Question

Researchers from the Hotchkiss Brain Institute would like to examine the calming effects of THC on patients suffering from PTSD, GAD, and OCD. At the end of the study, the researchers recorded the number of symptoms reported by the patients. Data from this study is shown below:

Does the effect of THC indicate any difference?

Test with α = .01.

Remember: state your hypothesis, identify the critical region, do your test, make a decision and state the results in the conventional manner

  OCD           GAD   PTSD    

                                  4                   1                    0

                                       6                   4                    2            G = 36

                                       3                   5                    0          

                                       7                   2                    2

_________________________________

                                  SS = 10       SS = 10          SS = 4

Answer 1

Here, we will apply one-way anova to see if there is significant difference in effect of THC on three illnesses

Null Hypothesis: The effect of THC is not significantly different for three illnesses.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ₁ = μ₂ = μ₃

Ha: Not all means are equal

The above hypotheses will be tested using an F-ratio for a One-Way ANOVA.

Critical region: alpha = 0.01 (when p-value < 0.01, we will reject Ho)

or  Rejection Region

Based on the information provided, the significance level is α=0.01, and the degrees of freedom are df1​=2 and df2​=2, therefore, the rejection region for this F-test is R={F:F>Fc​=8.022}

Test:

Group 1 :  OCD , Group 2: GAD, Group3: PTSD

Now,

The total sample size is N=12. Therefore, the total degrees of freedom are:

dftotal​=12−1=11

Also, the between-groups degrees of freedom are dfbetween​=3−1=2, and the within-groups degrees of freedom are:

dfwithin​=dftotal​−dfbetween​=11−2=9

First, we need to compute the total sum of values and the grand mean. The following is obtained

i,j∑​Xij​=20+12+4=36

Also, the sum of squared values is

i,j∑​Xij2​=110+46+8=164

Based on the above calculations, the total sum of squares is computed as follows

SStotal​=i,j∑​Xij2​−N1​(i,j∑​Xij​)2=164−12362​=56

The within sum of squares is computed as shown in the calculation below:

SSwithin​=∑SSwithingroups​=10+10+4=24

Now that sum of squares are computed, we can proceed with computing the mean sum of squares:

MSbetween​=SSbetween​​/dfbetween​=32/​2=16

MSwithin​=dfwithin​SSwithin​​=24/9​=2.667

Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows:

Test Statistics

F = MSbetween​​/MSwithin​ =16​/2.667 =6

Decision about the null hypothesis

Since it is observed that F=6≤Fc​=8.022, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.0221, and since p=0.0221≥0.01, it is concluded that the null hypothesis is not rejected.

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that not all 3 population means are equal, at the α=0.01 significance level.

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