You can use the following syntax to check your answers. Note the answers you get in R will not be EXACTLY the same you get by hand but they should be pretty close.

If your \(x = 542\) (ie the number of “successes” and your \(n = 3611\), this is how you can have R calculate a 90% Confidence Interval for you.

binom.test(x = 542, n = 3611,  conf.level = .90)[["conf.int"]]
## [1] 0.1403939 0.1602214
## attr(,"conf.level")
## [1] 0.9

25.

  1. The point estimate would be .15 (15%)
  2. nPhat(1-Phat) = 460.04 >/= 10, which means the sample size is less than 5% of the population
  3. Lower bound: .14 and upper bound: .16
  4. There is a 90% confidence that the proportion of adults 18+ who used their smartphone to make a purchase is between 14 and 16.

26.

  1. Point estimate would be .43
  2. nPhat(1-Phat)= 282.60 >/= 10, and the samlpe size is less than %5 of the population
  3. Lower bound: .42 and upper bound: .44
  4. There is a 95% confidence that the proportion of workers and retirees in the US 25+ who have less thn $10,000 in savings is between 42 and 44.

27.

  1. The point estimate is .519
  2. nPhat(1-nPhat) = 250.34 >/= 10, and the sample size is less than 5% of the population
  3. Lower bound is .488 and upper bound is .550
  4. Yes because it is possible that the actual proportion is not portrayed in the confidence interval, which can indicate that it is not likely.
  5. Lower bound is .450 and upper bound is .512.

28.

  1. The point estimate is .75.
  2. nPhat(1-nPhat) = 192 >/= 1, and the sample size is less than .5% of the population.
  3. Lower bound is .709 and the upper bound is .790
  4. Yes because it is possible that the actual proportion is not portrayed in the confidence interval.
  5. There is a 90% confidence that the proportion of workers and retirees in the US.

29.

  1. The point estimate is .540
  2. nPhat(1-nPhat) = 434.20 >/= 10, and the sample size is less than 5% of the population
  3. Lower bound is .520 and the upper bound is .560
  4. Lower bound is .509 and the upper bound is .571
  5. Yes because it is possible that the true porportion is not portrayed in the confidence interval, which makes it not likely.