Question

A poll conducted by GfK Roper Public Affairs and Corporate Communications asked a sample of 1007 adults in the United States, "As a child, did you ever believe in Santa Claus, or not?" Of those surveyed, 84% said they had believed as a child. Consider the sample as an SRS. We want to estimate the proportion p of all adults in the United States who would answer that they had believed to the question "As a child, did you ever believe in Santa Claus, or not?"

(a) Find a 90% confidence interval (±±0.0001) for p based on this sample.

The 90% confidence interval is from __ to ___

b) Find the margin of error (±±0.0001) for 90%.

The margin error is ___

(c) Suppose we had an SRS of just 100 adults in the United States.

What would be the confidence interval (±±) for 95% confidence?

The 50 % confidence interval (+) is from __ to __

(d) How does decreasing the sample size change the confidence interval when the confidence level remains the same?

a. Decreasing the sample size creates a wider interval

b.Decreasing the sample size creates a less wide interval.

Answer 1

a)

sample proportion, = 0.84
sample size, n = 1007
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.84 * (1 - 0.84)/1007) = 0.0116

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 * 0.0116
ME = 0.019

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.84 - 1.64 * 0.0116 , 0.84 + 1.64 * 0.0116)
CI = (0.821 , 0.859)

The 90% CI is from 0.8210 to 0.8590

b)
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.84 * (1 - 0.84)/1007) = 0.0116

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 * 0.0116
ME = 0.0190

c)

sample proportion, = 0.84
sample size, n = 100
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.84 * (1 - 0.84)/100) = 0.0367

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 * 0.0367
ME = 0.0602

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.84 - 1.64 * 0.0367 , 0.84 + 1.64 * 0.0367)
CI = (0.7798 , 0.9002)

The 95% CI is from 0.7798 to 0.9002

d)

Decreasing the sample size creates a wider interval