Question

Suppose you have developed a coronavirus vaccine and wish to verify that it is effective in reducing the probability of being infected with the disease. You obtain two random samples. The control group of 90 people is given a placebo, and after 1 year, 8 of them had been infected. The experimental group received the vaccine. Of the 300 people in the experimental group, 8 of them had been infected after 1 year. (SHOW WORK!)

The alternative hypothesis for this test is :  

The distribution (choose from Z, T, X^2, or F) for this test is  

The P-value of this test is:

Based on this information, we (choose reject or fail to reject)   the null hypothesis

Based on this we (choose from do or do not)   have enough evidence to say that this vaccine shows effective protection from the virus at the 5% significance level. (We assume it was safe, so 5% is a reasonable significance level)

Answer 1

Given that,
sample one, x1 =8, n1 =90, p1= x1/n1=0.089
sample two, x2 =8, n2 =300, p2= x2/n2=0.027
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.089-0.027)/sqrt((0.041*0.959(1/90+1/300))
zo =2.61
| zo | =2.61
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =2.61 & | z α | =1.96
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.6101 ) = 0.0091
hence value of p0.05 > 0.0091,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 2.61
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.0091
we have enough evidence to support the claim that this vaccine shows effective protection from the virus