SSC CGL Statistics – Statistical Inference – Estimation & Testing of Hypotheses – Exam-Level Drill
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df = classes − 1 − parameters estimated = 7 − 1 − 1 = 5.
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SSC CGL Statistics – Statistical Inference – Estimation & Testing of Hypotheses – Exam-Level Drill – 13 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.
Data tableThe expected frequency for Branch A – Plan X under independence is:
Directions (Q 36–38): 200 customers of two branches were asked which of three service plans they prefer.
| Branch | Plan X | Plan Y | Plan Z | Total |
|---|---|---|---|---|
| A | 30 | 20 | 50 | 100 |
| B | 20 | 30 | 50 | 100 |
| Total | 50 | 50 | 100 | 200 |
Correct Answer: C [25]
Explanation:
E = row total × column total/N = 100 × 50/200 = 25.
2.
(Use the directions above – Q 36–38)
Data tableThe degrees of freedom of the χ² test of independence are:
Data tableThe degrees of freedom of the χ² test of independence are:
Correct Answer: B [2]
Explanation:
Answer: 2. (2 − 1)(3 − 1) = 2.
3.
(Use the directions above – Q 36–38)
Data tableThe χ² value and the conclusion at 5% (χ²₀.₀₅, 2 df = 5.991) are:
Data tableThe χ² value and the conclusion at 5% (χ²₀.₀₅, 2 df = 5.991) are:
Correct Answer: C [χ² = 4; preference is independent of branch]
Explanation:
Four cells give 25/25 each, the Z cells give 0: χ² = 4 < 5.991.
4.
Two independent samples of 8 each have means 20 and 17 and variances 10 and 6 (divisor n − 1). With t₀.₀₂₅(14) = 2.145, the pooled t-test gives:
Correct Answer: A [t ≈ 2.12; not significant at 5%]
Explanation:
s_p² = (70 + 42)/14 = 8; SE = √(8 × 2/8) = √2; t = 3/1.414 ≈ 2.12 < 2.145 – a close call.
5.
x̄₁ − x̄₂ = 5 with σ₁ = 6, n₁ = 36 and σ₂ = 8, n₂ = 64. The 95% confidence interval for μ₁ − μ₂ is:
Correct Answer: D [2.23 to 7.77]
Explanation:
SE = √(36/36 + 64/64) = √2; 5 ± 1.96 × 1.414 = 5 ± 2.77.
6.
Sample 1: n = 9, Σ(x − x̄)² = 96. Sample 2: n = 7, Σ(x − x̄)² = 24. With F₀.₀₅(8, 6) = 4.15, the test of equal variances gives:
Correct Answer: B [F = 3; do not reject H₀]
Explanation:
s₁² = 96/8 = 12, s₂² = 24/6 = 4 → F = 3 < 4.15.
7.
Var(T₁) = σ²/n and Var(T₂) = πσ²/(2n), both unbiased for μ. The efficiency of T₂ relative to T₁ is about:
Correct Answer: C [0.637]
Explanation:
Efficiency = Var(T₁)/Var(T₂) = 2/π ≈ 0.637 (mean vs median for a normal population).
8.
T = (X₁ + 2X₂ + 3X₃)/k is an unbiased estimator of μ (X's from one population). The value of k is:
Correct Answer: A [6]
Explanation:
E(T) = (1 + 2 + 3)μ/k = μ → k = 6.
9.
A sample 2, 4, 6, 8 is drawn from N(μ, σ²). The maximum likelihood estimate of σ² is:
Correct Answer: D [5]
Explanation:
MLE divides by n: Σ(x − 5)²/4 = 20/4 = 5 (the unbiased estimate 20/3 divides by n − 1).
10.
H₀: μ = 50, σ = 10, n = 25. The rule is: reject H₀ if x̄ > 53. The probability of a Type I error is:
Correct Answer: A [0.0668]
Explanation:
SE = 10/5 = 2; α = P(x̄ > 53 | μ = 50) = P(Z > 1.5) = 0.0668.
11.
For the same rule, if actually μ = 54, the power of the test is:
Correct Answer: D [0.6915]
Explanation:
Power = P(x̄ > 53 | μ = 54) = P(Z > −0.5) = 0.6915 (β = 0.3085).
12.
A two-tailed z-test gives z = 2.33. The p-value is about:
Correct Answer: B [0.0198]
Explanation:
p = 2[1 − Φ(2.33)] = 2 × 0.0099 = 0.0198.
13.
A Poisson distribution, with m estimated from the data, is fitted to 7 frequency classes. The χ² goodness-of-fit test has how many degrees of freedom?
Correct Answer: C [5]
Explanation:
df = classes − 1 − parameters estimated = 7 − 1 − 1 = 5.
➤ 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
📘 SSC CGL 2026 – Statistics (Tier II Paper II)
Here you get the answer with a one-line explanation. 657 original questions with worked answers · notes, formulas, figures and statistical tables (z, t, χ², F) · practice sets + 3 mock tests
Get the book – ₹149 More quizzes
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