Question

A population consists of the following amounts: 9, 9, 10, 11, 11

1. Indicate total number possible samples of size 3

2. Compute the sampling distribution of sample means

3. Compare the population mean with the mean of the sampling

   distribution of sample means

4. Compare the dispersion in the population with that of the sample

   means.

Answer 1

Here population consists of the amount : 9,9, 10, 11, 11

sample size = 3

Total possible number of samples of size 3 = ⁵C₃ = 10

These samples are

{ (9,9,10), (9,9,11), (9,9,11), (9.10.11),(9,10,11),(9,11,11),(9,10,11),(9,10,11),(9,11,11),(10,11,11)}

Here we would found the sample mean for each sample.

Total sample size = 10

x1	x2	x3	Sample mean
9	9	10	9.33
9	9	11	9.67
9	9	11	9.67
9	10	11	10
9	10	11	10
9	11	11	10.33
9	10	11	10
9	10	11	10
9	11	11	10.33
10	11	11	10.67

sampling distribution of sample means

p() = 1/10 ; p = 9.33

= 2/10 ; p = 9.67

= 4/10 ; p = 10

= 2/10 ; p =10.33

= 1/10 ; p = 10.67

Here the distribution is given above.

Here population mean = (9 + 9 + 10 + 11 + 11)/5 = 10

mean of sampling distribution = 1/10 * 9.33 + 2/10 * 9.67 + 4/10 * 10 + 2/10 * 10.33 + 1/10 * 10.67 = 10

so here we can see that population mean is equal to mean of the sampling distribution of sample means

Now we have to compare the standard deviation or dispersion of population mean

Variance of population mean = 1/5 [(9 - 10)² + (9 - 10)² + (10 - 10)² + (11 - 10)² + (11 - 10)²] = 4/5 = 0.8

Now we will find the variance of sampling distribution of sample means

mean of sampling distribution = E[] = 10

VaR[] = E[]² -E[]²

E[]² = 1/10 * 9.33² + 2/10 * 9.67² + 4/10 * 10² + 2/10 * 10.33² + 1/10 * 10.67² = 100.133333

VaR[] = 100.133333 - 10² = 0.1333

so here as we can see that

Variance of sample means = Variance of population/sample size