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Insurance companies are interested in knowing the population percent of drivers who always buckle up before...

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 420 drivers and find that 286 claim to always buckle up. Construct a 84% confidence interval for the population proportion that claim to always buckle up. Use interval notation, for example, [1,5]

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Answer #1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 286 / 420 = 0.681

1 - = 0.319

Z/2 = 1.405

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.405 * (((0.681 * 0.319) / 420)

= 0.032

A 84% confidence interval for population proportion p is ,

- E < p < + E

0.681 - 0.032 < p < 0.681 + 0.032

0.649 < p < 0.713

The 84% confidence interval for the population proportion p is : [0.649 , 0.713]

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