Answer:
a) 1.29
b) 1.29
c) 1.28
d) 1.23
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 60
Population standard Deviation = 10
a) Standard error with infinite population
[tex]\text{Standard error} = \displaystyle\frac{\sigma}{\sqrt{n}} = \frac{10}{\sqrt{60}} = 1.29[/tex]
For finite population with size N,
[tex]\text{Standard error} = \sqrt{\displaystyle\frac{N-n}{N-1}}\times \displaystyle\frac{\sigma}{\sqrt{n}}[/tex]
b) N = 60,000
[tex]\text{Standard error} = \sqrt{\displaystyle\frac{60000-50}{60000-1}}\times \displaystyle\frac{10}{\sqrt{60}} = 1.29[/tex]
c) N = 6,000
[tex]\text{Standard error} = \sqrt{\displaystyle\frac{6000-50}{6000-1}}\times \displaystyle\frac{10}{\sqrt{60}} = 1.28[/tex]
d) N = 600
[tex]\text{Standard error} = \sqrt{\displaystyle\frac{600-50}{600-1}}\times \displaystyle\frac{10}{\sqrt{60}} = 1.23[/tex]
