Question

An economist is interested in studying the incomes of consumers in a particular country. The population standard deviation is known to be $1,000. A random sample of 49 individuals resulted in a mean income of $16,000. Construct 99% confidence interval interval for the average income.

Answer the following questions

a. i) Sample mean = ?    a. ii) critical value Z = ?    a. iii) Standard error = ?    a. iv) Margin of error = ?

a. v) Lower limit = ?    a. vi) upper limit = ?  

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 16000

Population standard deviation =    = 1000

Sample size = n =49

ii) critical value Z =

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576   ( Using z table )

. iii) Standard error = ( /n) = ( 1000/ 49) =142.86

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

= 2.576 * ( 1000/ 49)

= 368

At 99% confidence interval is,

- E < < + E

16000-368 < < 16000+368

15632< < 16368

v) Lower limit =15632    a. vi) upper limit = 16368