Question

The Corporate Lawyer, a magazine for corporate lawyers, would like to report the mean amount earned by lawyers in their area of specialization. How large a sample is required if the 95 percent level of confidence is used and the estimate is to be within $5,000? The population standard deviation is $20,000.

Answer 1

standard deviation =s =   = 20000

Margin of error = E = 5000

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96 ( Using z table ( see the 0.025 value in standard normal (z) table corresponding z value is 1.96 )

sample size = n = [Z_/2* / E] ²

n = ( 1.96* 20000/ 5000)²

n =61.4656

Sample size = n =62