A third team planned a case-control study. For this study, investigators selecte

Question

A third team planned a case-control study. For this study, investigators selected an equal number of individuals with and without an adverse event and then compared differences in the proportion in each of these groups that received Medication A. Based on the above they expect that for individuals with an adverse event that 84.2% used medication A and for individuals without an event that 79.5% used medication A. What sample size is needed to have 80% power to detect a significant difference for these rates? Note that this method has an equal number in each group (adverse and no adverse events).

Explanation / Answer

Define

P1: Proportion of individuals with an adverse event taking Medication A.

P2: Proportion of individuals without an adverse event taking Medication A.

The corresponding observed sample proportions are p1=0.845 p2=0.795

Let ‘n’ denotes the sample size from both groups each.

Set up the null hypothesis and alternative hypothesis as follows;

H0: P1=P2 Vs HA: P1P2

Let P=(n1p1+n2p2)/(n1+n2) =(0.845+0.795)/2 = 0.82

The test statistic to test H0 is

Z = (p1-p2)/(p(1-p)(2/n) ~ N(0,1)

Here we wish to find sample size ‘n’ such that power of the test is 0.80.

Power of the test = 1- P[Type II error]

P[Type II error] = P[Accept H0 | HA] = P[|Z| < Z table] = 0.20

P[|Z| < 1.28155] = 0.20 or P[Z > 1.28155] = 0.40

Therefore now let us solve the inequality;

(p1-p2)/(p(1-p)(2/n) > 1.28155

(0.845-0.795)/(0.82*0.18*2/n) > 1.28155

0.092*n > 1.28155

n > (1.28155/0.092)2 = 194

Hence the sample size in each group must be at least 195.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.