Question

What is the minimum sample size required to estimate a population mean with 90% confidence if the population standard deviation is estimated to be 30 and the desired margin of error is 2?

Answer 1

Solution :

Given that,

standard deviation = $\sigma$ = 30

margin of error = E = 2

At 90% confidence level the z is ,

$\alpha$ = 1 - 90% = 1 - 0.90 = 0.10

$\alpha$ / 2 = 0.10 / 2 = 0.05

Z_$\alpha$/2 = Z_0.05 = 1.645

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

= ((1.645 * 30) / 2 )²

= 609

Sample size = 609