SOLUTION: 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 a

Question 1141454: 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?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
90% confidence level give you tails of .05 on either side of the normal distribution curve.

z-score for an area under the normal distribution curve to the left of it equal to .05 is equal to -1.645.

since the normal distribution curve is symmetric about the mean, then the z-score for an area under the normal distribution curve to the right of it equal to .05 is equal to 1.645.

therefore your critical z-score at 90% confidence level is equal to -1.645 to 1.645.

in a proportion type study, the mean is the desired proportion.

that's called p.

the standard error is equal to s which is equal to sqrt(p * q / n)

q is equal to 1 - p.

you are looking for a martin of error of 1% which is equal to .01.

the formula for z-score is (x - m) / s

if you want your margin of error to be plus or minus 1%, and your mean is .22, then (x - m) must be equal to .01 * .22 = 0022.

your z-score formula becomes z = .0022 / s

z is equal to plus or minus 1.645.

since the normal distribution curve is symmetric about the mean, if you solve for one side of the normal distribution curve, you have automatically solved for the other side of it.

that's because the same value of s applies to both.

we'll work with the plus 1.645 z-score.

z = .0022 / s becomes 1.645 = .0022 / s

for formula for s given above is s = sqrt (p * q) / n.

this becomes s = (.22 * .78) / s) which becomes s = sqrt(.1716 / n)

the formula of 1.645 = .0022 / s becomes 1.645 = .0022 / sqrt(.1716)/n)

multiply both sides of this formula by sqrt(.1716) / n) to get:

1.645 * sqrt(.1716/n) = .0022

square both sides of this formula to get:

1.645^2 * .1716 / n = .0022^2

solve for n to get:

n = 1.645^2 * .1716 / .0022^2 = 95940.88656.

since s = sqrt(p * q / n), then s = sqrt( .1716 / 95940.88656) = .001337366

you now have:

m = .22
s = .001337366
z = plus or minus 1.645

the z-score formula is, once again;

z (x - m) / s

on the high side, you get 1.645 = (x - .22) / .001337366.

solve for x to get x = .001337366 * 1.645 + .22 = .2222

on the low side, you get -1.645 = (x - .22) / .001337366.

solve for x to get x = .001337366 * -1.645 + .22 = .2178

.2222 / .22 = 1.01 which is 1% above the mean.

.2178 / .22 = .99 which is 1% below the mean.

when n = 95940.88656, the required margin of error of plus or minus 1% of the mean is satisfied.

rounded to the nearest integer, the answer should be 95941.

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