Question

a survey of 1,700 licensed drivers revealed that 33% have received at least 1 speeding ticket in the last 2 years. construct a 99% confidence interval for the population mean of drivers that have received at least 1 speeding ticket in the last 2 years. ci = (30.06%, 35.94%) ci = (31.12%, 34.88%) ci = (31.67%, 34.23%) ci = (30.76%, 35.24%)

Explanation:

Step1: Identify the formula for confidence interval for proportion

The formula for a confidence interval for a population proportion \( p \) is:
\( \hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \)
where \( \hat{p} \) is the sample proportion, \( n \) is the sample size, and \( z_{\alpha/2} \) is the critical value from the standard normal distribution.

Step2: Determine the values

• Sample proportion \( \hat{p} = 0.33 \) (33%)
• Sample size \( n = 1700 \)
• For a 99% confidence interval, \( \alpha = 1 - 0.99 = 0.01 \), so \( \alpha/2 = 0.005 \). The critical value \( z_{0.005} \approx 2.576 \) (from standard normal table).

Step3: Calculate the standard error

Standard error \( SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.33(1 - 0.33)}{1700}} = \sqrt{\frac{0.33 \times 0.67}{1700}} \)
\( SE = \sqrt{\frac{0.2211}{1700}} \approx \sqrt{0.00013006} \approx 0.0114 \)

Step4: Calculate the margin of error

Margin of error \( ME = z_{\alpha/2} \times SE = 2.576 \times 0.0114 \approx 0.0294 \)

Step5: Calculate the confidence interval

Lower bound: \( \hat{p} - ME = 0.33 - 0.0294 = 0.3006 \) (30.06%)
Upper bound: \( \hat{p} + ME = 0.33 + 0.0294 = 0.3594 \) (35.94%)

Answer:

CI = (30.06%, 35.94%)