Question

A psychologist is interested in constructing a 95% confidence interval for the proportion of people who accept the theory that a person's spirit is no more than the complicated network of neurons in the brain. 71 of the 811 randomly selected people who were surveyed agreed with this theory. Round answers to 4 decimal places where possible.

a. With 95% confidence the proportion of all people who accept the theory that a person's spirit is no more than the complicated network of neurons in the brain is between and .

b. If many groups of 811 randomly selected people are surveyed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population proportion of all people who accept the theory that a person's spirit is no more than the complicated network of neurons in the brain and about  percent will not contain the true population proportion.

c, Assume that a sample is used to estimate a population proportion p. Find the 90% confidence interval for a sample of size 246 with 71 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.    < p <

Answer 1

a)

sample proportion, = 0.0875
sample size, n = 811
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.0875 * (1 - 0.0875)/811) = 0.0099

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 * 0.0099
ME = 0.0194

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.0875 - 1.96 * 0.0099 , 0.0875 + 1.96 * 0.0099)
CI = (0.0681 , 0.1069)

b)
If many groups of 811 randomly selected people are surveyed, then a different confidence interval would be produced from each group. About 95 percent of these confidence intervals will contain the true population proportion of all people who accept the theory that a person's spirit is no more than the complicated network of neurons in the brain and about 5 percent will not contain the true population proportion.

c)

sample proportion, = 0.2886
sample size, n = 246
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.2886 * (1 - 0.2886)/246) = 0.0289

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

Margin of Error, ME = zc * SE
ME = 1.64 * 0.0289
ME = 0.0474

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.2886 - 1.64 * 0.0289 , 0.2886 + 1.64 * 0.0289)
CI = (0.241 , 0.336)

0.241 < p < 0.336