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Out of a sample of 1,450 people, 1,000 said they prefer to drive themselves to work rather than carpool. Construct a 95% confidence interval for the population mean of people who would rather drive themselves than carpool.

CI = (65.84%, 72.09%)
CI = (66.58%, 71.35%)
CI = (65.27%, 72. 13%)
CI = (66.97%, 70.96%)

Respuesta :

The required confidence interval for the population mean is; B: CI = (66.58%, 71.35%)

What is the Confidence Interval?

We are given;

Sample size; n = 1450

Sample number; x = 1000

Thus,sample proportion is;

p^ = x/n = 1000/1450

p^ = 0.69

We are not given the population proportion but we will assume; p = 0.5

Formula for confidence interval is;

CI = p^ ± z√(p(1 - p)/n)

Where z for CL of 95% is; z = 1.96

Thus;

CI = 0.69 ± 1.96√(0.5(1 - 0.5)/1450)

CI = 0.69 ± 0.026

CI = (66.58%, 71.35%)

Read more about Confidence intervals at; https://brainly.com/question/26658887

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