Question

In: Statistics and Probability

A clinical trial is planned to compare an experimental medication designed to lower blood pressure to...

  1. A clinical trial is planned to compare an experimental medication designed to lower blood pressure to a placebo. Patients are enrolled and randomized to either the experimental medication or the placebo. The data shown below are data collected at the end of the study after 6 weeks on the assigned treatment.

Experimental (n=100)

Placebo (n=100)

Mean (SD) Systolic Blood Pressure

120.2 (15.4)

131.4 (18.9)

% Hypertensive

14%

22%

% With Side Effects

6%

8%

Is there a significant difference in the proportions of hypertensive patients between groups? Use α = 0.05.

Step 1.           Set up hypotheses and determine level of significance

           

Step 2.            Select the appropriate test statistic

                                        

    

Step 3.            Set up decision rule

                       

Step 4.            Compute the test statistic

step 5. conclusion

Solutions

Expert Solution

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P1 = P2
Alternative hypothesis: P1 P2

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p1 * n1 + p2 * n2) / (n1 + n2)
p = 0.18
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.05433
z = (p1 - p2) / SE

z = - 1.47

zcritical = + 1.96

Rejection region is - 1.96 > z > 1.96

Reject H0, - 1.96 > z > 1.96

where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than -1.47 or greater than 1.47.

Thus, the P-value = 0.142

Interpret results. Since the P-value (0.142) is greater than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that there a significant difference in the proportions of hypertensive patients between groups.


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