Question

In: Statistics and Probability

Choose a variable that will allow the use of dependent samples. For example, you might wish...

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.

Solutions

Expert Solution

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


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