Question

6. Trick or Treat In a survey of a random sample of 35 households in the Cherry Creek neighborhood of Denver, it was found that 11 households turned out the lights and pretended not to be home on Halloween.

1. Compute a 90% confidence interval for p, the proportion of all housholds in Cherry Creek that pretend not to be home on Halloween.
2. What assumptions are necessary to calculate the confidence interval of part (a)?
3. Interpretation The national proportion of all households in the United States that turn out the lights and pretend not to be home on Halloween is 0.28. Is 0.28 in the confidence interval you computed? Based on your answer, does it seem that the Cherry Creek neighborhood is much different (either a higher or a lower proportion) from the population of all U.S. households? Explain.

7. Marketing: Customer Loyalty In a marketing survey, a random sample of 730 women shoppers revealed that 628 remained loyal to their favorite supermarket during the past year (i.e., did not switch stores) (Source: Trends in the United States: Consumer Attitudes and the Supermarket, The Research Department, Food Marketing Institute).

1. Let p represent the proportion of all women shoppers who remain loyal to their favorite supermarket. Find a point estimate for p.
2. Find a 95% confidence interval for p. Give a brief explanation of the meaning of the interval.
3. Interpretation 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? What is the margin of error based on a 95% confidence interval?

Answer 1

Solotion 6) Tsick ox Treat: Given that Sample sige (n) - 35 Mean () a) 35 P0.31 Find the 901. of Confdente ndewal CT 0.90 Significance level (A ): 1- Confidence 1-0.90 0. 0.05 Crilicol Valve of Z Zp.os : l64 E: 1.64 0.31 (1-B.31)

.6uo.0061 0.128 E0.13 The Confidence intenal jn PEPEPIE 0 31-0.13 P< 0.31 +0.13 O 18< P< 044 * 90 of Cfor p 0. 18 to o u4 b)- Conditions yequired ^by a valid large- Sample infefnce foy Proportion from a binomial populadion selected >A andom Sample The Somple Size' ^ n s large C) * Yes, The CT 028, Since 90% Confidence inteaval doY p to ouu 0.18 *No, the chenry Creek neighbothood s neithey nov lawer proportion dom the population of all hasehobls. higher ppaon Manikeling Given hat Somple si3e (n)- 730 Mean (x) 628

a) Find he pont estimate for p 628 730 P 0.8602 b) 951. of confidence Tutenel- c. 0.95 Significonce level (<): 1- Confidence 21-0.9S 0.05 :0.025 E Zc P(1-F) 0.8602 (1-0.8602) 1.96 730 : .960.000l647 E o.02515 The Confidence lenal -E P <P+E 0.8602-0.025/5< p< 08602 0.02515 0.8350 P 0-8853

0 83 P 0.88 951. of CI fox pM o83 to o-88 Tf we had Conputed the intevol or differeut types of many 730 women Shoppers, we would have foord that 951, of the intenais actually Contained P. The Population propoxtion of all uomen shoppers . we kon 95 Confident that ouy interval ^ one of the inlexval thot Contain the Unknown Value P. c) The masimal margin erroy based on a 951. CI 0.8602 (o 1393) q-1-P 730 1-0-8602 o.1398 1.60.000167 E = o02515 The margin of erroy wan 2.5 pecentoge points