The following problems are copied from the chapter 16 exercises from Introduction to Modern Statistics First Edition by Mine Çetinkaya-Rundel and Johanna Hardin (https://openintro-ims.netlify.app/inference-one-prop.html)

Show as much work as possible. Indicate what values you use, n =, z =, p =, when appropriate.

  1. Coupons driving visits. A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they’d received in the mail. Using a mathematical model, construct a 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they’d received in the mail.

ANSWER: .235 + or - 1.96*.017= (.203, .271). I am 95% confident that the proportion of shoppers who visit a certain shop because they received a coupon in the mail lies in the interval (.203, .271).

1-sample proportions test without continuity correction

data:  142 out of 603, null probability 0.5
X-squared = 168.76, df = 1, p-value < 2.2e-16
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
 0.2033633 0.2709640
sample estimates:
        p 
0.2354892
  1. Will the coronavirus bring the world closer together? An April 2021 YouGov poll asked 4,265 UK adults whether they think the coronavirus bring the world closer together or leave us further apart. 12% of the respondents said it will bring the world closer together. 37% said it would leave us further apart, 39% said it won’t make a difference and the remainder didn’t have an opinion on the matter. (YouGov 2021)
  1. Calculate, using a mathematical model, a 90% confidence interval for the proportion of UK adults who think the coronavirus will bring the world closer together, and interpret the interval in context.

ANSWER: sqrt(p(1-p)/n))=sqrt(.12(1-.12)/4265)= .0050. I am 90% Confident that the proportion of UK adults who think coronovirus will bring the world closer together lies within the interval (.112,.128)

prop.test(x = 512, n = 4265, correct=FALSE, conf.level = .90)
1-sample proportions test without continuity correction

data:  512 out of 4265, null probability 0.5
X-squared = 2462.9, df = 1, p-value < 2.2e-16
alternative hypothesis: true p is not equal to 0.5
90 percent confidence interval:
 0.1121008 0.1284747
sample estimates:
        p 
0.1200469
  1. Suppose we wanted the margin of error for the 90% confidence level to be about 0.5%. How large of a sample size would you recommend for the poll? Hint: Use algebra to solve for n \(ME = z^*\sqrt{\frac{p(1-p)}{n}}\)

ANSWER: \(.005 = 1.645^*\sqrt{\frac{.12(1-.12)}{n}}\)= 11430.3 or 11430 people as these are people and one cannot have .3 of a person.

Date and time completed: Mon Oct 10 09:05:04 2022