In a clinical trial of a vaccine,14,000 children were randomly divided into two
ID: 3268309 • Letter: I
Question
In a clinical trial of a vaccine,14,000 children were randomly divided into two groups. The subjects in group 1 (the experimental group) were given the vaccine while the subjects in group 2 (the control group) were given a placebo. Of the 7,000 children in the experimental group, 72 developed the disease. Of the 7,000 children in the control group,112 developed the disease.
Determine whether the proportion of subjects in the experimental group who contracted the disease is less than the proportion of subjects in the control group who contracted the disease at the =0.05 level of significance.
Determine the test statistic.
z0=........ (Round to two decimal places as needed.)
The P-value is..... (Round to three decimal places as needed.)
What is the result of this hypothesis test?
A. Do not reject the null hypothesis because there is sufficient evidence to conclude that the proportion of subjects in the experimental group who contracted the disease is less than the proportion of subjects in the control group at alphaequals=0.05.
B. Reject the null hypothesis because there is not sufficient evidence to conclude that the proportion of subjects in the experimental group who contracted the disease is less than the proportion of subjects in the control group at alphaequals=0.05.
C. Reject the null hypothesis because there isis sufficient evidence to conclude that the proportion of subjects in the experimental group who contracted the disease is less than the proportion of subjects in the control group at alphaequals=0.05.
D. Do not reject the null hypothesis because there is notis not sufficient evidence to conclude that the proportion of subjects in the experimental group who contracted the disease is less than the proportion of subjects in the control group at alphaequals=0.05.
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: PExperimental> PControl
Alternative hypothesis: PExperimental < PControl
Note that these hypotheses constitute a one-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 = 0.01314
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.001925
z = (p1 - p2) / SE
z = - 2.97
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 one-tailed test, the P-value is the probability that the z-score is less than -2.97. We use the Normal Distribution Calculator to find P(z < -2.97) = 0.0015.
Thus, the P-value = 0.0015
Interpret results. Since the P-value (0.0015) is less than the significance level (0.05), we have to reject the null hypothesis.
C. Reject the null hypothesis because there isis sufficient evidence to conclude that the proportion of subjects in the experimental group who contracted the disease is less than the proportion of subjects in the control group at alpha equals=0.05.