1 n1 + 1 n2 = −0.41 2.49q 1 10 + 1 10 = −0.41 1.11 = −0.369 Since t0.025,18 = 2.101, do not reject H0 : µA = µB with the p-value = 0.72. There is no significant evidence that the two materials are different. (b) (10 points) Let’s apply a paird t-test for testing H0 : µA = µB against H1 : µA 6= µB. From the given data set, we have nd = 10 pairs, ¯d = 1 nd Pnd i=1 di = 1 10 (4.1) = 0.41, s 2 d = 1 nd − 1 Xnd i=1 (di − ¯d) 2 = Pnd i=1 d 2 i − ( Pnd i=1 di) 2 nd nd − 1 = 3.030 − 1.681 9 = 0.149, and, therefore, sd = p s 2 d = 0.386 and sd¯ = √ sd nd = 0.122. The value of the test statistic t0 associated with the null hypothesis H0 : µd = 0 against the alternative hypothesis H1 : µd 6