Question

construct the requested confidence interval.

a survey of 400 non - fatal accidents revealed that 268 involved a distracted driver using some kind of electronic device.

construct a 95% confidence interval for the proportion of non - fatal accidents involving a distracted driver using some kind of electronic device.

(0.60, 0.74)
(0.62, 0.72)
(0.65, 0.69)
(0.64, 0.70)
(0.66, 0.68)

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}$ is calculated as the number of successes (accidents with distracted drivers) divided by the sample size $n$. So $\hat{p}=\frac{268}{400}=0.67$.

Step2: Determine z-score for 95% CI

For a 95% confidence interval, the z-score ($z_{\alpha/2}$) is 1.96 (from standard normal distribution).

Step3: Calculate standard error

The standard error (SE) for a proportion is $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substituting values: $\sqrt{\frac{0.67(1 - 0.67)}{400}}=\sqrt{\frac{0.67\times0.33}{400}}=\sqrt{\frac{0.2211}{400}}\approx\sqrt{0.00055275}\approx0.0235$.

Step4: Calculate margin of error

Margin of error (ME) is $z_{\alpha/2}\times SE = 1.96\times0.0235\approx0.046$.

Step5: Calculate confidence interval

Lower bound: $\hat{p}-ME = 0.67 - 0.046 = 0.624$; Upper bound: $\hat{p}+ME = 0.67 + 0.046 = 0.716$. Rounding to two decimal places, the interval is approximately (0.62, 0.72).

Answer:

(0.62, 0.72)