Answered

Discover how IDNLearn.com can help you learn and grow with its extensive Q&A platform. Ask your questions and receive accurate, in-depth answers from our knowledgeable community members.

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.

Sagot :

Abidemiokinidn

A 95% confidence interval for the proportion of non-fatal accidents involving a distracted driver using some kind of electronic device is 62.5%<x<71.61%

The 95% confidence interval for the proportion is expressed according to the formula;

[tex]CI=p\pm z\sqrt{\frac{p(1-p)}{n} }[/tex]

p is the proportion = 268/400 = 0.67

z is the 95% z-score = 1.96

n is the sample space = 400

Substitute the given parameters into the formula to have:

[tex]CI=0.67\pm 1.96\sqrt{\frac{0.67(1-0.67)}{400} }\\CI=0.67\pm 1.96\sqrt{\frac{0.67(0.33)}{400} }\\CI=0.67\pm 1.96(0.0235)\\CI=0.67 \pm 0.0461\\CI=(0.67-0.0461, 0.67+0.0461)\\CI=(0.624, 0.7161)\\CI = (62.4 \%, 71.61 \%)[/tex]

Hence a 95% confidence interval for the proportion of non-fatal accidents involving a distracted driver using some kind of electronic device is 62.5%<x<71.61%

Learn more on confidence interval here: https://brainly.com/question/17212516

Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for choosing IDNLearn.com. We’re committed to providing accurate answers, so visit us again soon.