Question

In the year 2001 Youth Risk Behavior survey done by the U.S. Centers for Disease Control, 747 out of 1168 female 12th graders said they always use a seatbelt when driving. Please construct a 98% confidence interval for the proportion of 12th-grade females in the population who always use a seatbelt when driving. 1) Use R to find the score CI for the proportion of 12th-grade females in the population who always use a seatbelt when driving (please round to 4 decimals).

Answer 1

98% of confidence interval of Population proportion is determined by

(0.60683 ,0.67217)

Explanation:

Given the sample size n = 1168

Given 747 out of 1168 female 12th graders said they always use a seatbelt when driving

The sample proportion 'p' =
`(x)/(n) = (747)/(1168) = 0.6395`

98% of confidence interval of Population proportion is determined by

`p^(-) -Z_{(\alpha )/(2) } \sqrt{(p^(-) (1-p^-))/(n) }, p^- +Z_{(\alpha )/(2) } \sqrt{(p^-(1-p^-))/(n) } }`

The tabulated value Z₀.₀₂ score = 2.326

`(0.6395 -2.326 \sqrt{(0.6395(1-0.6395))/(1168) }, 0.6395 +2.326 \sqrt{(0.6395(1-0.6395))/(1168) } })`

on calculation , we get

(0.6395 -0.03267 , 0.6395 +0.03267)

(0.60683 ,0.67217)