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.

[21.9, 22.9] in thousands of dollars

Step-by-step 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

javeir earned 56 dollors more than kyle so kyle and javier had 14w but javier made 56 dollors more so 14w plus 56 equals 70w so kyle has 14w

step-by-step explanation: hope i plz give me brainliest with the points so this is not constant cause it says last week so its clearly not every week