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 387 drivers and find that 287 claim to always buckle up. Construct a 91 % confidence interval for the population proportion that claim to always buckle up. Use interval notation, for example, [1,5] ___________

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Expert Solution

Sample proportion = 287 / 387 = 0.742

= 1 - 0.91 = 0.09

Z/2 = Z0.09/2 = Z0.045 = 1.6954

91% confidence interval for p is

- Z/2 * sqrt [ ( 1 - ) / n ] < p < + Z/2 * sqrt [ ( 1 - ) / n ]

0.742 - 1.6954 * sqrt [ 0.742 * ( 1 - 0.742) / 387 ] < p < 0.742 + 1.6954 * sqrt [ 0.742 * ( 1 - 0.742) / 387 ]

0.704 < p < 0.780

91% CI is ( 0.704 , 0.780 )

orchestra answered 1 year ago

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