Question

An article in the New England Journal of Medicine described a randomized experiment that investigated the effectiveness of two medications for nausea in patients undergoing chemotherapy treatments for cancer. In the experiment, 157 patients were divided at random into two groups. One group of 78 patients were given a standard anti-nausea drug called prochlorperazine, while the other group of 79 patients received delta-9-tetrahydrocannabinol (i.e., THC, the active ingredient in marijuana). Both medications were delivered orally and no patients were told which of the two drugs they were taking. The observed response was whether or not the patient experienced relief from nausea when undergoing chemotherapy. 16 of the patients taking prochlorperazine experienced relief from nausea, while 36 of the patients taking THC experienced relief from nausea. Conduct a significance test to determine if there is evidence that there is a difference in the effectiveness of the two drugs

Answer 1

Proportion of patients getting relief after taking prochlorperazine, p1 = 16/78 = 0.2051282

Proportion of patients getting relief after taking THC, p2 = 36/79 = 0.4556962

Let P1 and P2 be the population proportion of patients getting relief after taking prochlorperazine and THC medication respectively.

Null hypothesis H0: P1 = P2

Alternative hypothesis H1: P1 P2

For this analysis, we assume the significance level is 0.05. The test method is a two-proportion z-test.

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)

where n1 and n2 are the sample sizes (number of patients) taking prochlorperazine and THC medication.

p = (0.2051282 * 78 + 0.4556962 * 79) / (78 + 79) = 0.3312102

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

= sqrt{ 0.3312102 * ( 1 - 0.3312102 ) * [ (1/78) + (1/79) ] }

= 0.0751252

z = (p1 - p2) / SE = (0.2051282 - 0.4556962) / 0.0751252 = -3.34

Since we have a two-tailed test, the P-value is the probability that the z-score is less than -3.34 or greater than 3.34.

We use the Normal Distribution Calculator to find P(z < -3.34) = 0.0004, and P(z > 3.34) = 0.0004. Thus, the P-value = 0.0004 + 0.0004 = 0.0008.

Since the P-value (0.0008) is less than the significance level (0.05), we reject the null hypothesis and conclude that there is significant evidence that P1 P2 and there is a difference in the effectiveness of the two drugs.