Question

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]

Answer 1

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]