Answer to Question #135344 in Statistics and Probability for Muhammad Moin

Question

Draw all possible random samples of size 2 with replacement from the
population 2, 3, 4, 5 and 7. Similarly draw all possible random samples of size
2 with replacement from other population 1, 2, 2, 4 and 5.Find the sampling
distribution of difference b/w two sample means and calculate its Mean,
Variance and Standard Error. Also find the Mean, Variance and Standard
Deviation of two Populations and Verify the Results.

Answer 1

solution

when drawn with replacement, the total number of combinations that can be obtained is:

$$n^r$$

Where r is the number of elements chosen and n the number of elements to Choose from.

$$=5^2=25$$

From the population $X= (2, 3, 4, 5, 7)$ each of the samples composed of 2 Numbers has a mean of:

$$\bar x_i = \frac{a_i+b_i}{2}$$

Where $\bar x_i$ Is the mean of the $i^{th}$ Combination. A series of sample means is formed i.e.

$$X_{means} = \bar x_1, \bar x_2, \bar x_3,...,\bar x_{25}$$

From the population$Y=(1,2,2,4,5)$ each of the samples composed of 2 Numbers has a mean of:

$$\bar y_i = \frac{a_i+b_i}{2}$$

Where $\bar y_i$ Is the mean of the $i^{th}$ Samples. A series of sample means is formed i.e

$$Y_{means} = \bar y_1, \bar y_2, \bar y_3,...,\bar y_{25}$$

The distribution of differences between the 2 means is given by:

$$differences =X_{means}-Y_{means}$$

The distribution of mean differences is obtained as:

$$(difference,\ frequency)=\ (1,9),\ (1.5,12),\ (2,4)$$

1. Answers for the distribution of differences in means

Mean:

$$\bar {difference} =\frac{ \sum (differences * frequency)}{n}$$ $$=\frac{(1*9)+(1.5*12)+(2*4)}{25} = 1.4$$

mean =1.4

Variance:

$$\sigma^2 =\frac{ \sum frequency * (differences -\bar {difference})}{n-1} = 0.29167$$

variance = 0.125

standard error:

$$SE = \sqrt {\sigma^2} = \sqrt{0.125} = 0.3536$$

standard error = 0.5401

2. answers for the distribution of X and Y

Mean of the population X:

$$\frac{\sum X}{n}=\frac{2+3+4+5+7}{5}=4.2$$

mean of X = 4.2

Variance of the distribution of X:

$$\sigma ^2 = \frac{\sum( X - \bar X)^2 }{n-1} = 3.7$$

variance of X = 3.7

Standard error of the distribution of means of X:

$$\sqrt {\sigma^2} = \sqrt{3.7} = 1.924$$

standard deviation of X= 1.924

The mean of the population Y:

$$\frac{\sum Y}{n}=\frac{1+2+2+4+5}{5}=2.8$$

mean of Y = 2.8

Variance of the distribution of means of Y:

$$\sigma ^2 = \frac{\sum( Y- \bar Y)^2 }{n-1} =2.7$$

variance of Y = 2.7

Standard error of the distribution of means of Y:

$\sqrt{\sigma^2}=\sqrt{2.7}= 1.6432$

standard deviation of Y = 1.6432

The difference between the mean on X and mean of Y:

$\bar X- \bar Y = 4.2-2.8 = 1.4$ . This is equal to the mean of the distribution of differences in means ($\bar {difference}$ )