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 sample is less than 5% of population
  3. lower bound: .168, upper bound: .194
  4. we are 90% confident that the proportion of of americans 18 years and older who have donated blood in the past 2 years is between .168 and .194

26.

  1. .43
  2. np (1-p)= 282, greater than 10
  3. lower fence: .401 upper fence .459
  4. we have 95 percent confidence that that the proportion of workers and retirees in the US 25 years of age and older that have less than 10000 in savings are between .401 and .459

27.

  1. .519
  2. np(1-p)= 250.12 which is greater than 10
  3. lower bound= .488 upper bound= .550
  4. It is possible. it is possible that the true proportion is not captured in confidence interval however its unlikely.

  5. lower bound: .450 upper bound: .512 28.

  6. .45
  7. np (1-p)=192
  8. upper:.715 lower: .785
  9. its possible, because its possible that the confidence interval does not represent correctly, although its unlikely.

  10. upper .215: lower:.285

29.

  1. lower: .07 upper: .15
  2. lower: .058 upper: .164
  3. as the level of confidence increases, the margin of error increases