Question

4. A physician wants to compare the blood pressures of six patients before and after treatment with a drug. The blood pressures are as follows: Patient 1 2 3 4 5 6 Before Drug 168 171 182 167 174 170 After Drug 171 170 180 173 178 172 The physician wants to test if there is a significant change of the blood pressure before and after taking the drug at 0.05 level of significance. The absolute value of the test statistic and the absolute critical value of the test are, respectively: a. 0.7192 and 1.812 b. 0.7192 and 2.228 c. 1.6151 and 2.015 d. 1.6151 and 2.571.

Answer 1

Answer Date: 12/12/2019 To test the hypothesis is that there is a significant change of the blood pressure before and after taking drug at 5% significance level. The null and alternative hypothesis is, H:lla = 0 H:d0 The t-test statistics is, First, compute the difference then find mean and standard deviation for computing the t- test statistics.

The mean difference table is, Difference -3 Before 168 171 182 167 174 170 172 After 171 170 180 173 178 172 174 Mean | From the table, the mean difference is -2. The standard deviationis, (d, -)? Sav n-1 [(-3) – (-2)] + [(1) – (-2)] +.... +[(-2)-(-2)]? +[(-2) – (-2)]? 6-1 = 3.0332

The standard deviation is 3.0332. The t-test statistics is, - 2 (3.0332) 16 = -1.6151 = 1.6151 The t-test statistics is 1.6151 - The absolute t critical value at 5% significance level is, The sample size is small and two-tailed test. Look in the column headed a = 0.05 and the row headed in the t distribution table by using degree of freedom is,

d.f.=n-1 = 6-1 The absolute t critical value at 5% significance level is 2.571). Decision The conclusion is that the t value corresponds to sample statistics is not fall in the critical region, so the null hypothesis is not rejected at 5% level of significance. There is insufficient evidence to indicate that there is a significant change of the blood pressure before and after taking drug. The result is not statistically significant. The answer is d.