Question

In a certain small city, the Mayor would like to know the average income of 25000 families. A survey is conducted and took a simple random sample of 900 families. The average income of these families is 64,500 dollars. The SD of the 900 incomes is $32,500. Thus, the Mayor concludes that the average income of the city is...... give or take …...

The 95 percent confident interval for the average incomes of all the families is from..... dollars to...…. dollars. (Hint: Use the approximate 2 SDs above and below the average instead of the exact 1.96 from the z-table.)

Answer 1

Solution :

Given that,

= $64500

= $32500

n = 900

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

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

= 1.96 * (32500 / 900)

= 2123.3333

At 95% confidence interval estimate of the population mean is,

- E < < + E

64500 - 2123.3333 < < 64500 + 2123.3333

62376.6667< < 66623.3333

(62376.6667 , 66623.3333)