SSC CGL Statistics – Analysis of Variance – Exam-Level Drill

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SSC CGL Statistics – Analysis of Variance – Exam-Level Drill – 15 questions, subject: Statistics (Tier II Paper II). Every question has the correct answer with a one-line explanation; tap "Show Answer". Free practice for TNPSC, UPSC, SSC, Bank and RRB exams – Shanmugam IAS Academy, Coimbatore.

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1.  

Directions (Q 36–40): Output of three machines run by four operators (one run each):

Machine Op 1 Op 2 Op 3 Op 4
M1 19 16 24 17
M2 10 14 12 14
M3 16 18 21 17

5% values: F(2, 6) = 5.14, F(3, 6) = 4.76.

Data tableThe correction factor G²/N is:

Correct Answer: C [3267]

Explanation:

G = 198, N = 12: 198²/12 = 39,204/12 = 3,267.

2.  
(Use the directions above – Q 36–40)
Data tableThe sum of squares between machines is:

Correct Answer: D [98]

Explanation:

Row totals 76, 50, 72: (76² + 50² + 72²)/4 − 3,267 = 3,365 − 3,267 = 98.

3.  
(Use the directions above – Q 36–40)
Data tableThe sum of squares between operators is:

Correct Answer: A [27]

Explanation:

Column totals 45, 48, 57, 48: (45² + 48² + 57² + 48²)/3 − 3,267 = 27.

4.  
(Use the directions above – Q 36–40)
Data tableGiven total SS = 161, the error SS and its mean square are:

Correct Answer: C [SSE = 36, MSE = 6]

Explanation:

SSE = 161 − 98 − 27 = 36; df = 2 × 3 = 6; MSE = 6.

5.  
(Use the directions above – Q 36–40)
Data tableAt the 5% level the conclusions are:

Correct Answer: B [Machines differ significantly; operators do not]

Explanation:

F(machines) = 49/6 ≈ 8.17 > 5.14; F(operators) = 9/6 = 1.5 < 4.76.

6.  
Three teaching methods give scores (12, 14, 16), (15, 17, 19) and (20, 22, 24). With F₀.₀₅(2, 6) = 5.14, the ANOVA shows:

Correct Answer: A [F = 12.25; methods differ]

Explanation:

Totals 42, 51, 66, G = 159, CF = 2,809: SSB = (42² + 51² + 66²)/3 − 2,809 = 98; SSE = 8 × 3 = 24; F = (98/2)/(24/6) = 12.25.

7.  
Reverse questionAn incomplete one-way ANOVA table: Treatments df 4, SS 64; Error SS 90, MS 4.5. The number of observations N and the F ratio are:

Correct Answer: C [N = 25, F = 3.56]

Explanation:

Error df = 90/4.5 = 20 → N = 20 + 4 + 1 = 25; MSB = 16; F = 16/4.5 ≈ 3.56.

8.  
Two groups (10, 12, 14, 12) and (15, 17, 16, 16) are compared by one-way ANOVA, giving F = 19.2. The pooled two-sample t statistic (in size) is:

Correct Answer: B [4.38]

Explanation:

With two groups F = t², so |t| = √19.2 ≈ 4.38.

9.  
Under H₀ in one-way ANOVA, the expected values of MSB and MSE are:

Correct Answer: D [Both equal σ²]

Explanation:

E(MSE) = σ² always and E(MSB) = σ² + Σnᵢαᵢ²/(k − 1), which is σ² when all αᵢ = 0; F is then close to 1 on average.

10.  
A randomised block experiment has 5 treatments in 4 blocks. The treatment F ratio has which degrees of freedom?

Correct Answer: A [(4, 12)]

Explanation:

Treatments 5 − 1 = 4; error (5 − 1)(4 − 1) = 12.

11.  
In a one-way layout, the treatment model is xᵢⱼ = μ + αᵢ + eᵢⱼ. The ANOVA null hypothesis is:

Correct Answer: D [α₁ = α₂ = … = αₖ = 0]

Explanation:

No treatment effects ⇔ all treatment means equal μ.

12.  
A Latin square experiment compares 5 treatments. The degrees of freedom for error are:

Correct Answer: B [12]

Explanation:

(t − 1)(t − 2) = 4 × 3 = 12; total 24 = 4 + 4 + 4 + 12.

13.  
One-way ANOVA with 4 groups of 4 (error df 12) gives MSE = 3; t₀.₀₂₅(12) = 2.179. Group means are 20, 23, 21.5 and 25. Which pair does NOT differ significantly by the critical-difference method?

Correct Answer: C [20 and 21.5]

Explanation:

CD = 2.179 × √(2 × 3/4) ≈ 2.67; only the gap 1.5 is below it.

14.  
A randomised block analysis gives SS(treatments) = 30 (df 3), SS(blocks) = 40 (df 2), SS(error) = 8 (df 6). If the blocks are wrongly ignored (one-way analysis), the error mean square becomes:

Correct Answer: D [6]

Explanation:

Block SS and df join the error: (40 + 8)/(2 + 6) = 6 instead of 8/6 ≈ 1.33; the treatment F falls from 7.5 to about 1.67.

15.  
In a two-way layout with 2 observations per cell, the cell means are A-Low 10, A-High 13, B-Low 11, B-High 20. What does the pattern suggest?

Correct Answer: B [Interaction: the fertiliser effect is larger for B]

Explanation:

Gain from Low to High: A +3, B +9 – unequal gains (non-parallel lines) mean interaction.

➤ Full notes, shortcuts and every answer explained – SSC CGL 2026 Statistics (Tier II Paper II) book (₹149): www.shanmugamiasacademy.in/study-materials/ssc-cgl-statistics

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