In: Statistics and Probability
Choose a variable that will allow the use of dependent samples. For example, you might wish to see if a persons’ proficiency at softball has changed/improved after a training camp. Do not use variables that are presented in the course in order to illustrate the concept. Select a sample of data (10 to 20) value pairs (e.g. before and after), and then complete the following:
i. Show the computations of the sample statistic, the standard deviation of the
differences in pairs, the test statistic.
j. State the decision. A training camp has changed/improved a person’s proficiency.
at softball, or has not changed/improved a person’s proficiency.
at softball.
k. Summarize the results.
l. Construct a confidence interval. State the conclusion. Summarize the results.
We can use football training sample test where the number of goals the players made in the pre-training and in the post-training games. Then their comparison can be made. Here since the same players are used to compare the results we call it a dependent sample test.
Sr. | Pre | Post | D (Pre - post) | D^2 |
1 | 5 | 7 | -2 | 4 |
2 | 7 | 9 | -2 | 4 |
3 | 10 | 11 | -1 | 1 |
4 | 7 | 8 | -1 | 1 |
5 | 5 | 6 | -1 | 1 |
6 | 8 | 8 | 0 | 0 |
7 | 7 | 6 | 1 | 1 |
8 | 10 | 9 | 1 | 1 |
9 | 9 | 10 | -1 | 1 |
10 | 8 | 12 | -4 | 16 |
11 | 8 | 11 | -3 | 9 |
12 | 5 | 8 | -3 | 9 |
13 | 5 | 6 | -1 | 1 |
14 | 5 | 8 | -3 | 9 |
15 | 7 | 6 | 1 | 1 |
Total | -19 | 59 | ||
Mean | -1.267 | |||
SD | 1.580 |
Mean =
SD =
Test
(There is no difference between number of goals in pre and post training )
(There are more number of goals after the training than before.)
Since we have d = pre - post are favourable would be -d
Test Stat =
= .......where null difference = 0
Test Stat : -3.1057
Taking level of significance = 0.05
Since this is a one tailed test
Critical value =
=
= 2.1448
Since |Test Stat| > C.V.
We reject the null hypothesis at 0.05 and conclude that training significantly improved the players game play.
(1 - )% confidence interval for the mean difference
Again taking the level of significance = 0.05