Question

A poll reported that only 1283 out of a total of 1964 adults in a particular region said they had a​ "great deal of​ confidence" or​ "quite a lot of​ confidence" in the military. Assume the conditions for using the CLT are met. Complete parts​ (a) through​ c below.

A. Find a 95​% confidence interval for the proportion that express a great deal of confidence or quite a lot of confidence in the​ military, and interpret this interval.Answer (____, ____) (3 decimal places)

Interpret this interval. Select the correct choice below and fill in the answer boxes to complete your choice. ​(Type integers or decimals rounded to three decimal places as​ needed.)

We are 95% confident that the population proportion of adults having a great deal or quite a lot of confidence in the military is between Answer ____ and _____

b. Find an 80% confidence interval. Interpret it.

The 80% confidence interval for the proportion that express a great deal of confidence or quite a lot of confidence in the military is Answer ____ and _____(Round to 3 decimal places as​ needed.)

Interpret this interval. Select the correct choice below and fill in the answer boxes to complete your choice. ​(Type integers or decimals rounded to three decimal places as​ needed.)

We are 80​% confident that the population proportion of adults having a great deal or quite a lot of confidence in the military is between Answer _____ and _____

c. Which interval is​ wider? The width of the 95% confidence interval is 0.058and the width of the 80% confidence interval is Answer _____The 99% interval is wider. ​(Round to three decimal places as​ needed.)

Answer 1

a)

sample proportion, = 0.6533
sample size, n = 1964
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.6533 * (1 - 0.6533)/1964) = 0.0107

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 * 0.0107
ME = 0.021

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.6533 - 1.96 * 0.0107 , 0.6533 + 1.96 * 0.0107)
CI = (0.632 , 0.674)

We are 95% confident that the population proportion of adults having a great deal or quite a lot of confidence in the military is between Answer 0.632 and 0.674

b)

sample proportion, = 0.6533
sample size, n = 1964
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.6533 * (1 - 0.6533)/1964) = 0.0107

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 * 0.0107
ME = 0.0137

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.6533 - 1.28 * 0.0107 , 0.6533 + 1.28 * 0.0107)
CI = (0.64 , 0.667)
We are 80​% confident that the population proportion of adults having a great deal or quite a lot of confidence in the military is between Answer 0.640 and 0.667

c)

width of the 80% confidence interval is 0.027