Question

Question 18 ? of 20 Step 1 of 1 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 $23.4. and the variance is known to be $129.96. How large of a sample would be required in order to estimate the mean per capita income at the 98 % level of confidence with an error of at most $0.55? Round your answer up to the next integer. Answer(How to Enter) 2 Points Keypad : ?Tables

Answer 1

Solution :

Given that,

Varinace = $\sigma$ ² = 129.96

standard deviation = $\sigma$ = 11.4

margin of error = E = 0.55

At 98% confidence level the z is ,

$\alpha$ = 1 - 98% = 1 - 0.98 = 0.02

$\alpha$ / 2 = 0.02 / 2 = 0.01

Z_$\alpha$/2 = Z_0.01 = 2.326

Sample size = n = ((Z_$\alpha$/2 * $\sigma$ ) / E)²

= ((2.326 * 11.4) / 0.55)²

= 2324.36 = 2325

Sample size = 2325