A researcher wants to test the effectiveness of an experimental treatment for de
ID: 3225281 • Letter: A
Question
A researcher wants to test the effectiveness of an experimental treatment for depression which involves exercise and classical music. He randomly selects n = 50 patients taking depression medication from several clinics and invites them to his lab where he assesses their levels of depression with the Beck Depression Inventory (rated 0-63, with higher scores indicating more severe levels of depression). He finds that their mean depression score is M = 24.3 with a standard deviation of s = 14.47. He then has them come to the lab four times a week to exercise on an elliptical for 30 minutes while listening to Johann Sebastian Bach over the course of one month. After the month, the researcher re-assesses the patients with the Beck Depression Inventory and finds that their mean depression score is M = 17.08 with a standard deviation of s = 20.02. Explain why the researcher cannot conclude that the treatment had an effect at this point (just based on the difference in means before and after the treatment)? What numeric value must be taken into consideration?Explanation / Answer
Solution:-
tate the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: Before - after < 0
Alternative hypothesis: Before - after > 0
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[ (s12/n1) + (s22/n2) ]
S.E = 3.493
DF = n - 1 = 50 -1
D = 49
t = [ (x1 - x2) - D ] / SE
t = 2.067
where di is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 49 degrees of freedom is more than 2.067
Thus, the P-value = 0.022
Interpret results. Since the P-value (0.022) is greater than the significance level (0.01), we cannot reject the null hypothesis.
From the above test we do not have sufficient evidence to conclude that Depression score has decreased.
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