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. .1808
  2. 341.84 >/= 10; sample is less than 5% of the population
  3. (0.1676,.1939)
  4. We are 90% confident that the population proportion will be between 0.1676 and 0.1939

26.

  1. 0.4302
  2. 282.6 >/= 10; sample is less than 5% of the population
  3. (.4016, .4588)
  4. We are 95% confident that the population proportion will be between .4588 and .4016

27.

  1. 0.5194
  2. 250.4 >/= 10, sample is less than 5% of the population
  3. (.4885, 0.5503)
  4. yes, but it is not likely
  5. (.4497, .5115)

28.

  1. .75
  2. 192 >/= 10, sample is less than 5% of the population
  3. (0.7152, .7848)
  4. yes but it is not likely
  5. (0.2152, .2848)

29.

  1. (.0708, .1514)
  2. (0.0582, 0.1640)
  3. Margin of error went from .0403 to to 0.0529.