Question

osteoporosis is a condition in which people experience decreased bone mass and an increase in the risk of bone fracture. Actonel is a drug that helps combat osteoporosis in postmenopausal women. In clinical trials, 1374 postmenopausal women were randomly divided into experimental and control groups. The subjects in the experimental group were administered 5 milligrams of Actonal, while the subjects in the control group were administered a placebo. The number of women who experienced a bone fracture over the course of one year was recorded. Of the 696 women in the experimental group, 27 experienced a fracture during the course of the year. Of the 678 women in the control group, 49 experienced a fracture during the course of the year.

a. Does the sample evidence suggest the drug is effective in preventing bone fractures? Use the × = 0.01 level of significance.

b. Construct a 95% confidence interval for the difference between the two population proportions, p exp - p control.

Answer 1

Solution-A:

Ho:p1=p2

Ha:p1<p2

alpha=0.01

p1^=x1/n1= 27/696=0.0387931

p2^=x2/n2=49/678= 0.07227139

Us eti 83 calculator to get z stat and p value

STAT>TESTS>2-PropZtest

z=-2.7141

p=0.0033

p<0.01

Reject null hypothesis

Accept Alternative Hypothesis

Conclusion:
There is suffcient statistical evidence at 1% level of significance to conlcude that

the drug is effective in preventing bone fractures

Solution-B:

to get 95% confidence interval for difference in proportions

STAT>TESTS>2-PropZint

we get

95% confidence interval for the difference between the two population proportions, p exp - p control

=(-0.0577,-0.0093)