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In: Statistics and Probability

An economist wants to estimate the mean per capita income (in thousands of dollars) for a...

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 $22.6$ for a random sample of 2692 people. Assume the population standard deviation is known to be $12.4$. Construct the 98% confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal place.

Solutions

Expert Solution

Given that,

= 22.6

= 12.4

n =2692

At 98% confidence level the z is ,

  = 1 - 98% = 1 - 0.98= 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01= 2.326

Margin of error = E = Z/2* ( /n)

=2.326 * (12.4 / 2692)

= 0.6

At 98% confidence interval estimate of the population mean is,

- E < < + E

22.6 - 0.6 < < 22.6 + 0.6

22.0< < 23.2

(22.0,23.2 )

orchestra answered 2 years ago
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