You can use the following syntax to check your answers. Note the answers you get in R will not be EXACTLY the same you get by hand but they should be pretty close.

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. phat= 542/3611= 0.15
  2. (3611)(0.15)(0.85)=460.4 greater than/equal to 10 and 3611 is less than/equal to 0.05N
  3. alpha=0.1 z0.05= 1.645; Lower Bound= 0.140; Upper Bound= 0.160
  4. We are 90% confident that the proportion of Americans 18 years and older that have used a smartphone to make a purchase is between 0.140 and 0.160

26.

  1. phat=496/1153= 0.430
  2. 1153(0.43)(0.57)=282.6 is greater than/equal to 10 and 1153 is less than/equal to 0.05N
  3. Lower Bound = 0.401; Upper Bound = 0.459
  4. we are 95% confident that the proportion of US workers/retirees 25 years and older that have less than 10,000 in savings is between 0.401 and 0.459

27.

  1. phat=521/1003= 0.52
  2. 1003(0.52)(0.48)=250.3 is greater than/equal to 10 and 1003 is less than/equal to 0.05N
  3. Lower Bound = 0.489; Upper Bound= 0.551
  4. yes it is possible because the true proportion may not be captured in the confidence interval but it is not likely
  5. Lower Bound = 0.449; Upper Bound = 0.511

28.

  1. phat= 768/1024= 0.75
  2. 1024(0.75)(0.25) is greater than/equal to 10 and 1024 is less than/equal to 0.05N
  3. Lower Bound = 0.715; Upper Bound = 0.785
  4. Yes it is possible but it is not likely at all
  5. Lower Bound = 0.215; Upper bound = 0.285

29.

  1. 944/1748= 0.54
  2. 1748(0.54)(0.46)= 434.2 is greater than/equal to 10 and 1748 is less than/equal to 0.05N
  3. Lower Bound = 0.520; Upper Bound = 0.560
  4. Lower Bound = 0.509; Upper Bound = 0.571
  5. Increasing the level of confidence widens the interval