Question

Question #4:

1. 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. Interpret your result.

2. 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.      

Answer 1

a.

Confidence interval :

p = 420/1000

sd = (p*(1-p))^0.5

= ((420/1000)*(1-(420/1000)))^0.5

= 0.49

CI : [ 420/1000 - 1.96*0.49/(1000^0.5) , 420/1000 + 1.96*0.49/(1000^0.5) ]

= [ 0.39 , 0.45 ]

95% confidence interval for the proportion of people : [ 0.39 , 0.45 ]

b.

for 99%

z = 2.576

sample proportion of the current customer accounts is within .03 of the true proportion

0.03 >= 2.576*[(0.7*(1-0.7))^0.5]/(n^0.5)

n^0.5 >= [2.576*[(0.7*(1-0.7))^0.5]] / 0.03

n >= (39.349)^2 = 1548.34

n is integer so to be witihin 0.03 we will take integer just greater than 1548.34

n >= 1549

P.S. (please upvote if you find the answer satisfactory)