To test the effectiveness of a vaccine, 150 experimental animals were given the
ID: 3233322 • Letter: T
Question
To test the effectiveness of a vaccine, 150 experimental animals were given the vaccine and 150 were not. All 300 were then exposed to the disease. Among those vaccinated, 10 contracted the disease. Among the control group (i.e., those not vaccinated), 30 contracted the disease. Can we conclude that the vaccine is effective in reducing the infection rate for this disease? Use a significance level of .025. 1. H_0: p_1 - p_2 [a] 0 vs. H_1: p_1 - P_2 [b] 0 2. Test statistic: z = [c] 3. with alpha = .025, reject the null hypothesis if z [e] [1]. 4. [g] reject the null hypothesis since z_obs, [h][i]. 5. There [j] sufficient evidence to conclude that the vaccine is effective in reducing the infection rate for this disease.Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: PVaccinated> PControl
Alternative hypothesis: PVaccinated < PControl
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.025. 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 = 0.1333
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.03925
z = (p1 - p2) / SE
z = - 3.397
zcritical = - 1.96
We will reject the null hypothesis if z test is smaller than zcritical.
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 - 3.397. We use the Normal Distribution Calculator to find P(z < - 3.397) = Less than 0.0001
Interpret results. Since the P-value (Almost 0) is less than the significance level (0.025), we have to reject the null hypothesis.
From the above test we have sufficient evidence in the favor of the claim that vaccine is effective in reducing the infection rate for dieases.
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