ISS PREVIOUS YEAR 2016 PAPER-2 SOLUTION SET-A Q.NO. 11
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ISS PREVIOUS YEAR 2016 PAPER-2 SOLUTION SET-A
One-Way ANOVA Problem (Training Methods Comparison)
Question 11
A company used three different methods to train its employees. The number of units of output produced by employees trained under the three methods are:
Method A: 50, 45, 55, 44Method B: 64, 48, 52, 56, 44Method C: 46, 42, 48, 45, 57, 42
Test whether the three training methods are equally satisfactory at 5% level of significance.
Given: F(2, 12) at 0.05 = 3.89
Also, identify correct hypotheses:
H0: μA = μB = μC
H0: μA ≠ μB ≠ μC
H1: μA, μB, μC are not all equal
H1: μA = μB = μC
Solution
This is a One-Way ANOVA problem.
Step 1: Sample Sizes
n1 = 4 (Method A)n2 = 5 (Method B)n3 = 6 (Method C)
Total observations:
N = 4 + 5 + 6 = 15
Number of groups:
k = 3
Step 2: Hypotheses
Null hypothesis:
H0: μA = μB = μC
Alternative hypothesis:
H1: Means are not all equal
Correct statements: 1 and 3
Correct answer: (a) 1 and 3
Step 3: Compute Means
Method A
TA = 50 + 45 + 55 + 44 = 194Mean A = 194 / 4 = 48.5
Method B
TB = 64 + 48 + 52 + 56 + 44 = 264Mean B = 264 / 5 = 52.8
Method C
TC = 46 + 42 + 48 + 45 + 57 + 42 = 280Mean C = 280 / 6 = 46.67
Grand Total and Mean
T = 194 + 264 + 280 = 738
Grand Mean:
X̄ = 738 / 15 = 49.2
Step 4: Correction Factor
C.F. = T² / NC.F. = 738² / 15C.F. = 544644 / 15 = 36309.6
Step 5: Total Sum of Squares
First compute ΣX²:
Method A:50² + 45² + 55² + 44² = 9486
Method B:64² + 48² + 52² + 56² + 44² = 14176
Method C:46² + 42² + 48² + 45² + 57² + 42² = 13222
Total:
ΣX² = 9486 + 14176 + 13222 = 36884
Now,
SST = ΣX² − C.F.SST = 36884 − 36309.6 = 574.4
Step 6: Sum of Squares Between Groups
SSB = (TA²/n1 + TB²/n2 + TC²/n3) − C.F.
= (194²/4 + 264²/5 + 280²/6) − 36309.6
= (9409 + 13939.2 + 13066.67) − 36309.6
= 36414.87 − 36309.6 = 105.27
Step 7: Sum of Squares Within Groups
SSE = SST − SSBSSE = 574.4 − 105.27 = 469.13
Step 8: Degrees of Freedom
Between groups:
dfB = k − 1 = 3 − 1 = 2
Within groups:
dfE = N − k = 15 − 3 = 12
Total:
dfT = 14
Step 9: Mean Squares
Mean square between:
MSB = SSB / 2 = 105.27 / 2 = 52.64
Mean square within:
MSE = SSE / 12 = 469.13 / 12 = 39.09
Step 10: F Statistic
F = MSB / MSEF = 52.64 / 39.09 ≈ 1.35
ANOVA Summary
Between groups: SS = 105.27, df = 2, MS = 52.64
Within groups: SS = 469.13, df = 12, MS = 39.09
Total: SS = 574.40, df = 14
F = 1.35
Step 11: Decision Rule
Critical value:
F(2,12) = 3.89
Since:
1.35 < 3.89
Fail to reject H0
Final Conclusion
There is no significant difference between the three training methods.
Hence, all three methods are equally satisfactory.
Final Answers
Correct hypotheses: (a) 1 and 3
Conclusion:The three training methods are statistically equally effective at 5% level
ISS PREVIOUS YEAR 2016 PAPER-2 SOLUTION SET-A
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