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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.

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abidemiokin

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

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