Question

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 386 drivers and find that 310 claim to always buckle up. Construct a 95% confidence interval for the population proportion that claim to always buckle up.

Answer 1

The 95% confidence interval for the population proportion that claim to always buckle up based on a sample of 386 drivers, with 310 affirming they always buckle up, is (0.7645, 0.8407).

Step-by-step explanation:

To construct a 95% confidence interval for the population proportion that claim to always buckle up, we use the sample data provided. We have 310 out of 386 drivers claiming to always buckle up.

The sample proportion (p') is calculated as:

p' = X/n

= 310/386

= 0.8026

We then use the standard error of the proportion which is calculated as:

SE = sqrt(p'(1-p')/n)

= sqrt(0.8026*(1-0.8026)/386)

= 0.0194

Next, we find the z-score for a 95% confidence level, which is approximately 1.96. The margin of error (ME) is:

ME = z*SE

= 1.96*0.0194

= 0.0381

Finally, the confidence interval is:

p' ± ME

= 0.8026 ± 0.0381

= (0.7645, 0.8407)

So the 95% confidence interval for the population proportion is (0.7645, 0.8407).