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. .181
  2. np(1-p)= 341.8>=10, and the sample is less than 5% of the population
  3. lower bound: .168, upper bound: .194
  4. we are 90% confident that the proportion of adult americans 18 years and older who have donated blood in the past 2 years is between .168 and .194

26.

  1. Type answer here.
  2. Type answer here.
  3. Type answer here.
  4. Type answer here.

27.

  1. .519
  2. np(1-p)= 250.39>=, and the sample is less than 5% of the population
  3. lower bound: .488, upper bound: .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. lower bound: .450, upper bound: .512

28.

  1. Type answer here.
  2. Type answer here.
  3. Type answer here.
  4. Type answer here.
  5. Type answer here.

29.

  1. lower bound: .071, upper bound: .194
  2. lower bound: .058, upper bound: .164
  3. as the level of confidence increases, the margin of error increases.