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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 $⁢22.4 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.

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Expert Solution

Solution :

Given that,

Point estimate = sample mean =   = =22.4

Population standard deviation =    = 9.3

Sample size = n =2128

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 ( Using z table    )

Margin of error = E = Z/2    * ( /n)

= 2.326 * ( 9.3 /  2128 )

= 0.5
At 98% confidence interval estimate of the population mean
is,

- E < < + E

22.4- 0.5 <   < 22.4+ 0.5

(21.9, 22.9)

orchestra answered 2 years ago
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