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 22% 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 1%? (Do not round the intermediate calculations. Round the final answer to the nearest whole number.)
Sample size
Solution,
Given that,
= 22% = 0.22
1 - = 1 - 0.22 = 0.78
margin of error = E = 1% = 0.01
At 90% confidence level
= 1 - 90%
= 1 - 0.90 = 0.10
/2
= 0.05
Z/2
= Z0.05 = 1.645
sample size = n = (Z / 2 / E )2 * * (1 - )
= ( 1.645 / 0.01 )2 * 0.22 * 0.78
= 4643.54
Sample size = n = 4644
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