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. Point estimate p hat is 417/2306= 0.181
  2. n is a random sample smaller than 5% of adult Americans aged 18+, and 341.8 is greater than 10.
  3. (0.168,0.194).
  4. We are 90% confident that the proportion of adult Americans 18 years+ who have donated blood in the past 2 years is between 0.168 and 0.194.

26.

  1. Point estimate p hat is 496/1153=0.430
  2. n is a random sample smaller than 5% of the total population size, and 282.6003> 10.
  3. (0.401,0.459).
  4. We are 95% confident that the proportion of workers and retirees in the U.S. 25 years old+ who have less than $10,000 in savings is between 0.401 and 0.459.

27.

  1. Point estimate p hat is 521/1003= 0.519.
  2. n is a random sample smaller than 5% of the total population size, and 250.388>10.
  3. (0.488,0.550) We are 95% confident that the proportion of adult Americans who believe that televisions are a luxury they could do without is between 0.488 and 0.550.
  4. Yes, it is possible that a supermajority of Americans believe that television is a luxury they could do without because it’s possible it’s not in the confidence interval, although it is unlikely.
  5. (0.450, 0.512).

28.

  1. Point estimate p hat is 768/1024= 0.75.
  2. n is a random sample smaller than 5% of total population size and 192>10.
  3. (0.715, 0.785) We are 99% confident that the proportion of adult Americans aged 18+ for which the issue of family values is extremely or very important in determining their vote for president is between 0.715 and 0.785.
  4. It is possible that it is not in the confidence interval, however it is very unlikely.
  5. (0.215,0.285).

29.

  1. (0.071,0.151).
  2. (0.058,0.164).
  3. The level of confidence increases and the margin of error increase simultaneously.