Question

In a trial designed to test the effectiveness of aspirin in preventing heart disease, 11,037 male physicians were treated with aspirin and 11,034 male physicians were given placebos. Among the subjects in the aspirin treatment group, 139 experienced myocardial infarctions (heart attacks). Among the subjects given placebos, 239 experienced myocardial infarctions (based on data from “Final Report on the Aspirin Component of the Ongoing Physicians' Health Study," New England Journal of Medicine, Vol. 321: 129-135). Use a 0.01 significance level to test the claim that aspirin has no effect on myocardial infarctions.

Please include:

Null and alternative hypotheses

critical t or z

standardized test statistic

decision about the null

Answer 1

Given that,
sample one, x1 =139, n1 =11037, p1= x1/n1=0.013
sample two, x2 =239, n2 =11034, p2= x2/n2=0.022
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
reject Ho, if zo < -2.576 OR if zo > 2.576
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.013-0.022)/sqrt((0.017*0.983(1/11037+1/11034))
zo =-5.191
| zo | =5.191
critical value
the value of |z α| at los 0.01% is 2.576
we got |zo| =5.191 & | z α | =2.576
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -5.1907 ) = 0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: -5.191
critical value: -2.576 , 2.576
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that aspirin has no effect on myocardial infractions