Question

A poll conducted by GfK Roper Public Affairs and Corporate Communications asked a sample of 1001 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 p based on this sample. The 90% confidence interval is from to (b) Find the margin of error (± ± 0.0001) for 90%. The margin of error is (c) Suppose we had an SRS of just 150 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? Decreasing the sample size creates a less wide interval Decreasing the sample size creates a wider interval

Answer 1

a)

sample proportion, = 84% = 0.84 =
sample size, n = 1001
Standard error ,SE =

SE = = 0.0116

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Z_c = Z(α/2) = 1.645

Margin of Error, ME = Z_c * SE
ME = 1.645 * 0.0116
ME = 0.019

Confidence interval (CI) =( - ME , + ME )
CI = (0.84 - 0.019 , 0.84 + 0.019)
CI = (0.821 , 0.859)

The 90% CI is from 0.821 to 0.859

b)

Margin of Error = 0.019

c)

sample proportion, = 0.84
sample size, n = 150
Standard error, SE =  
SE = = 0.0299

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 * 0.0299
ME =0.049

CI = ( - ME , + ME)
CI = (0.84 - 0.049 , 0.84 + 0.049)
CI = (0.791 , 0.889)

The 95% CI is from 0.791 to 0.889

d)

Decreasing the sample size creates a wider interval