Question

A researcher wishes to​ estimate, with 99​% confidence, the population proportion of adults who eat fast food four to six times per week. Her estimate must be accurate within 55​% 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 42​% 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?

Please answer all the questions.

Answer 1

Solution,

Given that,

a) =  1 - = 0.5

margin of error = E = 0.05

At 99% confidence level

= 1 - 99%  

= 1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z0.005  = 2.576

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

= (2.576 / 0.05 )2 * 0.5 *0.5

= 663.57

sample size = n = 664

b) = 0.42

1 - = 1 - 0.42 = 0.58

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

= (2.576 / 0.05 )2 * 0.42 *0.58

= 646.59

sample size = n = 647

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