Question

2. Basic Computation: Confidence Interval for p Consider n = 200 binomial trials with r = 80 successes.

1. Check Requirements Is it appropriate to use a normal distribution to approximate the p̂ distribution?
2. Find a 95% confidence interval for the population proportion of successes p.
3. Interpretation Explain the meaning of the confidence interval you computed.

3. Basic Computation: Sample Size What is the minimal sample size needed for a 95% confidence interval to have a maximal margin of error of 0.1:

1. if a preliminary estimate for p is 0.25?
2. if there is no preliminary estimate for p?

4. Basic Computation: Sample Size What is the minimal sample size needed for a 99% confidence interval to have a maximal margin of error of 0.06:

1. if a preliminary estimate for p is 0.8?
2. if there is no preliminary estimate for p?

5. Myers-Briggs: Actors Isabel Briggs Myers was a pioneer in the study of personality types. The following information is taken from MBTI Manual: A Guide to the Development and Use of the Myers-Briggs Type Indicator by Myers and McCaulley (Consulting Psychologists Press). In a random sample of 62 professional actors, it was found that 39 were extroverts.

1. Let p represent the proportion of all actors who are extroverts. Find a point estimate for p.
2. Find a 95% confidence interval for p. Give a brief interpretation of the meaning of the confidence interval you have found.
3. Check Requirements Do you think the conditions np > 5 and nq > 5 are satisfied in this problem? Explain why this would be an important consideration.

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

Ñ - 80 - 0.40 200 * . a) to approximate binomial using normal distribution * np>, 10 & n(1-p)>, 10 * everytrial should be independent of the other following conditions are satisfied. b) CIE ſ + Zaplano 0.40 for 95% confidence Za = 1.96 (Ie 0.40 $ 1.9650.40(1-0.40) 1 = 0.40 I 0.07 CI -(0.33, 0.17) We are as% confident that the population proportion/ will lie within the interval calculated above (0.33,0.47) 3) Mangin of ersoor (ME) - Za Plume) 1. for 95% confidence Za = 1.96 a) if p=0.25 ME - 1.96 0.25(1-0.25) 0.1 = 1,96 0.25*0.75

M = (1.96 )ÃO.25 * 0.75 ~ 73 (sounded up Lol . b) if there is no pre-estimate for p than assume p=0.5 then n= _*0.5*0.5 ñ 97(rounded