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 = 496, n = 1153,  conf.level = .95)[["conf.int"]]
## [1] 0.4013800 0.4593413
## attr(,"conf.level")
## [1] 0.95

25.

  1. .18
  2. yes
  3. (.168, .194)
  4. 90% sure that proportion who donated between .168 and .194

26.

  1. .43
  2. np(1-p) < 10
  3. .401, .459
  4. 95% sure that proportion is between 40 and 46%

27.

  1. .181
  2. yes
  3. .488, .55
  4. possible but not likely
  5. .45, .512

28.

  1. .75
  2. nP(1-P) = 192
  3. .714, .784
  4. yes, because confidence interval is very wide.
  5. .216, .287

29.

  1. .071, .151
  2. .058, .164
  3. level of confidence = greater margin of error