Question

For this problem, carry at least four digits after the decimal
in your calculations. Answers may vary slightly due to
rounding.

In a marketing survey, a random sample of 734 women shoppers
revealed that 640 remained loyal to their favorite supermarket
during the past year (i.e., did not switch stores).

(a) Let *p* represent the proportion of all women
shoppers who remain loyal to their favorite supermarket. Find a
point estimate for *p*. (Round your answer to four decimal
places.)

(b) Find a 95% confidence interval for *p*. (Round your
answers to three decimal places.)

lower limit | |

upper limit |

Give a brief explanation of the meaning of the interval.

- 95% of all confidence intervals would include the true proportion of loyal women shoppers.

- 5% of the confidence intervals created using this method would include the true proportion of loyal women shoppers.

- 5% of all confidence intervals would include the true proportion of loyal women shoppers.

- 95% of the confidence intervals created using this method would include the true proportion of loyal women shoppers.

(c) As a news writer, how would you report the survey results
regarding the percentage of women supermarket shoppers who remained
loyal to their favorite supermarket during the past year?

Report the margin of error.Report
*p̂*. Report *p̂* along with
the margin of error.Report the confidence interval.

What is the margin of error based on a 95% confidence interval?
(Round your answer to three decimal places.)

Answer #1

a)

sample proportion, = 0.8719

b)

sample size, n = 734

Standard error, SE = sqrt(pcap * (1 - pcap)/n)

SE = sqrt(0.8719 * (1 - 0.8719)/734) = 0.0123

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

CI = (pcap - z*SE, pcap + z*SE)

CI = (0.8719 - 1.96 * 0.0123 , 0.8719 + 1.96 * 0.0123)

CI = (0.848 , 0.896)

lower limit = 0.848

Upperlimit = 0.896

95% of the confidence intervals created using this method would include the true proportion of loyal women shoppers.

c)

Margin of Error, ME = zc * SE

ME = 1.96 * 0.0123

ME = 0.024

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