Question

An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in California. He believes that the mean income is $21.3, and the variance is known to be $32.49. How large of a sample would be required in order to estimate the mean per capita income at the 99% level of confidence with an error of at most $0.39? Round your answer up to the next integer.

Answer 1

Solution:

Given ,

2 = 32.49

E = 0.39 Margin of error

c = 99% = 0.99 ...confidence level

Find sample size required.

c = 0.99

= 1- c = 1- 0.99 = 0.01

  /2 = 0.005

Using Z table ,

= 2.576

Now, sample size (n) is given by,

n = ()2 * (2)/(E2)

= [(2.576)2 * 32.49)]/[0.392]

= 1417.46457751

= 1418 ..(round to the next whole number)

Answer : Required Sample size is n = 1418