In: Statistics and Probability
Question #4:
Answer: In a survey of 1,000 people, 420 are opposed to an income tax increase. Construct a 95% confidence interval for the proportion of people in the population opposed to this tax increase.
Solution:
n = 1000
= 420/1000 = 0.42
At 95% confidence interval,
= 0.05
Z/2 = 1.96
So, 95% CI:
CI = Z/2 * √ (1-)/n
CI = 0.42 1.96 * √ 0.42 (1-0.42)/1000
CI = 0.42 1.96 * 0.015608
CI = 0.42 0.0306
CI = (0.3894, 0.4506)
Therefore, at 95% confidence interval for the proportion of people in the population opposed to this tax increase is between 0.3894 and 0.4506
Answer: The internal auditing staff of a local lawn-service company performs a sample audit each quarter to estimate the proportion of accounts that are current (between 0 and 60 days after billing). The historical records show that over the past 8 years 70 percent of the accounts have been current. Determine the sample size needed in order to be 99 percent confident that the sample proportion of the current customer accounts is within .03 of the true proportion of all current accounts for this company.
Solution:
= 0.70
At 99% confidence interval,
Z/2 = 2.576
E = 0.03
n = (1-) * (Z/2 / E)2
n = 0.7*0.3*(2.576/0.03)2
n = 0.21 * 7373.08
n = 1548.3468
Therefore, the sample size needed is 1548 in order to be 99 percent confident that the sample proportion of the current customer accounts is within .03 of the true proportion of all current accounts for this company.