Question

An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in California. Suppose that the mean income is found to be $22.4 for a random sample of 2128 people. Assume the population standard deviation is known to be $9.3. Construct the 98% confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal place.

Answer 1

[21.9, 22.9] in thousands of dollars

Explanation:

The formula for Confidence Interval

= Mean ± z score × standard deviation/√n

Mean = $22.4

Standard deviation = $9.3

n = 2128

z score of 98% confidence interval = 2.326

Confidence Interval = 22.4 ± 2.326 × 9.3/√2128

= 22.4 ± 0.469

Confidence Interval =

22.4 - 0.469

= 21.931

≈ 21.9 in thousand of dollars

22.4 + 0.469

= 22.869

≈ 22.9 in thousands of dollars

Hence, the 98% confidence interval =

[21.9, 22.9] in thousands of dollars