Question

3- A recent Gallup poll of a random sample of 1,015 US adults reported that a 95% confidence interval for the population proportion of adults who frequently worry about being the victim of identity theft is (0.32, 0.40).

(a) We are 95% confident that what parameter is contained in this interval? (Circle your answer.) Can we say that there is a probability of 95% that this parameter is actually contained in this interval?

A) sample proportion B) population proportion C) sample size D) margin of error

(b) What statistic is guaranteed to be in this interval? (Circle your answer.) What is the value of this statistic?

A) sample proportion B) population proportion C) sample size D) margin of error

(c) What do we actually mean when we say we are 95% confident?

(d) What is the margin of error, m, for this confidence interval?

(e) What type of distribution was used to obtain this confidence interval? What is the value of the critical value, z, that was used to obtain this confidence interval?

(f) According to this confidence interval, is it plausible that the true percentage of US adults who frequently worry about being the victim of identity theft is 60%?

(g) If you had been planning this study (with no initial knowledge or estimates of the value of the population proportion), what would the minimum sample size needed to guarantee a margin of error of 5%?

Answer 1

a) Option - B) population population proportion

b) Option - A) Sample proportion

c) We are 95% confident that the true population proportion lies between the interval limits 0.32 and 0.4.

d) Margin of error = (0.4 - 0.32)/2 = 0.04

e) We should use z-distribution.

At 95% confidence level, the critical value is z0.025 = 1.96

f) No, it is not plausible, because 0.60 doesn't lie in the confidence interval.

g) Margin of error = 0.05

Or, z0.025 * sqrt(p(1 - p)/n) = 0.05

Or, 1.96 * sqrt(0.5(1 - 0.5)/n) = 0.05

Or, n = (1.96 * sqrt(0.5 * 0.5)/0.05)^2

Or, n = 385