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ANSWERS 👉 1-5 6-10 11-17 18-27 NUMERICAL QUESTIONS
NUMERICAL QUESTION 1: One-Way ANOVA (10 marks)
Question (Model)
Three groups of students were taught using three different teaching methods. Their test scores are:
- Method A: 8, 9, 6
- Method B: 4, 5, 3
- Method C: 1, 2, 3
Test by one-way ANOVA whether there is a significant difference among the three methods at 0.05 level.
Step 1: Arrange the data
Group | Scores | n |
A | 8, 9, 6 | 3 |
B | 4, 5, 3 | 3 |
C | 1, 2, 3 | 3 |
Total N = 3 + 3 + 3 = 9, Number of groups k = 3
Step 2: Compute group means and grand mean
- Mean of A:
XˉA=8+9+63=233≈7.67\bar{X}_A = \frac{8 + 9 + 6}{3} = \frac{23}{3} \approx 7.67XˉA=38+9+6=323≈7.67
- Mean of B:
XˉB=4+5+33=123=4\bar{X}_B = \frac{4 + 5 + 3}{3} = \frac{12}{3} = 4XˉB=34+5+3=312=4
- Mean of C:
XˉC=1+2+33=63=2\bar{X}_C = \frac{1 + 2 + 3}{3} = \frac{6}{3} = 2XˉC=31+2+3=36=2
- Grand Mean (GM)
All scores = 8, 9, 6, 4, 5, 3, 1, 2, 3 → Sum = 41
GM=419≈4.56GM = \frac{41}{9} \approx 4.56GM=941≈4.56
Step 3: Compute Sum of Squares Between Groups (SSB)
Formula:
SSB=∑ni(Xˉi−GM)2SSB = \sum n_i(\bar{X}_i - GM)^2SSB=∑ni(Xˉi−GM)2
- For A:
3(7.67−4.56)2≈3(3.11)2≈3(9.67)=29.013(7.67 - 4.56)^2 \approx 3(3.11)^2 \approx 3(9.67) = 29.013(7.67−4.56)2≈3(3.11)2≈3(9.67)=29.01
- For B:
3(4−4.56)2=3(−0.56)2=3(0.31)=0.933(4 - 4.56)^2 = 3(-0.56)^2 = 3(0.31) = 0.933(4−4.56)2=3(−0.56)2=3(0.31)=0.93
- For C:
3(2−4.56)2=3(−2.56)2=3(6.55)=19.653(2 - 4.56)^2 = 3(-2.56)^2 = 3(6.55) = 19.653(2−4.56)2=3(−2.56)2=3(6.55)=19.65
SSB≈29.01+0.93+19.65≈49.59 (≈ 49.56)SSB \approx 29.01 + 0.93 + 19.65 \approx 49.59 \ (\text{≈ }49.56)SSB≈29.01+0.93+19.65≈49.59 (≈ 49.56)
Step 4: Compute Sum of Squares Within Groups (SSW)
Formula:
SSW=∑∑(Xij−Xˉi)2SSW = \sum \sum (X_{ij} - \bar{X}_i)^2SSW=∑∑(Xij−Xˉi)2
Group A (mean = 7.67)
- (8 − 7.67)² ≈ 0.11
- (9 − 7.67)² ≈ 1.78
- (6 − 7.67)² ≈ 2.78
→ SSA ≈ 0.11 + 1.78 + 2.78 = 4.67
Group B (mean = 4)
- (4 − 4)² = 0
- (5 − 4)² = 1
- (3 − 4)² = 1
→ SSB(within) = 2
Group C (mean = 2)
- (1 − 2)² = 1
- (2 − 2)² = 0
- (3 − 2)² = 1
→ SSC(within) = 2
SSW=4.67+2+2=8.67 (≈ 8.67)SSW = 4.67 + 2 + 2 = 8.67 \ (\text{≈ }8.67)SSW=4.67+2+2=8.67 (≈ 8.67)
Step 5: Degrees of Freedom
- Between groups:
dfbetween=k−1=3−1=2df_{between} = k - 1 = 3 - 1 = 2dfbetween=k−1=3−1=2
- Within groups:
dfwithin=N−k=9−3=6df_{within} = N - k = 9 - 3 = 6dfwithin=N−k=9−3=6
Step 6: Mean Squares
MSB=SSBdfbetween=49.562≈24.78MSB = \frac{SSB}{df_{between}} = \frac{49.56}{2} \approx 24.78MSB=dfbetweenSSB=249.56≈24.78 MSW=SSWdfwithin=8.676≈1.44MSW = \frac{SSW}{df_{within}} = \frac{8.67}{6} \approx 1.44MSW=dfwithinSSW=68.67≈1.44
Step 7: F-ratio
F=MSBMSW=24.781.44≈17.15F = \frac{MSB}{MSW} = \frac{24.78}{1.44} \approx 17.15F=MSWMSB=1.4424.78≈17.15
Step 8: Decision
For df(2,6) at 0.05 level, table value of F ≈ 5.14 (you can mention this directly in exam).
Since
Fcalculated(17.15)>Fcritical(5.14)F_{calculated} (17.15) > F_{critical} (5.14)Fcalculated(17.15)>Fcritical(5.14)
✅ Reject the null hypothesis.
Conclusion (One sentence)
There is a significant difference among the three teaching methods. At least one method is more effective than the others.
🔹 NUMERICAL QUESTION 2: Chi-square Test (6 marks)
Question (Model)
A psychologist wants to know whether preference for counselling depends on gender. Data from 70 people are:
Prefer Counselling (Yes) | Do Not Prefer (No) | Row Total | |
Males | 10 | 20 | 30 |
Females | 30 | 10 | 40 |
Column Total | 40 | 30 | 70 |
Test by Chi-square (χ²) whether preference is independent of gender at 0.05 level.
Step 1: State Hypotheses
- H₀: Gender and preference for counselling are independent (no association).
- H₁: Gender and preference for counselling are associated.
Step 2: Compute Expected Frequencies
Formula:
E=(Row Total)(Column Total)Grand TotalE = \frac{(\text{Row Total}) (\text{Column Total})}{\text{Grand Total}}E=Grand Total(Row Total)(Column Total)
- Male–Yes:
EMY=30×4070=120070≈17.14E_{MY} = \frac{30 \times 40}{70} = \frac{1200}{70} \approx 17.14EMY=7030×40=701200≈17.14
- Male–No:
EMN=30×3070=90070≈12.86E_{MN} = \frac{30 \times 30}{70} = \frac{900}{70} \approx 12.86EMN=7030×30=70900≈12.86
- Female–Yes:
EFY=40×4070≈22.86E_{FY} = \frac{40 \times 40}{70} \approx 22.86EFY=7040×40≈22.86
- Female–No:
EFN=40×3070≈17.14E_{FN} = \frac{40 \times 30}{70} \approx 17.14EFN=7040×30≈17.14
Step 3: Apply Chi-square Formula
χ2=∑(O−E)2E\chi^2 = \sum \frac{(O - E)^2}{E}χ2=∑E(O−E)2
Calculate cell-wise:
1. Male–Yes:
(10−17.14)217.14=(−7.14)217.14=50.9817.14≈2.98\frac{(10 - 17.14)^2}{17.14} = \frac{(-7.14)^2}{17.14} = \frac{50.98}{17.14} \approx 2.9817.14(10−17.14)2=17.14(−7.14)2=17.1450.98≈2.98
2. Male–No:
(20−12.86)212.86=7.14212.86=50.9812.86≈3.96\frac{(20 - 12.86)^2}{12.86} = \frac{7.14^2}{12.86} = \frac{50.98}{12.86} \approx 3.9612.86(20−12.86)2=12.867.142=12.8650.98≈3.96
3. Female–Yes:
(30−22.86)222.86=7.14222.86=50.9822.86≈2.23\frac{(30 - 22.86)^2}{22.86} = \frac{7.14^2}{22.86} = \frac{50.98}{22.86} \approx 2.2322.86(30−22.86)2=22.867.142=22.8650.98≈2.23
4. Female–No:
(10−17.14)217.14≈2.98\frac{(10 - 17.14)^2}{17.14} \approx 2.9817.14(10−17.14)2≈2.98
Now sum:
χ2≈2.98+3.96+2.23+2.98≈12.15\chi^2 \approx 2.98 + 3.96 + 2.23 + 2.98 \approx 12.15χ2≈2.98+3.96+2.23+2.98≈12.15
Step 4: Degrees of Freedom
df=(r−1)(c−1)=(2−1)(2−1)=1df = (r - 1)(c - 1) = (2 - 1)(2 - 1) = 1df=(r−1)(c−1)=(2−1)(2−1)=1
Step 5: Decision
Table value of χ² at df = 1, 0.05 level ≈ 3.84.
Since
χcalculated2(12.15)>χcritical2(3.84)\chi^2_{calculated} (12.15) > \chi^2_{critical} (3.84)χcalculated2(12.15)>χcritical2(3.84)
✅ Reject H₀.
Conclusion
There is a significant association between gender and preference for counselling. Preference is not independent of gender.
🔹 NUMERICAL QUESTION 3: Mann–Whitney U Test (6 marks)
Question (Model)
Two groups of students are given different types of motivation training. Their performance scores are:
- Group A: 5, 7, 9
- Group B: 1, 3, 6
Test whether there is a significant difference between the groups using Mann–Whitney U test.
Step 1: Combine and Rank All Scores
Scores combined:
- Group A: 5, 7, 9
- Group B: 1, 3, 6
Order them from smallest to largest and assign ranks:
Score | Group | Rank |
1 | B | 1 |
3 | B | 2 |
5 | A | 3 |
6 | B | 4 |
7 | A | 5 |
9 | A | 6 |
Step 2: Sum of Ranks
- Group A ranks: 3, 5, 6 →
R1=3+5+6=14R_1 = 3 + 5 + 6 = 14R1=3+5+6=14
- Group B ranks: 1, 2, 4 →
R2=1+2+4=7R_2 = 1 + 2 + 4 = 7R2=1+2+4=7
n₁ = n₂ = 3
Step 3: Compute U₁ and U₂
Formulas:
U1=n1n2+n1(n1+1)2−R1U_1 = n_1 n_2 + \frac{n_1(n_1+1)}{2} - R_1U1=n1n2+2n1(n1+1)−R1U2=n1n2+n2(n2+1)2−R2U_2 = n_1 n_2 + \frac{n_2(n_2+1)}{2} - R_2U2=n1n2+2n2(n2+1)−R2
- For Group A (U₁):
U1=3×3+3(3+1)2−14=9+6−14=1U_1 = 3 \times 3 + \frac{3(3+1)}{2} - 14 = 9 + 6 - 14 = 1U1=3×3+23(3+1)−14=9+6−14=1
- For Group B (U₂):
U2=3×3+3(3+1)2−7=9+6−7=8U_2 = 3 \times 3 + \frac{3(3+1)}{2} - 7 = 9 + 6 - 7 = 8U2=3×3+23(3+1)−7=9+6−7=8
Take the smaller U:
U=1U = 1U=1
Step 4: Decision (small-sample table)
For n₁ = 3, n₂ = 3, critical U at 0.05 (two-tailed) ≈ 2 (you can state this).
Since
Ucalculated(1)<Ucritical(2)U_{calculated} (1) < U_{critical} (2)Ucalculated(1)<Ucritical(2)
✅ Reject H₀.
Conclusion
There is a significant difference between the two motivation training methods.
🔹 NUMERICAL QUESTION 4: Spearman’s Rank Correlation (ρ) (6 marks)
Question (Model)
A psychologist measures motivation scores (X) and test performance (Y) of 6 students:
Student | X (Motivation) | Y (Performance) |
A | 10 | 1 |
B | 20 | 2 |
C | 30 | 13 |
D | 40 | 7 |
E | 50 | 21 |
F | 60 | 10 |
Compute Spearman’s Rank Correlation (ρ).
Step 1: Rank X and Y
X is already in ascending order → ranks for X:
10, 20, 30, 40, 50, 60 → 1, 2, 3, 4, 5, 6
Now rank Y (smallest = rank 1):
Y values and ranks:
- 1 → rank 1
- 2 → rank 2
- 7 → rank 3
- 10 → rank 4
- 13 → rank 5
- 21 → rank 6
Match with students:
Student | X | Rank X (Rₓ) | Y | Rank Y (Rᵧ) |
A | 10 | 1 | 1 | 1 |
B | 20 | 2 | 2 | 2 |
C | 30 | 3 | 13 | 5 |
D | 40 | 4 | 7 | 3 |
E | 50 | 5 | 21 | 6 |
F | 60 | 6 | 10 | 4 |
Step 2: Compute d and d²
d=Rx−Ryd = R_x - R_yd=Rx−Ry
Student | Rₓ | Rᵧ | d = Rₓ − Rᵧ | d² |
A | 1 | 1 | 0 | 0 |
B | 2 | 2 | 0 | 0 |
C | 3 | 5 | -2 | 4 |
D | 4 | 3 | 1 | 1 |
E | 5 | 6 | -1 | 1 |
F | 6 | 4 | 2 | 4 |
∑d2=0+0+4+1+1+4=10\sum d^2 = 0 + 0 + 4 + 1 + 1 + 4 = 10∑d2=0+0+4+1+1+4=10
n = 6
Step 3: Apply Spearman’s Formula
ρ=1−6∑d2n(n2−1)\rho = 1 - \frac{6 \sum d^2}{n(n^2 - 1)}ρ=1−n(n2−1)6∑d2ρ=1−6×106(62−1)=1−606×35=1−60210\rho = 1 - \frac{6 \times 10}{6(6^2 - 1)} = 1 - \frac{60}{6 \times 35} = 1 - \frac{60}{210}ρ=1−6(62−1)6×10=1−6×3560=1−21060ρ=1−0.2857≈0.71\rho = 1 - 0.2857 \approx 0.71ρ=1−0.2857≈0.71
Conclusion
Spearman’s rank correlation (ρ) ≈ 0.71, which indicates a strong positive relationship between motivation and performance.
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ANSWERS 👉 1-5 6-10 11-17 18-27 NUMERICAL QUESTIONS
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