Question

The game and fish commission would like to estimate the proportion of fishermen fishing without a license. How many fishermen must they sample to be 90% confident that their estimate is within 0.025 of the population proportion? 

Answer 1

Margin of error = e = 0.025

Confidence level = c = 0.90

Proportion of fishermen fishing without a license is not given.

So we consider it as p = 0.5

z critical value for (1+c)/2 = (1+0.90)/2 = 0.95 is

zc = 1.645 (From statistical table of z values, average of 1.64 and 1.65, (1.64+1.65)/2 = 1.645)

Sample size (n) :

n = 1082.41

n = 1082 (Round to nearest integer)

1082 fishermen must they sample to be 90% confident that their estimate is within 0.025 of the population proportion.