Question

  The addition of bran to the diet has been reported to benefit patients with diverticulosis. Several different bran preparations are available, and a public health researcher wants to test the efficacy of two of them on patients, since favorable claims have been made for each. Among the consequences of administering bran that requires testing is the transit time through the alimentary canal. Does it differ in the two groups of patients taking these two preparations? The null hypothesis is that the two groups come from the same population. By random allocation the researcher selects two groups of patients aged 40-64 with diverticulosis of comparable severity. Sample 1 consists of 35 patients who are given treatment A, and sample 2 consists of 32 patients who are given treatment B. The transit times of food through the gut (in hours) are measured by a standard technique with marked pellets and the results are shown below

Treatment A s=16.47,n=35 Treatment B s=17.63,n=32

The assumptions are: that transit times are normally distributed, that the two samples come from distributions that may differ in their mean value, but not in the standard deviation, and that transit times are independent of each other. The test statistic is thus t = (x1 − x2) − (µ1 − µ2) sp r 1 n1 + 1 n2 where t is a t-distribution with df = n1 +n2 −2. The pooled standard deviation is given by sp = s (n1 − 1)s 2 1 + (n2 − 1)s 2 2 n1 + n2 − 2 The larger the difference, x1−x2, is, the more likely the null hypothesis is rejected. What is the smallest value of x bar1 − x bar2 that allows you to reject the null hypothesis at α = 0.05?

Answer 1

We have hole. two tramite therefore we Sample- A Sample B N = 35 N = 32 S, = 16 . Persarior the t-tert where the Samples are from Normal TE Xi - X2 - (M.-M2) S2 = 17 We two Now we have Null hypothein that Hoi Mi-M2=0 vs Hi Mp-lz to & we reject Ho for Hi 4 1T1 > tiap NitN2-2 (N, -1) 577 (2 - 2 5 Spa NitN2-2 (35-1) (16)² + (32-1) (1772 35+32-2 = 27.738 M 271.74 1/2

T= x - 72 771.74 N2=32 N= 35 sp=271.74 Sp=7271.74 + žz Xi - X2 16.48 × 0.244 4.080 Now from the Biometrica takle we ti-, NatNy-2 - ti- 0:05, Nitz-62 t0,975, 35132-2 = t 0.975, 65 & 2 ITI ÿ t 1-2, NIIN2-2 - X-X2 > 2x4.030=8.06 X1 X2 > 8.06 the leart value X - X we need to hove is 8.061 to reject the Null hypothers + 2/2