Question

At 90% confidence, how large a sample should be taken to obtain a margin of error of 0.011 for the estimation of a population proportion? Assume that past data are not available for developing a planning value for p. Round up to the next whole number

Answer 1

Margin of error = 0.041

confidence interval = 90% = 0.90

We don't know population proportion. So we assume it is 50% that is $\hat{p}$ = 0.5

$\hat{q}$ = 1 - $\hat{p}$ = 1 - 0.5 = 0.5

α = 1 - 0.90 = 0.10

P value = 1 - (α /2)

= 1 - (0.10/2)

= 1 - 0.05

= 0.95

From z score table, we have  

Critical value Zα/₂ = 1.65

Zα/₂ $\sqrt{[\hat{p}\hat{q}/n]}$ = 0.041

1.65 $\sqrt{([0.5)(0.5)/n]}$ = 0.041

$\sqrt{([0.5)(0.5)/n]}$ = 0.041 / 1.65

$\sqrt{([0.5)(0.5)/n]}$ = 0.02484

Squaring on both the sides, we get;

0.25/n = 0.00061703

n = 0.25 / 0.00061703

n = 405.1695  

n = 405