Question

Suppose N= 10,000, n= 2,000, and s= 50.

a. Compute the standard error of xusing the finite populationcorrection factor.

b. Repeat part a assuming n= 4,000.

c. Repeat part a assuming n= 10,000.

d. Compare parts a, b, and c and describe what happens to the standard error of xas nincreases.

e. The answer to part c is 0. This indicates that there is no sampling error in this case. Explain.

Short answer

a. The standard error is 1

b. The standard error is 0.6124

c. The standard error is 0

d. The standard error decrease as the sample size increase

e. In part(c) there is no sampling error because the sample size is equal to the population size

Step-by-step solution

Given information

Suppose N= 10,000, n= 2,000, and s= 50.

Calculating the standard error

a.

For N= 10,000, n= 2,000, and s= 50.

The standard error is

$\begin{array}{c}{\widehat{\sigma }}_{\overline{x}}=\frac{s}{\sqrt{n}}\sqrt{\frac{N-n}{N}}\\ =\frac{50}{\sqrt{2000}}\sqrt{\frac{10000-2000}{10000}}\\ =1\end{array}$

Calculating the standard error

b.

For N= 10,000, n= 4,000, and s= 50.

The standard error is

$\begin{array}{c}{\widehat{\sigma }}_{\overline{x}}=\frac{s}{\sqrt{n}}\sqrt{\frac{N-n}{N}}\\ =\frac{50}{\sqrt{4000}}\sqrt{\frac{10000-4000}{10000}}\\ =0.6124\end{array}$

 Step 4: Calculating the standard error

c.

For N= 10,000, n= 10,000, and s= 50.

The standard error is

$\begin{array}{c}{\widehat{\sigma }}_{\overline{x}}=\frac{s}{\sqrt{n}}\sqrt{\frac{N-n}{N}}\\ =\frac{50}{\sqrt{10000}}\sqrt{\frac{10000-10000}{10000}}\\ =0\end{array}$

Step 5:

d.

The standard error decrease as the sample size increase

Explanation for zero sampling error

e.

In part(c) there is no sampling error because the sample size is equal to the population size. i.e, all population units are taken that is why the sampling error is zero.