In a randomly selected group of 650 automobile deaths,180 were alcohol related.Construct a 95 percent confidence interval for the true proportion of all automobile accidents caused by alcohol.
A) $[ .243 \quad .311 ]$
B) $\left[ \begin{array} { l l }
.262 & .292
\end{array} \right]$
C) $[ .259 \quad .294 ]$
D) $\left[ \begin{array} { l l }
.263 & .291
\end{array} \right]$

Question

In a randomly selected group of 650 automobile deaths,180 were alcohol related.Construct a 95 percent confidence interval for the true proportion of all automobile accidents caused by alcohol.
A) $[ .243 \quad .311 ]$
B) $\left[ \begin{array} { l l } 
.262 & .292
\end{array} \right]$
C) $[ .259 \quad .294 ]$
D) $\left[ \begin{array} { l l } 
.263 & .291
\end{array} \right]$

Quizplus · Accepted Answer

The correct answer is A because the confidence interval is calculated as:sample proportion +/- margin of errorwhere the margin of error is:z * sqrt((sample proportion * (1 - sample proportion)) / sample size)where z is the z-score corresponding to the desired confidence level. For a 95% confidence level, z = 1.96.Plugging in the values, we get:0.18 +/- 1.96 * sqrt((0.18 * (1 - 0.18)) / 650)= 0.18 +/- 0.034= [0.243, 0.311]

In a Randomly Selected Group of 650 Automobile Deaths,180 Were [.243.311][ .243 \quad .311 ][.243.311]

In a Randomly Selected Group of 650 Automobile Deaths,180 Were $[ .243 \quad .311 ]$