Question

Finite Population Correction Factor For Formulas 7-2 and 7-3 we assume that the population is infinite or very large and that we are sampling with replacement. When we sample without replacement from a relatively small population with size N, we modify E to include the finite population correction factor shown here, and we can solve for n to obtain the result given here. Use this result to repeat part (b) of Exercise 38, assuming that we limit our population to a county with 2500 women who have completed the time during which they can give birth.

$E=z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}\hat{q}}{n}}\sqrt{\frac{N-n}{N-1}}$

$n=\frac{N\hat{p}\hat{q}{\left(z_{\frac{\alpha }{2}}\right)}^2}{\hat{p}\hat{q}{\left(z_{\frac{\alpha }{2}}\right)}^2+\left(N-1\right)E^2}$

Short answer

The sample size required is equal to 1238.

Step-by-step solution

Given information

It is given that for a finite population of 2500 women, 82% of them give birth.

The confidence level is given to be equal to 99%, and the margin of error is equal to 0.02 (two percentage points).

Find the critical value

The level of significance is equal to 0.01.

The value of $z_{\frac{\alpha }{2}}$ from the standard normal distribution table is equal to 2.5758.

Determine the sample size

Use the given formula to compute the sample size:

$n=\frac{N\times \hat{p}\hat{q}\times {\left(z_{\frac{\alpha }{2}}\right)}^2}{\hat{p}\hat{q}\times {\left(z_{\frac{\alpha }{2}}\right)}^2+\left(N-1\right)E^2}$

Here,

$\hat{p}$ is the sample proportion of women who have given birth and is equal to 0.82.

$\hat{q}$ is the sample proportion of women who have not given birth.

N is the population size and is equal to 2500.

E is the margin of error is given to be equal to 0.02.

$z_{\frac{\alpha }{2}}$is equal to 2.5758.

Substitute all the values in the given formula as follows:

The formula for sample size,

$$\begin{gathered} n=\frac{N\times \hat{p}\hat{q}\times {\left(z_{\frac{\alpha }{2}}\right)}^2}{\hat{p}\hat{q}\times {\left(z_{\frac{\alpha }{2}}\right)}^2+\left(N-1\right)E^2} \\ =\frac{2500\times \left(0.82\times \left(1-0.82\right)\right)\times {\left(2.5758\right)}^2}{\left(0.82\times \left(1-0.82\right)\right)\times {\left(2.5758\right)}^2+\left(2500-1\right)\times {\left(0.02\right)}^2} \\ \approx 1238 \end{gathered}$$

Therefore, the sample size required is equal to 1238.