Question

A researcher wishes to​ estimate, with 95​% ​confidence, the population proportion of adults who eat fast food four to six times per week. Her estimate must be accurate within 1​% of the population proportion.

​(a) No preliminary estimate is available. Find the minimum sample size needed.

​(b) Find the minimum sample size​ needed, using a prior study that found that 40​% of the respondents said they eat fast food four to six times per week.

​(c) Compare the results from parts​ (a) and​ (b).

​(a) What is the minimum sample size needed assuming that no prior information is​ available?

n=________​(Round up to the nearest whole number as​ needed.)

​(​b) What is the minimum sample size needed using a prior study that found that 40​% of the respondents said they eat fast food four to six times per​ week?

n=_________​(Round up to the nearest whole number as​ needed.)

​(c) How do the results from​ (a) and​ (b) compare?

A.Having an estimate of the population proportion raises the minimum sample size needed.

B.Having an estimate of the population proportion has no effect on the minimum sample size needed.

C.Having an estimate of the population proportion reduces the minimum sample size needed.

Answer 1

Solution,

Given that,

a) =  1 - = 0.5

margin of error = E = 0.01

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025  = 1.96

sample size = n = (Z / 2 / E )2 * * (1 - )

= (1.96 / 0.01)2 * 0.5 *0.5

sample size = n = 9604

b) = 0.40

1 - = 1 - 0.40 = 0.60

sample size = n = (Z / 2 / E )2 * * (1 - )

= (1.96 / 0.01)2 * 0.40 * 0.60

= 9219.84

sample size = n = 9220

c) C.Having an estimate of the population proportion reduces the minimum sample size needed.