values for each test? Explain your reasoning. (4 marks) e. Why is there a differ

Question

values for each test? Explain your reasoning. (4 marks) e. Why is there a difference in the significance 17) A special education teacher was attempting to determine an appropriate program for children with attention deficit disorder (ADD) so she read some recent research. One article reported results from a study implementing a drug intervention program with students who had ADD. A second article reported the results of a study that used a behaviour modification program with students who had ADD. In both studies, distractibility was measured in each child before and after the program was implemented; the same standardized measure was used in both studies. Lower distractibility scores were associated with better behavour in class. Both studies used a t-test to analyze their data·The t-tests examining the reduction in distractibility scores between pre-program and post-program (including p values) and other important information for each study are listed below. Drug Intervention n = 4520 Behavioural Modification n = 32 sd = 20 effect size 0.85 p = .03 sd 20 effect size 0.1 5 p=.001 If a student with ADD scored 68 on the distractibility measure before the program was implemented, what would the child's final distractibility score be after the implementation of each program, assuming her improvements (i.e., reduction in distractibility score) were typical or average (show your calculations)? (4 marks) If you were the special education teacher, which program would you select for your ADD students? Explain your reasoning. (2 marks)

Explanation / Answer

Let us denote "effect size" by "e".

Here, let us take the before treatment mean score of distractibility, ie, the initial score before applying any treatment as "mi".

Let after treatment or final distractibility be "mf".

We know, for a particular treatment, e = (mi-mf)/sd.

For the particular student, initial score, mi = 68 and her improvements or reduction in score is average.

We have, for Drug Intervention, e=0.15, sd=20.

Hence, from the above formula, mf=mi-(sd×e) =68-(20 × 0.15)

So for drug intervention treatment, the child's final score would be 65.

Again, for Behavioural Modification, e=0.85, sd=20.

Using the same formula, we get mf=68-(20×0.85)

Thus with the second treatment, the child's final score is 51.

Now let us consider the second part of the problem. There are a few points to be remembered in case of a clinical trial:

Considering the above 2 points, we can say that Behavioural Modification is the better choice among the 2 treatments mentioned here.

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