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. 341.84 is greater or equal to 10 – the sample is less than 5% of the population
  3. .181 +/- 1.645 x root ((.181(1-.181))/2306) = .168 and .194
  4. 90% confident that the proportion of the population of Amercians aged 18+ who have donated blood in the last two years is between .168 andd .194
binom.test(x = 417, n = 2306,  conf.level = .90)[["conf.int"]]
## [1] 0.1677377 0.1945310
## attr(,"conf.level")
## [1] 0.9

26.

  1. 496/1153 = .430
  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. .430 +/- 1.96 x root ((.430(1-.430))/1153) = .402 and .495
  4. 95% confident that the proportion of the population of workers and retirees aged 25+ who have less than $10,000 in savings is between .402 and .495
binom.test(x = 496, n = 1153,  conf.level = .95)[["conf.int"]]
## [1] 0.4013800 0.4593413
## attr(,"conf.level")
## [1] 0.95

27.

  1. 529/1003 = .519
  2. n x p hat (1-p hat) = 250.39 which is greater than or equal to 10 and the sample is less than 5% of the population.
  3. .519 +/- 1.96 x root ((.519(1-.519))/1003) = .488 and .550
  4. It is possible because the true proportion could not be captured in our confidence interval, but it is unlinkely because .6 is outside of our interval.
  5. lower bound -> .488 upper bound -> .550
binom.test(x = 529, n = 1003,  conf.level = .95)[["conf.int"]]
## [1] 0.4959804 0.5586938
## attr(,"conf.level")
## [1] 0.95

28.

  1. 768/1024 = .75
  2. n x p hat (1-p hat) = 192 which is greater than or equal to 10 and the sample is less than 5% of the population.
  3. .75 +/- 2.575 x root ((.75(1-.75))/1024) = .715 and .785
  4. It is possible because it is just outside of our confidence interval but not very likely due to our high confidence interval
  5. lower bound -> .215 upper bound -> .285
binom.test(x = 768, n = 1024,  conf.level = .99)[["conf.int"]]
## [1] 0.7135318 0.7841063
## attr(,"conf.level")
## [1] 0.99

29.

  1. 26/234 = .111 .111 +/- 1.96 x root ((.111(1-.111))/234) = lower .071 and upper .151
  2. .111 +/- 2.575 x root ((.111(1-.111))/234) = lower .058 and upper .164
  3. an increase in confidence level increases the m.o.e.
binom.test(x = 26, n = 234,  conf.level = .95)[["conf.int"]]
## [1] 0.07387792 0.15855410
## attr(,"conf.level")
## [1] 0.95
binom.test(x = 768, n = 1024,  conf.level = .99)[["conf.int"]]
## [1] 0.7135318 0.7841063
## attr(,"conf.level")
## [1] 0.99