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Population 1,2,4,5,8 · Draw all possible sample of size 2 W.O.R · Sampling distribution of Proportion of even No. · Verify the results

Sagot :

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]

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