Question

Suppose a simple random sample of size 50 is selected from a population with σ=10σ=10. Find the value of the standard error of the mean in each of the following cases (use the finite population correction factor if appropriate).
a. The population size is infinite.
b. The population size is N=50,000.N=50,000.
c. The population size is N=5000.N=5000.
d. The population size is N=500.N=500.

Answer 1

Answer:

a) $\sigma_{\bar x} = 1.414$

b) $\sigma_{\bar x} = 1.414$

c) $\sigma_{\bar x} = 1.414$

d) $\sigma _{\bar x} = 1.343$

Step-by-step explanation:

Given that:

The random sample is of size 50 i.e the population standard deviation  =10

Size of the sample n = 50

a) The population size is infinite;

The standard error is determined as:

$\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}$

$\sigma_{\bar x} = 1.414$

b) When the population size N= 50000

n/N = 50/50000 = 0.001 < 0.05

Thus ; the finite population of the standard error is not applicable in this scenario;

Therefore;

The standard error is determined as:

$\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}$

$\sigma_{\bar x} = 1.414$

c)  When the population size N= 5000

n/N = 50/5000 = 0.01 < 0.05

Thus ; the finite population of the standard error is not applicable in this scenario;

Therefore;

The standard error is determined as:

$\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}$

$\sigma_{\bar x} = 1.414$

d) When the population size N= 500

n/N = 50/500 = 0.1 > 0.05

So; the finite population of the standard error is applicable in this scenario;

Therefore;

The standard error is determined as:

$\sigma _{\bar x} = \sqrt{\dfrac{N-n}{N-1} }\dfrac{\sigma}{\sqrt{n} } }$

$\sigma _{\bar x} = \sqrt{\dfrac{500-50}{500-1} }\dfrac{10}{\sqrt{50} } }$

$\sigma _{\bar x} = 1.343$