Question

In: Statistics and Probability

A study of randomized controlled trial of animals facilitated therapy with dolphins in the treatment of...

A study of randomized controlled trial of animals facilitated therapy with dolphins in the treatment of (mild and moderate) depression collected the data as follows:

Depression Improved
Treatment Yes No Total
Dolphin 10 5 15
Control 3 12 15
Total 13 17 30

Is the distribution of patients' depression significantly different between dolphin treatment and control treatment? Is there a treatment effect or could such a difference be explained by chance? Use the 8 step procedure to answer this question.

Solutions

Expert Solution

Solution:

Here, we have to use chi square test for independence of two categorical variables.

Null hypothesis: H0: The distribution of patients' depression is significantly different between dolphin treatment and control treatment.

Alternative hypothesis: Ha: The distribution of patients' depression is significantly different between dolphin treatment and control treatment.

We assume/given level of significance = α = 0.05

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

E = row total * column total / Grand total

We are given

Number of rows = r = 2

Number of columns = c = 2

Degrees of freedom = df = (r – 1)*(c – 1) = 1*1 = 1

α = 0.05

Critical value = 3.841459

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies

Depression Improved

Treatment

Yes

No

Total

Dolphin

10

5

15

Control

3

12

15

Total

13

17

30

Expected Frequencies

Depression Improved

Treatment

Yes

No

Total

Dolphin

6.5

8.5

15

Control

6.5

8.5

15

Total

13

17

30

Calculations

(O - E)

3.5

-3.5

-3.5

3.5

(fo-fe)^2/fe

1.884615

1.441176

1.884615

1.441176

Test Statistic = Chi square = ∑[(O – E)^2/E] = 6.651584

χ2 statistic = 6.651584

P-value = 0.009907

(By using Chi square table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is not sufficient evidence to conclude that the distribution of patients' depression is significantly different between dolphin treatment and control treatment.


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