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. The requirements for constructing a confidence interval are met because the sample is less than 5% of the population.
  3. LB: .168 UB: .194
  4. We are 90% confident that the population proportion of adult American who are 18 and older who hae donated in the last two years is between .168 and .194.

26.

  1. .430
  2. The requirements for constructing a confidence interval are met because the sample is less than 5% of the population.
  3. LB: .402 UB: .459
  4. We are 95% confident that the population proportion of workers and retirees who have less then $10,000 in savings is between .402 and .459.

27.

  1. .5194
  2. The requirements for constructing a confidence interval are met because the sample is less than 5% of the population.
  3. We are 95% confident that the proportion of adult Americans who believe TV is a luxury is between .489 and .550.
  4. It is possible that the population proportion is above 60% because the true proportion may not be accounted for. However, it is quite unlikely because .6 is outside the confidence interval.
  5. LB: .450 UB: .512

28.

  1. .75
  2. The requirements for constructing a confidence interbal are met because the sample is less tha 5% of the population.
  3. We are 99% confident that the proportion of adult Americans 18 and older for which family values are significant in determining votes is between .715 and .785.
  4. It is possible for the population proportion to be above 70% because the true proportion may not be accounted for. However, .7 does not lie inside the confidence interval, so it is unlikely.
  5. LB: .215 UB: .285

29.

  1. LB: .071 UB: .151
  2. LB: .058 UB: .164
  3. As the confidence level increases, the margin of error increases as well.