Aphysical 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 she wishes the estimate to be within two percentage points with 99% confidence, assuming that (a) she uses the estimates of 21.2% male and 19.7% female from a previous year? (b) she does not use any prior estimates?

no of sample within two percentage points  is 5409

no of sample not use prior estimates is 8321

Step-by-step explanation:

Given data

E = 2% = 0.02

male m = 21.2 % = 0.212

female f = 19.7 % = 0.197

to find out

no of sample within two percentage points and if does not use any prior estimates

solution

first we calculate no of sample within two percentage points

we know Z critical value for 99% is 2.58

so no of sample will be

n = (Z / E )² × m(1-m) + f ( 1-f )

put all the value we get no of sample

n = (2.58 / 0.02 )² × 0.212 ( 1 - 0.212 ) + 0.197 ( 1 - 0.197 )

n =  16641 ×0.3250

no of sample within two percentage points  is 5409

in 2nd part no of sample not use prior estimates

here than f and m will be 0.5

than

n = (Z / E )² × m(1-m) + f ( 1-f )

put all value

n = (2.58 / 0.02 )² × 0.5 ( 1 - 0.5 ) + 0.5 ( 1 - 0.5 )

n =  16641 ×  0.5

no of sample not use prior estimates is 8321

answer: for each 3 girls there is 2 boys . 3: 2

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