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 vg 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. 0.150
  2. 460.40>= 10, and the sample is less than 5% of the population.
  3. (0.140,0.160)
  4. We are 90% confident that the proportion of adult Americans 18 years and older who have used their smartphones to make a purchase is between 0.140 and 0.160.

26.

  1. 0.430
  2. 282.60>=10, and the sample is less than 5% of the population.
  3. (0.401,0.459)
  4. We are 95% confident that the proportion of workers and retirees in the United States 25 years of age and older who have less than $10,000 in savings is between 0.401 and 0.459.

27.

  1. 0.519
  2. 250.39>=10, and the sample is less than 5% of the population.
  3. (0.488,0.550)
  4. Yes. It is possible that the population proportion is more than 60%, because it is possible that the true proportion is not captured in the confidence interval. It is not likely.
  5. (0.450,0.512)

28.

  1. 0.75
  2. 192.00>=10, and the sample is less than 5% of the population.
  3. (0.715,0.785)
  4. Yes. It is possible that the population proportion below 70%, because it is possible that the true proportion is not captured in the confidence interval. It is not likely.
  5. (0.215,0.285)

29.

  1. 0.540
  2. 434.20>=10, and the sample is less than 5% of the population.
  3. (0.520,0.560)
  4. (0.509,0.571)
  5. Increasing the level of confidence widens the interval.