A recent survey indicates that the proportion of season ticket holders for the school hockey team who renew their seats is about 0.80. Using a 95% confidence interval and a margin of error of 0.025, what is the approximate size of the sample needed to estimate the true proportion who plan to renew their seats?
A) 689
B) 697
C) 984
D) 1179

Question

A recent survey indicates that the proportion of season ticket holders for the school hockey team who renew their seats is about 0.80. Using a 95% confidence interval and a margin of error of  0.025, what is the approximate size of the sample needed to estimate the true proportion who plan to renew their seats?
A) 689 
B) 697 
C) 984 
D) 1179

Quizplus · Accepted Answer

The formula for the sample size needed to estimate a population proportion with a specified margin of error and confidence level is:

$n = (z^2*p*q)/E^2$

where:

n = sample size
z = z-score corresponding to the desired confidence level (in this case, 1.96 for 95% confidence)
p = estimated proportion (in this case, 0.80)
q = 1 - p
E = margin of error (in this case, 0.025)

Substituting in the given values, we get:

$n = (1.96^2*0.80*0.20)/(0.025^2) \approx 983.78 \approx 984$

Therefore, we need a sample size of at least 984 to estimate the true proportion of season ticket holders who plan to renew their seats with a margin of error of 0.025 and 95% confidence. The nearest option is C (984), so that is the correct answer.

A Recent Survey Indicates That the Proportion of Season Ticket