Question

In spite of the potential safety hazards, some people would like to have an Internet connection in their car. A preliminary survey of adult Americans has estimated this proportion to be somewhere around 0.20. (a) Use the given preliminary estimate to determine the sample size required to estimate the proportion of adult Americans who would like an Internet connection in their car to within 0.04 with 95% confidence. (Enter your answer as a whole number.) (b) The formula for determining sample size given in this section corresponds to a confidence level of 95%. How would you modify this formula if a 99% confidence level was desired? n = p(1 − p) 1.645 B 2 n = p(1 − p) 2.33 B 2 n = p(1 − p) 2.05 B 2 n = p(1 − p) 2.58 B 2 n = p(1 − p) 1.96 B 2 Correct: Your answer is correct. (c) Use the given preliminary estimate to determine the sample size required to estimate the proportion of adult Americans who would like an Internet connection in their car to within 0.04 with 99% confidence. (Enter your answer as a whole number.)

Answer 1

Part a

We are given

p = 0.20, q = 1 – p = 1 – 0.20 = 0.80

Confidence level = 95%

Margin of error = E = 0.04

Critical Z value = 1.96

(by using z-table)

Sample size formula is given as below:

n = p*q*(Z/E)^2

n = 0.20*0.80*(1.96/0.04)^2

n = 384.16

n = 385

Required sample size = 385

Part b

WE have to modify above formula for sample size for confidence level = 99%

For 99% confidence level,

Critical Z value = 2.5758

Sample size formula is given as below:

n = p*q*(Z/E)^2

n = 0.20*0.80*(2.5758/0.04)^2

Part c

We are given

p = 0.20, q = 1 – p = 1 – 0.20 = 0.80

Confidence level = 99%

Margin of error = E = 0.04

Critical Z value = 2.5758

(by using z-table)

Sample size formula is given as below:

n = p*q*(Z/E)^2

n = 0.20*0.80*(2.5758/0.04)^2

n = 663.4746

n = 664

Required sample size = 664