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. 0.18
  2. 341.84 is greater or equal to 10, so the sample is less than 5% of the population.
  3. 0.193 and 0.168
  4. It’s 90% confident that the proportion of the population of Amercians aged 18+ who have donated blood in the last two years is between 0.168 andd 0.193.

26.

  1. 0.43
  2. n x phat (1-phat) = 282.6 which is greater than or equal to 10 and the sample is less than 5% of the population.
  3. 0.401, 0.459
  4. We are 95% confident that the population proportion of workers and retirees who have less then $10,000 in savings is between 0.401 and 0.459.

27.

  1. 0.52
  2. 250.39 ??? 10. Sample is less than 5% of the population
  3. 0.489, 0.550.
  4. It is possible that the true proportion is not caputured in the confidence interval. It is not liukely because 0.6 is outside of the confidence interval.
  5. (0.450,0.512)

28.

  1. 0.75
  2. 192 ??? 10. The sample is less than 5% of the population
  3. (0.715,0.785)
  4. Possible because it is possible that the true proportion is not captured in the confidence interval. However, it is not likely because 0.70 is outside of the confidence interval.
  5. (0.215,0.285)

29.

  1. (0.07,0.151)
  2. (0.058,0.164)
  3. Increasing the confidence level increases the margin of error.