Question

The science of ergonomics studies the influence of “human factors” in technology, i.e. how human beings relate to and work with machines. With widespread use of computers for data processing, computer scientists and psychologists are getting together to study human factors. One typical study investigates the productivity of secretaries with different word processing programs. An identical task was given to 18 secretaries, randomly allocated to three groups. Group 1 used a menu-driven program. Group 2 used a command-driven program, and Group 3 used a mixture of both approaches. The secretaries all had about the same level of experience, typing speed, and computer skills. The time (in minutes) taken to complete the task was observed. The results were as follows:


Groep / Group 1

Groep / Group 2

Groep / Group 3

Totaal / Total

Menu-aangedrewe / Menu-driven

Opdrag-aangedrewe / Command-driven

Mengsel / Mixed



12

14

10



16

10

8



10

14

10



12

12

10



10

10

8



16

14

4



n1 =

n2 = A

n3 =

N = B

R1 =

R2 =

R3 = C

G = D










Die volgende gedeeltelik voltooide ANAVA tabel is vir die bostaande data verkry. / The following partially completed ANOVA table was obtained for the above data.

ANOVA table:

Source

SS

df

MS

Fcalc

Ftab

Treatments

H

J

34.8889

M

P

Error

84

15

L



Total

I

K






What is the value of P at 0.05 significance level in the ANOVA table? / Wat is die waarde van P met ‘n 0.05 betekenispeil in die ANAVA tabel.?

Answer 1

To fill in the missing values in the ANOVA table and compute the required P-value at a significance level of 0.05, we need to determine a few quantities based on the information provided.

### Step 1: Calculate the sample sizes, means, and other necessary sums.

**Groups:**
- Group 1 (Menu-driven): 6 observations
- Group 2 (Command-driven): 6 observations
- Group 3 (Mixed): 6 observations

From the given data, we can find the sum of the times (add each group's values):

- **Group 1**: 12 + 16 + 10 + 12 + 10 + 16 = 76
- **Group 2**: 14 + 10 + 14 + 12 + 10 + 14 = 84
- **Group 3**: 10 + 8 + 10 + 10 + 8 + 4 = 60

Now we compute the total sum and the grand mean:
- Total Sum = 76 + 84 + 60 = 220
- Overall Count (N) = 18
- Grand Mean (G) = Total Sum / N = 220 / 18 ≈ 12.2222

### Step 2: Calculate the sum of squares

1. **Treatment (Between Groups) Sum of Squares (SST)**:
\[
SST = \sum \frac{n_i (\bar{x}_i - G)^2}{n}
\]
Where:
- \( \bar{x}_1 = 76/6 \approx 12.6667 \)
- \( \bar{x}_2 = 84/6 \approx 14 \)
- \( \bar{x}_3 = 60/6 \approx 10 \)

Calculating \( SST \):
\[
SST = 6 \left( (12.6667 - 12.2222)^2 + (14 - 12.2222)^2 + (10 - 12.2222)^2 \right)
\]
\[
SST = 6 \left( (0.4445)^2 + (1.7778)^2 + (-2.2222)^2 \right)
\]
\[
SST = 6 \left( 0.1972 + 3.1605 + 4.9377 \right) \approx 6 \times 8.2954 \approx 49.7724
\]

### Step 3: Calculate the error (within groups) sum of squares (SSE)

The formula for the Within Groups Sum of Squares (SSE):
\[
SSE = \sum (x_{ij} - \bar{x}_i)^2
\]

Calculating within each group:

- For Group 1:
\[
(12 - 12.6667)^2 + (16 - 12.6667)^2 + (10 - 12.6667)^2 + (12 - 12.6667)^2 + (10 - 12.6667)^2 + (16 - 12.6667)^2
\]
≈ 2.7778 + 11.1111 + 7.1111 + 0.4444 + 7.1111 + 11.1111 = 39.6667

- For Group 2:
\[
(14 - 14)^2 + (10 - 14)^2 + (14 - 14)^2 + (12 - 14)^2 + (10 - 14)^2 + (14 - 14)^2
\]
≈ 0 + 16 + 0 + 4 + 16 + 0 = 36

- For Group 3:
\[
(10 - 10)^2 + (8 - 10)^2 + (10 - 10)^2 + (10 - 10)^2 + (8 - 10)^2 + (4 - 10)^2
\]
≈ 0 + 4 + 0 + 0 + 4 + 36 = 44

Total SSE = 39.6667 + 36 + 44 = 119.6667

### Step 4: Total Sum of Squares (SSTotal)
\[
SSTotal = SST + SSE \approx 49.7724 + 119.6667 = 169.4391
\]

### Step 5: Fill in the ANOVA table

- Treatments Degrees of Freedom (df) = k - 1 = 3 - 1 = 2
- Error Degrees of Freedom (df) = N - k = 18 - 3 = 15
- Total Degrees of Freedom (df) = N - 1 = 18 - 1 = 17

### Step 6: Calculate Mean Squares (MS)

- Treatments Mean Square (MST) = SST / df = 49.7724 / 2 ≈ 24.8862
- Error Mean Square (MSE) = SSE / df = 119.6667 / 15 ≈ 7.9786

### Step 7: Calculate Fcalc
\[
F_{calc} = \frac{MST}{MSE} = \frac{24.8862}{7.9786} \approx 3.12
\]

### Step 8: P-value determination
For F-distribution with (2, 15) degrees of freedom at a significance level of 0.05, we need to refer to F distribution tables or use statistical software to find P.

The critical value at 0.05 significance level (Ftab) is often around 4.0 for (2, 15) df. Since the calculated F is lower than the critical value, we fail to reject the null hypothesis.

Thus, the P-value would be greater than 0.05, making P not statistically significant.

Therefore, you can enter:
- \( P > 0.05 \)

Make sure you verify the F-value via a statistical calculator to properly get the exact P-value for your condition. It should closely align with general distributions!