Question:
A population consists 1, 2, 4, 5, 8. Draw all possible samples of size 2 without replacement from this population.
Verify that the sample mean is an unbiased estimate of the population mean.
Answer:
[tex]Samples: \{(1,2),(1,4),(1,5),(1,8),(2,4),(2,5),(2,8),(4,5),(4,8),(5,8)\}[/tex]
[tex]\hat p = \frac{3}{5}[/tex] --- proportion of evens
The sample mean is an unbiased estimate of the population mean.
Step-by-step explanation:
Given
[tex]Numbers: 1, 2, 4, 5, 8[/tex]
Solving (a): All possible samples of 2 (W.O.R)
W.O.R means without replacement
So, we have:
[tex]Samples: \{(1,2),(1,4),(1,5),(1,8),(2,4),(2,5),(2,8),(4,5),(4,8),(5,8)\}[/tex]
Solving (b): The sampling distribution of the proportion of even numbers
This is calculated as:
[tex]\hat p = \frac{n(Even)}{Total}[/tex]
The even samples are:
[tex]Even = \{2,4,8\}[/tex]
[tex]n(Even) = 3[/tex]
So, we have:
[tex]\hat p = \frac{3}{5}[/tex]
Solving (c): To verify
[tex]Samples: \{(1,2),(1,4),(1,5),(1,8),(2,4),(2,5),(2,8),(4,5),(4,8),(5,8)\}[/tex]
Calculate the mean of each samples
[tex]Sample\ means = \{1.5,2.5,3,4.5,3,3.5,5,4.5,6,6.5\}[/tex]
Calculate the mean of the sample means
[tex]\bar x = \frac{1.5 + 2.5 +3 + 4.5 + 4 + 3.5 + 5 + 4.5 + 6 + 6.5}{10}[/tex]
[tex]\bar x = \frac{40}{10}[/tex]
[tex]\bar x = 4[/tex]
Calculate the population mean:
[tex]Numbers: 1, 2, 4, 5, 8[/tex]
[tex]\mu = \frac{1 +2+4+5+8}{5}[/tex]
[tex]\mu = \frac{20}{5}[/tex]
[tex]\mu = 4[/tex]
[tex]\bar x = \mu = 4[/tex]
This implies that [tex]\bar x[/tex] is an unbiased estimate of the [tex]\mu[/tex]
