Question

Lipitor is a cholesterol-reducing drug. In clinical trials, 94 subjects were treated with the drug and 200 were given a placebo. In the 94 subjects that received the treatment, 8 developed infections while in the placebo group of 200 there were 12 that developed an infection. At α=.05, is there a greater risk of infection from using Lipitor? Assume that P₁ is the Percentage of Lipitor users that developed an infection.

1. What is the null hypothesis?

2. What is the alternate hypothesis?

3. At α=.05, what is the p-value rounded to 4 decimal places?

Answer 1

Given that,
sample one, x1 =8, n1 =94, p1= x1/n1=0.085
sample two, x2 =12, n2 =200, p2= x2/n2=0.06
null, Ho: p1 = p2
alternate, H1: p1 > p2
level of significance, α = 0.05
from standard normal table,right tailed z α/2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.085-0.06)/sqrt((0.068*0.932(1/94+1/200))
zo =0.797
| zo | =0.797
critical value
the value of |z α| at los 0.05% is 1.645
we got |zo| =0.797 & | z α | =1.645
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: right tail - Ha : ( p > 0.7973 ) = 0.21263
hence value of p0.05 < 0.21263,here we do not reject Ho
ANSWERS
---------------
1.
null, Ho: p1 = p2
2.
alternate, H1: p1 > p2
test statistic: 0.797
critical value: 1.645
decision: do not reject Ho
3.
p-value: 0.21263
we do not have enough evidence to support the claim that greater risk of infection from using Lipitor.