Question

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 391 drivers and find that 280 claim to always buckle up. Construct a 96% confidence interval for the population proportion that claim to always buckle up. Use interval notation

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Answer #1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 280 / 391 = 0.716

1 - = 1 - 0.716 = 0.284

Z/2 = 2.054

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

= 2.054 * (((0.716 * 0.284) / 391)

Margin of error = E = 0.047

A 96% confidence interval for population proportion p is ,

- E < p < + E

0.716 - 0.047 < p < 0.716 + 0.047

0.669 < p < 0.763

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