In: Statistics and Probability
A. what type of test should be used and why?
B. Compose a null hypothesis to accompany the test. Record both a generic version (through the use symbols) and an English version (using words) – for the generic version, it may be easier to insert an equation box into the word document and type via equation
C. Compose an alternative hypothesis to accompany the test. Record both a generic version (through the use of symbols) and an English version (using words) – for the generic version
D. Compose and type the results of the test. Do you “Reject” or “Fail to Reject” the null hypothesis?
id | Score | Training |
1 | 2 | 0.00 |
2 | 2 | 0.00 |
3 | 4 | 0.00 |
4 | 1 | 0.00 |
5 | 4 | 0.00 |
6 | 3 | 0.00 |
7 | 0 | 0.00 |
8 | 2 | 0.00 |
9 | 7 | 0.00 |
10 | 5 | 0.00 |
11 | 4 | 1.00 |
12 | 4 | 1.00 |
13 | 6 | 1.00 |
14 | 0 | 1.00 |
15 | 6 | 1.00 |
16 | 5 | 1.00 |
17 | 2 | 1.00 |
18 | 3 | 1.00 |
19 | 6 | 1.00 |
20 | 4 | 1.00 |
A. We have two groups here, who took the training (Training =
1.00) and those who didn't take any training (Training = 0.0).
Hence, we will test the difference in scores between the two
groups.
The test used should be independent samples t-test.
B. Null Hypothesis: μ1 = μ2
There is no difference in the average scores between the two groups
of students who took the training and those who didn't take the
training.
C. Alternative Hypothesis: μ1 =/ μ2
There is a significant difference in the average scores between the
two groups of students who took the training and who didn't take
the training.
D. The formula used to calculate the statistic is:
The descriptive statistics of the two groups are:
Training = 0 | Training = 1 | |
Mean | 3 | 4 |
Variance | 4.22 | 3.78 |
Observations | 10 | 10 |
We can assume equal variances.
Inputting all the values in the formula, we get:
t = -1.118
p-value = 0.278 (two-tailed)
Hence, we fail to reject the null hypothesis as the p-value>0.05. We conclude that there is no difference in the average scores between the two groups.