Question

1. You want to be able to estimate the average IQ of people in Winston-Salem. You would be satisfied knowing the average IQ plus-or-minus 5 IQ points (using a 90% confidence interval). The standard deviation for IQ tests is typically 15 points; so you’ll use that as a guideline. How many people would you need to survey in order to be able to estimate the average to your desired degree of accuracy?

Answer 1

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Solution

standard deviation =s =   =15

Margin of error = E = 5

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 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*15 / 5)2

n =34.5744

Sample size = n =35