Question

In: Statistics and Probability

A survey is conducted to estimate the average household income in a large metropolitan area. A...

A survey is conducted to estimate the average household income in a large metropolitan area. A random sample of 150 households in this area yielded an average household income of $55,000 with a standard deviation of $13,200. Construct a 95% confidence interval for the mean household income in this area. (Choose the BEST answer)

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Expert Solution

Solution :


Given that,

Point estimate = sample mean =     = 55000

Population standard deviation =    = 13200

Sample size n =150

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E =   Z/2    * ( /n)
= 1.96 * ( 13200 / 150)

= 2112.4
At 95% confidence interval
is,

- E < < + E

55000  - 2112.4 <   <55000 + 2112.4

52887.6 <   < 57112.4


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