Consider an example of what is often termed "pre/post" test data. Suppose you wish to test the effect of Abilify on the well-being of depressed individuals, using a standardized "well-being scale" that sums Likert-type items to obtain a score that could range from 0 to 20. Higher scores indicate greater well-being ( that is, Abilify is having a positive effect).

Data:

Moodpre: 3.00, .00, 6.00, 7.00, 4.00, 3.00, 2.00, 1.00, 4.00

Moodpost: 5.00, 1.00, 5.00, 7.00, 10.00, 9.00, 7.00, 11.00, 8.00

Difference: 2.00, 1.00, .00, 6.00, 6.00, 5.00, 10.00, 4.00

a) Which test should be used and why?

b) Is there a significant difference between the pre-test and the post-test results?

c) Are there any problems with this study? Give me one example

a)

The most appropriate test to use is "T-test for two dependent means "( also called Paired t-test ).

Because, here our two sets of data are directly related to each other as it is 'pre' and 'post' data. And we don't know the population parameters from which this samples are taken. So in this case the most appropriate test is Paired t-test.

b)

The value of test statistic is = 3.142857

Corresponding p-value = 0.01375

Here the p-value is less than the significance level of 0.05 (standard). That means our test is significant and null hypothesis got rejected and alternate hypothesis has to be considered.

So there is a significant difference between the pre-test and the post-test results

(moodpost have significantly higher mean well being score compared to moodpre.)

c)

The major problems with this study is sample size is small. Here the sample size is only 9, which is too less to statistically generalize the effect of a treatment in a population.

Step-by-step explanation

b)

Calculation of test statistic,

Difference Scores Calculations

Mean: 3.67

μ = 0

S² = SS⁄df  = 98/(9-1) = 12.25

S²_M = S²/N  = 12.25/9 = 1.36

S_M = √S²_M  = √1.36 = 1.17

T-value Calculation

t = (M - μ)/S_M 

= (3.67 - 0)/1.17

=  3.142857

So,

the value of test statistic is = 3.142857

Corresponding p-value = 0.01375 {From standard t-table, available on internet }

Taking the standard significance level of 0.05 (Since not specified in the question)

Please see the attached file for the complete solution