Question

An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in Texas. Suppose that the mean income is found to be $21.8 for a random sample of 1691 people. Assume the population standard deviation is known to be $9.8. Construct the 99% confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal place

Answer #1

Sample size = n = 1691

Sample mean = = 21.8

Population standard deviation = = 9.8

We have to construct 99% confidenc interval for the population mean.

Here population standard deviation is known so we have to use one sample z-confidence interval.

z confidence interval

Here E is a margin of error

Zc = 2.58 ( Using z table)

So confidence interval is ( 21.8 - 0.6149 , 21.8 + 0.6149) =
> **( 21.2 , 22.4)**

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An economist wants to estimate the mean per capita income (in
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