Answer:
a) [tex]\sigma_{\bar x} = 1.414[/tex]
b) [tex]\sigma_{\bar x} = 1.414[/tex]
c) [tex]\sigma_{\bar x} = 1.414[/tex]
d) [tex]\sigma _{\bar x} = 1.343[/tex]
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:
[tex]\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}[/tex]
[tex]\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}[/tex]
[tex]\sigma_{\bar x} = 1.414[/tex]
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:
[tex]\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}[/tex]
[tex]\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}[/tex]
[tex]\sigma_{\bar x} = 1.414[/tex]
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:
[tex]\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}[/tex]
[tex]\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}[/tex]
[tex]\sigma_{\bar x} = 1.414[/tex]
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:
[tex]\sigma _{\bar x} = \sqrt{\dfrac{N-n}{N-1} }\dfrac{\sigma}{\sqrt{n} } }[/tex]
[tex]\sigma _{\bar x} = \sqrt{\dfrac{500-50}{500-1} }\dfrac{10}{\sqrt{50} } }[/tex]
[tex]\sigma _{\bar x} = 1.343[/tex]
