Question

Test the significance of differences between the two samples using the following data
Sample Scores
First Sample 10 9 8 7 5 7 8 9 9 7 3
Second sample 16 14 11 10 6 5 12 13 15 11 3

Answer 1

Conditions

Test the significance of differences between the two samples using the following data

Sample Scores

First Sample

10 9 8 7 5 7 8 9 9 7 3

Second sample 16 14 11 10 6 5 12 13 15

11 3

Solution

This is a task for independent two-sampled test.

For this test, the null hypothesis is that the means of samples are equal:

$$H _ {0}: M _ {1} = M _ {2}$$ $$H _ {\alpha}: M _ {1} \neq M _ {2}$$ $$t = \frac {\bar {X} _ {1} - \bar {X} _ {2}}{S _ {X _ {1} X _ {2}} \sqrt {\frac {1}{n _ {1}} + \frac {1}{n _ {2}}}},$$ $$S _ {X _ {1} X _ {2}} = \sqrt {\frac {1}{2} \left(S _ {\bar {X} _ {1}} ^ {2} + S _ {\bar {X} _ {2}} ^ {2}\right)}$$ $$S _ {\bar {X} _ {1}} ^ {2} = \frac {\sum_ {i = 1} ^ {1 1} \left(X _ {1} - \bar {X} _ {1}\right) ^ {2}}{n}$$ $$S _ {\bar {X} _ {2}} ^ {2} = \frac {\sum_ {i = 1} ^ {1 1} \left(X _ {2} - \bar {X} _ {2}\right) ^ {2}}{n}$$

For this example:

$$t = 2.188$$

The degrees of freedom:

$$k = 11 + 11 - 2 = 20$$

For these degrees of freedom the t-criteria value is:

2.08600- for p=0.95

$$t = 2.188 > 2.086$$

We can make a conclusion, that with probability $95\%$ there is a difference between 2 groups. H0 is rejected.