Question

Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the proportions of boys and girls under 10 years old who are afraid of spiders. Assume the "worst case" scenario for the value of both sample proportions. We want a 99% confidence level and for the error to be smaller than 0.02

Answer 1

Answer:

The sample size n = 4225

Step-by-step explanation:

We will use maximum error formula = $\frac{z_{a} S.D}{\sqrt{n} }$

but we will find sample size "n"

$\sqrt{n} = \frac{z_{a} S.D}{maximum error}$

Squaring on both sides , we get

$n = (\frac{z_{a} S.D}{maximum error})^2$

Given 99% confidence interval (z value) = 2.56

given maximum error = 0.02

$n = (\frac{z_{a} p(1-p)}{maximum error})^2$

n≤ $(\frac{2.56X\frac{1}{2} }{0.02}) ^{2}$    ( here S.D = p(1-p) ≤ 1/2

on simplification , we get n = 4225

Conclusion:

The sample size of two samples is  n = 4225

verification:-

We will use maximum error formula = $\frac{z_{a} S.D}{\sqrt{n} }$ = $\frac{2.56 1/2}{\ \sqrt{4225} }$  = 0.0196

substitute all values and simplify we get maximum error is 0.02