Question

A doctor released the results of clinical trials for a vaccine to prevent a particular disease. In these clinical​ trials, 400 comma 000 children were randomly divided in 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 200 comma 000 children in the experimental​ group, 38 developed the disease. Of the 200 comma 000 children in the control​ group, 81 developed the disease. Does it appear to be the case that the vaccine was​ effective? Use the alphaequals0.01 level of significance.

Answer 1

Answer:

We conclude that the vaccine appears to be​ effective.

Step-by-step explanation:

We are given that a doctor released the results of clinical trials for a vaccine to prevent a particular disease.

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 200 comma 000 children in the experimental​ group, 38 developed the disease. Of the 200 comma 000 children in the control​ group, 81 developed the disease.

Let $p_1$ = proportion of subjects in the experimental​ group who developed the disease.

$p_2$ = proportion of subjects in the control​ group who developed the disease.

So, Null Hypothesis, $H_0$ : $p_1\geq p_2$       {means that the vaccine does not appears to be​ effective}

Alternate Hypothesis, $H_A$ : $p_1$       {means that the vaccine appears to be​ effective}

The test statistics that would be used here Two-sample z proportion statistics;

                       T.S. =  $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$  ~ N(0,1)

where, $\hat p_1$ = sample proportion of children in the experimental​ group who developed the disease = $\frac{38}{200,000}$ =  0.00019

$\hat p_2$ = sample proportion of children in the control​ group who developed the disease = $\frac{81}{200,000}$ =  0.00041

$n_1$ = sample of children in the experimental​ group = 200,000

$n_2$ = sample of children in the control​ group = 200,000

So, test statistics  =  $\frac{(0.00019-0.00041)-(0)}{\sqrt{\frac{0.00019(1-0.00019)}{200,000}+\frac{0.00041(1-0.00041)}{200,000} } }$

                              =  -4.02

The value of z test statistics is -4.02.

Now, at 0.01 significance level the z table gives critical values of -2.33 for left-tailed test.

Since our test statistics is less than the critical value of z as -4.02 < -2.33, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the vaccine appears to be​ effective.