Question

The article "Nine Out of Ten Drivers Admit in Survey to Having Done Something Dangerous" (Knight Ridder Newspapers, July 8,2005\()\) reported on a survey of 1,100 drivers. Of those surveyed, 990 admitted to careless or aggressive driving during the previous 6 months. Assume that the sample is representative of the population of drivers. Answer the four key questions (QSTN) to confirm that the suggested method in this situation is a confidence interval for a population proportion.

Short answer

The 95% confidence interval for the population proportion is (0.882, 0.918).

Step-by-step solution

Determine the Sample Proportion

First, find the sample proportion (p̂) by dividing the number of drivers admitting to careless or aggressive driving by the total number surveyed: p̂ = 990 / 1100 = \(0.9\) .

Confirm that the Sample Size Is Large Enough

Confidence intervals for population proportions require that the sample size be large enough to assume a normal distribution. This can be confirmed with the following checks: np̂ ≥ 10 and n(1 - p̂) ≥ 10. Here, np̂= 1100*0.9 = 990 and n(1 - p̂) = 1100*(1-0.9) = 110, both are greater than 10, so the assumption is Valid.

Calculate the Margin of Error

Typically, confidence intervals are calculated with a 95% level of confidence. The formula for the confidence interval is: \(p̂ ± Z * √((p̂*(1 - p̂))/n)\). Here Z is Z-value from the Z-table for the desired level of confidence. For a 95% confidence level, Z = 1.96. So, the margin of error = 1.96 * √((0.9*(1 - 0.9))/1100) = 0.018.

Calculate the Confidence Interval

Now, the confidence interval can be calculated using the sample proportion and margin of error: \(0.9 ± 0.018\). So, the 95% confidence interval for the population proportion is (0.882, 0.918).