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 $36.6 for a random sample of 1012 people. Assume the population standard deviation is known to be $8.5. Construct the 80% confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal place.

Answer 1

Solution :

Given that,

Point estimate = sample mean = =$36.6

Population standard deviation = = $8.5

Sample size = n = 1012

At 80% confidence level the z is ,

= 1 - 80% = 1 - 0.80 = 0.2

/ 2 = 0.2 / 2 = 0.1

Z/2 = Z0.1 = 1.282

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

= 1.282 * (8.5 / 1012)

= 0.34

At 80% confidence interval estimate of the population mean is,

- E < < + E

36.6 - 0.3< < 36.6 + 0.3

36.3 < < 36.9

lower endpoint = 36.3

upper endpoint = 36.9

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