Give step-by-step solution with explanation and final answer:5. Aclinical psychologist is interested in the effectiveness of cognitive behavioral therapy
for depression. He gives patients the Beck De

Question

Give step-by-step solution with explanation and final answer:5. Aclinical psychologist is interested in the effectiveness of cognitive behavioral therapy
for depression. He gives patients the Beck Depression Inventory three times, before therapy,
after therapy and a six-month follow up. Lower scores indicate less depression. The data is
presented below:
[ Subject: [ Pretherapy | PostTherapy | SixMonths |
[oo [6 T 6 [ 2 ]
[ [a 72 7 "3 1]
[5 7 3 © "1 1]
LL @ [6 3 [1 1]
TXyor=42 IX-21 X14 X=7
EXPror = 186 X= 113 IX =58 XP =15
Conduct the appropriate statistical analysis. Set alpha = .01
Step 1: State your hypotheses in words. (2 points)
Step 2: Set up the criteria for making the decision. That is, what is the critical value?
(1 point)
Step 3: Complete the following table: (Show all of your work!)
(7 points)
Source ss df Ms FE
Between (Group)
Subjects
Error (Within)
Total 39
Step 4: Based on your result, would it be appropriate to do a post-hoc test?
(1 point)
(Circle One) Yes No

VIPSolutions · Accepted Answer

Let's solve the problem step-by-step for a **repeated-measures ANOVA**.

---

### **Step 1: State your hypotheses in words.**

- **Null Hypothesis ($H_0$)**: There is no difference in depression scores across the three time points (pre-therapy, post-therapy, six months).
- **Alternative Hypothesis ($H_1$)**: At least one time point has a different mean depression score.

---

### **Step 2: Set up the criteria for making the decision.**

- **Alpha ($\alpha$)** = 0.01
- **Degrees of freedom** for Between (k-1): $ 3-1 = 2 $
- **Degrees of freedom** for Error (Within): $ (n-1)(k-1) = (4-1)(3-1) = 3 \times 2 = 6 $
- **Critical value**: Use the F-table for $ df_1 = 2 $, $ df_2 = 6 $, $ \alpha = 0.01 $.

From an F-table:  
  **Critical F = 10.92**

---

### **Step 3: Complete the following table (Show all your work)**

#### **Means for Each Condition:**
- Pre: $ \frac{21}{4} = 5.25 $
- Post: $ \frac{14}{4} = 3.5 $
- Six: $ \frac{7}{4} = 1.75 $

#### **Step 3A: Compute Sums of Squares**

**Total SS** (given): 39

**Formulae:**

$$
SS_{Total} = \sum X^2 - \frac{(\sum X)^2}{N}
$$
Where $ N = 12 $ (4 subjects × 3 times).

$$
\sum X = 42 \
\sum X^2 = 186 \
\frac{(\sum X)^2}{N} = \frac{42^2}{12} = \frac{1764}{12} = 147 \
SS_{Total} = 186 - 147 = 39 \quad (\text{matches given})
$$

**SS_Between (Groups):**
$$
SS_{Between} = \sum \frac{(\text{Column Total})^2}{n_{col}} - \frac{(\sum X)^2}{N}
$$
Each time point has 4 scores:

$$
= \frac{21^2}{4} + \frac{14^2}{4} + \frac{7^2}{4} - 147 \
= \frac{441}{4} + \frac{196}{4} + \frac{49}{4} - 147 \
= 110.25 + 49 + 12.25 - 147 \
= 171.5 - 147 = 24.5
$$

**SS_Subjects:**
$$
SS_{Subjects} = \sum \frac{(\text{Row Total})^2}{k} - \frac{(\sum X)^2}{N}
$$
Each subject has 3 scores:

Subject Totals:  
(1): $6+6+2 = 14$  
(2): $4+2+3 = 9$  
(3): $5+3+1 = 9$  
(4): $6+3+1 = 10$

$$
= \frac{14^2}{3} + \frac{9^2}{3} + \frac{9^2}{3} + \frac{10^2}{3} - 147 \
= \frac{196}{3} + \frac{81}{3} + \frac{81}{3} + \frac{100}{3} - 147 \
= 65.33 + 27 + 27 + 33.33 - 147 \
= 152.66 - 147 = 5.67
$$

**SS_Error (Within):**
$$
SS_{Error} = SS_{Total} - SS_{Between} - SS_{Subjects}
$$
$$
= 39 - 24.5 - 5.67 = 8.83
$$

#### **Step 3B: Degrees of Freedom (df)**
- Between (Group): $ k-1 = 2 $
- Subjects: $ n-1 = 3 $
- Error (Within): $ (k-1)(n-1) = 6 $
- Total: $ N-1 = 11 $

#### **Step 3C: Mean Squares (MS)**
- $ MS_{Between} = \frac{SS_{Between}}{df_{Between}} = \frac{24.5}{2} = 12.25 $
- $ MS_{Error} = \frac{SS_{Error}}{df_{Error}} = \frac{8.83}{6} = 1.47 $

#### **Step 3D: F Value**
- $ F = \frac{MS_{Between}}{MS_{Error}} = \frac{12.25}{1.47} = 8.33 $

---

### **Step 3: Fill in the Table**

| Source            | SS     | df | MS    | F    |
|-------------------|--------|----|-------|------|
| Between (Group)   | 24.5   | 2  | 12.25 | 8.33 |
| Subjects          | 5.67   | 3  |       |      |
| Error (Within)    | 8.83   | 6  | 1.47  |      |
| Total             | 39     | 11 |       |      |

---

### **Step 4: Should you do a post-hoc test?**

- Compare calculated F (8.33) to critical F (10.92).
- $ 8.33 < 10.92 $: **Not significant at $\alpha = .01$.**

**Answer:**  
No, it would **not** be appropriate to do a post-hoc test.

---

## **Final Answers**

**Step 1:**  
Null: No difference in depression across time points.  
Alternative: At least one time point differs.

**Step 2:**  
Critical F (2, 6, .01) = **10.92**

**Step 3:**  
(Table above)

**Step 4:**  
**No** (circle "No")

Let me know if you need any clarifications or further explanation!

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Study Help, Fast Answers

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Related Answers

Study Help, Fast Answers