Question

(Hypothetical) In a major national survey of crime victimization, the researchers found that 20.0% of Americans age 12 or older had been a victim of crime. The size of the sample was 144,400. (30 points possible.)

Estimate the percentage of Americans age 12 or older who were victims at the 95% confidence level. Write a sentence explaining the meaning of this confidence interval.

Estimate the percentage of victims at the 99% confidence level. Write a sentence explaining the meaning of this confidence interval.

Imagine that the Sample size was cut in half, but the survey found the same value of 20.0% for the percentage of victims.

Would the 95% confidence interval increase or decrease? By how much?

Would the 99% confidence interval increase or decrease? By how much?

Answer #1

a) At 95% confidence interval the critical value is
z_{0.025} = 1.96

The 95% confidence interval is

p +/- z_{0.025} * sqrt(p(1 - p)/n)

= 0.2 +/- 1.96 * sqrt(0.2 * 0.8/144400)

= 0.2 +/- 0.002

= 0.198, 0.202

We are 95% confident that the true population proportion lies in the above confidence interval.

At 99% confidence interval the critical value is
z_{0.005} = 2.58

The 99% confidence interval is

p +/- z_{0.005} * sqrt(p(1 - p)/n)

= 0.2 +/- 2.58 * sqrt(0.2 * 0.8/144400)

= 0.2 +/- 0.003

= 0.197, 0.203

We are 99% confident that the true population proportion lies in the above confidence interval.

c) If the sample size cut in half , the 95% confidence interval wil increase by times.

If the sample size cut in half , the 99% confidence interval wil increase by times.

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