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.
- 15%
- np(1-p) = 460
- .140 and .160
- We are 90% confident that the proportion of Americans 18+ who have bought something with their smartphone is between .140 and .160.
26.
- 43%
- np(1-p) = 283
- .401 and .459
- We are 95% confident that the proportion of 25+ workers/retirees that have less than 10k in savings is between .401 and .459.
27.
- 52%
- np(1-p) = 250
- .489 and .551
- Possible but super unlikely.
- .449 and .511
28.
- 75%
- np(p-1) = 192
- .715 and .785
- Possible, not super likely.
- .215 and .285
29.
- 54%
- np(p-1) = 434
- .520 and .560
- .509 and .571
- It widens it.