Question

The Manitoba Tourism Board plans to sample information centre visitors entering the province to learn the fraction of visitors who plan to camp in the province. Current estimates are that 33% of visitors are campers. How large a sample would you take to estimate at a 90% confidence level the population proportion with an allowable error of 4%? (Do not round the intermediate calculations. Round the final answer to the nearest whole number.)  

sample size:

Answer 1

Solution:

Given that,

= 0.33

1 - = 1 - 0.33 = 0.67

margin of error = E = 4% = 0.04

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z0.05 = 1.645

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

= ( 1.645 / 0.04)2 * 0.33 * 0.67

= 373.87

= 374

n = sample size = 374