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

Answer:

98% of confidence interval of Population proportion is determined by

(0.60683 ,0.67217)

Step-by-step 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' = $\frac{x}{n} = \frac{747}{1168} = 0.6395$

98% of confidence interval of Population proportion is determined by

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

The tabulated value Z₀.₀₂ score  = 2.326

$(0.6395 -2.326 \sqrt{\frac{0.6395(1-0.6395)}{1168} }, 0.6395 +2.326 \sqrt{\frac{0.6395(1-0.6395)}{1168} } })$

on calculation , we get

(0.6395 -0.03267 , 0.6395 +0.03267)

(0.60683 ,0.67217)