Question

1. The American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 31 31 who smoke. In an earlier study, the population proportion was estimated to be 0.2 0.2 . How large a sample would be required in order to estimate the fraction of Americans over 31 31 who smoke at the 95% 95% confidence level with an error of at most 0.03 0.03 ? Round your answer up to the next integer.

2.

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the eighth grade level. In an earlier study, the population proportion was estimated to be 0.230.23.

How large a sample would be required in order to estimate the fraction of tenth graders reading at or below the eighth grade level at the 80%80% confidence level with an error of at most 0.030.03? Round your answer up to the next integer.

Answer 1

Solution,

Given that,

= 0.2

1 - = 1- 0.2 = 0.8

margin of error = E = 0.03

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_0.025 = 1.96

Sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.96 / 0.03 )² * 0.2 * 0.8

= 682.95

Sample size = n = 693

2.

Given that,

= 0.23

1 - = 1 - 0.23 = 0.77

margin of error = E = 0.03

At 80% confidence level

= 1 - 80%  

= 1 - 0.80 = 0.2

/2 = 0.1

Z/2 = Z_0.1 = 1.282

Sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.282 / 0.03 )² * 0.23 * 0.77

= 323.41

Sample size = n = 324