Question

A physical therapist wants to determine the difference in the proportion of men and women who participate in regular sustained physical activity. What sample size should be obtained if he wishes the estimate to be within two percentage points with 95​% ​confidence, assuming that

​(a) he uses the estimates of 22.8​% male and 18.8​% female from a previous​ year?

​(b) he does not use any prior​ estimates?

Answer 1

Answer

(A) we have margin of error = 2% or 2/100 = 0.02

we have to use 95% confidence level, so z value that we need to use will be 1.96 (using z distribution table)

proportion for male is 22.8% = 22.8/100 = 0.228

So, we get p = 0.228

Using the formula for sample size n, we get

setting the given values, we get

rounding it off to next whole number, we get sample size n = 1691 for males

Calculation for females

proportion = p = 18.8% = 18.8/100 = 0.188

we have margin of error = 2% or 2/100 = 0.02

we have to use 95% confidence level, so z value that we need to use will be 1.96 (using z distribution table)

Using the formula for sample size n, we get

setting the given values, we get

rounding it off to next whole number, we get sample size n = 1467 for females

(b) when no prior estimates was given, then we have the following formula

sample size n =

where z = 1.96 for 95% confidence level and ME = margin of error = 2% = 2/100 = 0.02

setting the values, we get

sample size n =

So, required sample size is 2401