Question

In: Statistics and Probability

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 410 drivers and find that 283 claim to always buckle up. Construct a 91% confidence interval for the population proportion that claim to always buckle up.

Solutions

Expert Solution

Solution :

Given that,

n = 410

x = 283

Point estimate = sample proportion = = x / n = 283/410=0.690

1 -   = 1-0.690 =0.31

At 91% confidence level the z is ,

= 1 - 91% = 1 - 0.91 = 0.09

/ 2 = 0.09 / 2 = 0.045

Z/2 = Z0.045 = 1.695   ( Using z table )

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

= 1.695 (((0.69*0.31) /410 )

E = 0.039

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.69-0.039 < p < 0.69+0.039

0.651< p <0.729

91% confidence interval for the population proportion that claim to always buckle up (0.651 to 0.729)


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