Question

A poll asked for​ people's opinion on whether closing local newspapers would hurt civic​ life;428 of 1002 respondents said it would hurt civic life a lot. Complete parts a through d below.

a. Find the proportion of the respondents who said that closing local papers would hurt civic life a lot. Answer ______ ​(Round to three decimal places as​ needed.)

b. Find a​ 95% confidence interval for the population proportion who believed closing newspapers would hurt civic life a lot. Assume the poll used a simple random sample​ (SRS). (In​ fact, it used random​ sampling, but a more complex method than​ SRS.) Answer (______, and _____) ​(Round to three decimal places as​ needed.)

c. Find an​ 80% confidence interval for the population proportion who believed closing newspapers would hurt civic life a lot. Answer (______, and _____) ​(Round to three decimal places as​ needed.)

d. Which interval is wider A,B,C or D?

A.The​ 95% interval is wider. To get a higher degree of​ certainty, the interval needs to be widened just on the right side.

B.The​ 95% interval is wider. To get a higher degree of​ certainty, the interval needs to be widened just on the left side.

C.The​ 80% interval is wider. To get a higher degree of​ certainty, the interval needs to be widened.

D. The​ 95% interval is wider. To get a higher degree of​ certainty, the interval needs to be widened.

Answer 1

a)

sample proportion, = 0.427

b)

sample size, n = 1002
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.427 * (1 - 0.427)/1002) = 0.0156

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0156
ME = 0.0306

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.427 - 1.96 * 0.0156 , 0.427 + 1.96 * 0.0156)
CI = (0.396 , 0.458)

c)

sample proportion, = 0.427
sample size, n = 1002
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.427 * (1 - 0.427)/1002) = 0.0156

Given CI level is 80%, hence α = 1 - 0.8 = 0.2
α/2 = 0.2/2 = 0.1, Zc = Z(α/2) = 1.28

Margin of Error, ME = zc * SE
ME = 1.28 * 0.0156
ME = 0.02

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.427 - 1.28 * 0.0156 , 0.427 + 1.28 * 0.0156)
CI = (0.407 , 0.447)

d)

sample proportion, = 0.427
sample size, n = 1002
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.427 * (1 - 0.427)/1002) = 0.0156

Given CI level is 80%, hence α = 1 - 0.8 = 0.2
α/2 = 0.2/2 = 0.1, Zc = Z(α/2) = 1.28

Margin of Error, ME = zc * SE
ME = 1.28 * 0.0156
ME = 0.02

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.427 - 1.28 * 0.0156 , 0.427 + 1.28 * 0.0156)
CI = (0.407 , 0.447)