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 415 drivers and find that 286 claim to always buckle up. Construct a Invalid variable% confidence interval for the population proportion that claim to always buckle up.

Use interval notation, for example, [1,5]

Solutions

Expert Solution

Solution:

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

x = 286

n = 415

P = x/n = 286/415 = 0.689156627

Confidence level = 95% (by default)

Critical Z value = 1.96

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.689156627 ± 1.96* sqrt(0.689156627*(1 – 0.689156627)/415)

Confidence Interval = 0.689156627 ± 0.0445

Lower limit = 0.689156627 - 0.0445 = 0.6446

Upper limit = 0.689156627 + 0.0445 = 0.7337

Confidence interval = [0.6446, 0.7337]


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