Question

Draw all possible samples of size 3 without replacement from the finite population 8, 10, 12, 14, 16, 18. Find the sampling distribution of the sample means and calculate its mean, variance and standard error. Also, find the mean, variance and standard deviation of the population and verify the results.

Answer 1

We have population values 8, 10, 12, 14, 16 and 18. The population size is $M=6$, and the sample size is $m=3$.

Thus, the number of possible samples which can be drawn without replacement is

$$\binom{M}{m}=\binom{6}{3}=20$$

The following table gives $20$ possible combinations of values of size $3$ and their sample means.

Let's find the mean, variance and standard deviation for the sampling distribution:

Sample mean:

$$\mu_x= \frac{\sum f\cdot x}{n}=\frac{260}{20}= 13.$$

Sample variance is calculated as follows:

$$\sigma^2_{x}= \frac{\sum f\cdot x^2-\frac{\big(\sum f\cdot x\big)^2}{n}}{n-1}=\\=\frac{3426.7-\frac{260^2}{20}}{20-1}=\frac{46.7}{19}= 2.45$$

and the sample standard deviation is equal to the square root of the sample variance:

$$\sigma_x= \sqrt{2.45}=1.56.$$

The mean and the standard deviation of the population are computed as follows:

$$\mu = \frac{{\sum X}}{M}=$$ $$\frac{8+10+12+14+16+18}{6}=13,$$

$$\sigma = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \mu)^2} {M}}=$$ $$=\sqrt{\frac{(8-13)^2+(10-13)^2+(12-13)^2+(14-13)^2+(16-13)^2+(18-13)^2} {6}}=$$

$$=\sqrt{\frac{70}{6}}=\sqrt{11.66}=3.42.$$

Therefore the variance and standard error are equal $\sigma^2=11.66$ and $\frac{\sigma }{{\sqrt m }}\sqrt {\frac{{M – m}}{{M – 1}}} = \frac{{3.42}}{{\sqrt 3 }}\sqrt {\frac{{6 – 3}}{{6 – 1}}} = 1.529$ respectively.

Hence, the results differ by decimals.