In a random sample of 65 patients undergoing a standard surgical procedure, 12 required medication for postoperative pain. In a random sample of 90 patients undergoing a new procedure, only 14 Required medication.
Construct a 98% confidence interval for the difference in the proportions of patients needing pain Medication between the old and new procedures.
A) (-0.003, 0.061)
B) (0.185, 0.156)
C) (0.042, 0.328)
D) (-0.114, 0.172)

Question

Quizplus · Accepted Answer

We can use the formula for the confidence interval for the difference in proportions:

(point estimate) +/- (critical value) * (standard error of the difference)

where the point estimate is the difference of the sample proportions, the critical value is obtained from a t-distribution with (65+90-2) degrees of freedom and a significance level of 0.01, and the standard error of the difference is calculated as:

sqrt[(p1*q1/n1) + (p2*q2/n2)]

where p1 and p2 are the sample proportions, q1=1-p1 and q2=1-p2 are the complements, and n1=65 and n2=90 are the sample sizes.

Using the given data, we have:

point estimate = (12/65) - (14/90) = -0.057
p1 = 12/65, q1 = 53/65, n1 = 65
p2 = 14/90, q2 = 76/90, n2 = 90
standard error = sqrt[(p1*q1/n1) + (p2*q2/n2)] = sqrt[(0.181*0.819/65) + (0.156*0.844/90)] = 0.097

The critical value corresponding to a 98% confidence level and (65+90-2) = 153 degrees of freedom is obtained from a t-distribution table or calculator and is approximately 2.328.

Therefore, the 98% confidence interval for the difference in proportions of patients needing pain medication between the old and new procedures is:

-0.057 +/- 2.328 * 0.097 = (-0.114, 0.172)

The correct choice is D, which contains this interval.

In a Random Sample of 65 Patients Undergoing a Standard