91Ó°ÊÓ

Given

N=500,n=64,s=24

The size of the standard error, including the finite population correction factor, is given by,

${\widehat{\mathrm{σ}}}_{\stackrel{¯}{x}}=\frac{s}{\sqrt{n}}\sqrt{\frac{N-n}{N}}$

Therefore,

Also, the size of the standard error of without including the finite population correction factor is given by,

${\widehat{\mathrm{σ}}}_{\stackrel{¯}{x}}=\frac{s}{\sqrt{n}}$

Therefore,

$$\begin{gathered} {\widehat{\mathrm{σ}}}_{\stackrel{¯}{x}}=\frac{24}{\sqrt{64}} \\ =3 \end{gathered}$$

Hence, from the above results, the standard error is approximately the same with and without including the finite population correction factor.

Using part a, value of can be obtained. If it is greater than 0.05, then the finite population correction factor should be included in the standard error calculation.

Then,

Here, it is observed that

Thus, the finite population correction factor should not be included in the standard error calculation.

Using part b,$\frac{n}{N}$ value of can be obtained. If it is greater than 0.05 then the finite population correction factor should be included in the standard error calculation.

Then,

$$\begin{gathered} \frac{n}{N}=\frac{64}{5000} \\ =0.0128 \end{gathered}$$

role="math" localid="1663920493306" $\frac{n}{N}=0.0128<0.05$

Thus, the finite population correction factor should be included in the standard error calculation.

Using part b,$\frac{n}{N}$ value of can be obtained. If it is greater than 0.05 then the finite population correction factor should be included in the standard error calculation.

Then,$$" width="9" height="19" role="math">nN=4005000=0.08

Here, it is observed that $\frac{n}{N}=0.08>0.05$

Thus, the finite population correction factor should be included in the standard error calculation.