6 Assuming that we have no way to estimate the population pr

6. Assuming that we have no way to estimate the population proportion, find the sample size that would be required to determine a 95% confidence interval for the true proportion of all teams that have a season winning percentage better than 0.500. We want to be within 0.10 of the true population proportion, that is, we want the margin of error, e, to not exceed 0.10. 69 90 79 80 95 71 86 63 97 74 67 91 96 81 94 89 71 79 73 56

Solution

Margin of Error = + 0.10

alpha = 0.05

Standard Error = sqrt(p (1 - p)/n) = sqrt(0,5 * (1 - 0.5)/n)

true proportion p = winning percentage = 0.5

Critical value for alpha = 0.05 is 1.96

Margin of Error = Critical value * Std Error

+ 0.10 = + 1.96 * sqrt (0.5 * (1 - 0.5)/n)

+ 0.10 = + 1.96 * sqrt (0.25/n)

n = 96.04

Hence a sample of size n = 97 will be required (rounding off 96.04 to a next number since sample cannot be in fraction)