Question

An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in Texas. He believes that the mean income is $21.9, and the variance is known to be $31.36. How large of a sample would be required in order to estimate the mean per capita income at the 90% level of confidence with an error of at most $0.26? Round your answer up to the next integer.

Answer #1

Solution :

Given that,

variance =
^{2} = 31.36

standard deviation = = 5.6

margin of error = E = 0.26

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_{/2}
= Z_{0.05} = 1.645

Sample size = n = ((Z_{/2}
*
) / E)^{2}

= ((1.645 * 5.6) / 0.26)^{2}

= 1255.33 = 1256

Sample size = 1256

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