SSC CGL Statistics – Statistical Inference – Estimation & Testing of Hypotheses – Level 2 · Exam Level · Paper II Sprint

← Back

SSC CGL Statistics – Statistical Inference – Estimation & Testing of Hypotheses – Level 2 · Exam Level · Paper II Sprint – 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.

Get daily current affairs & exam alerts – follow us YouTube WhatsApp Channel Telegram Channel Facebook Page Instagram
Share with friends & study groups
Link to this page from your website / blog (HTML code)
1.  
In a sample of 400 items, 64 are defective. The 95% confidence interval for the proportion defective is about:

Correct Answer: B [0.1241 to 0.1959]

Explanation:

p = 0.16, SE = √(0.16 × 0.84/400) ≈ 0.0183; 0.16 ± 1.96 × 0.0183.

2.  
Sample 1: n = 64, x̄ = 68, σ = 8. Sample 2: n = 36, x̄ = 65, σ = 6. Testing H₀: μ₁ = μ₂ (two-tailed), the result is:

Correct Answer: A [z ≈ 2.12; significant at 5% but not at 1%]

Explanation:

SE = √(64/64 + 36/36) = √2 ≈ 1.414; z = 3/1.414 ≈ 2.12; 1.96 < 2.12 < 2.576.

3.  
n = 9, x̄ = 48, s = 6 (divisor n − 1). Test H₀: μ = 52 at 5% (two-tailed; t₀.₀₂₅ for 8 df = 2.306). The decision is:

Correct Answer: D [t = −2; do not reject H₀]

Explanation:

t = (48 − 52)/(6/3) = −2; |t| < 2.306.

4.  
Genetic theory predicts a 1 : 2 : 1 ratio. In 200 plants the counts are 40, 110 and 50. The χ² value and the decision at 5% (χ²₀.₀₅, 2 df = 5.991) are:

Correct Answer: C [χ² = 3; theory fits]

Explanation:

Expected 50, 100, 50: 100/50 + 100/100 + 0 = 3 < 5.991.

5.  
A 2 × 2 table has cells a = 25, b = 15, c = 15, d = 25 (N = 80). The χ² statistic (without Yates' correction) is:

Correct Answer: D [5]

Explanation:

χ² = N(ad − bc)²/[(a+b)(c+d)(a+c)(b+d)] = 80 × 400²/40⁴ = 5.

6.  
Sample A: n = 8, s² = 30; sample B: n = 11, s² = 10. For H₀: σ₁² = σ₂² at 5% (F₀.₀₅(7, 10) = 3.14), the conclusion is:

Correct Answer: C [F = 3; do not reject H₀]

Explanation:

F = larger/smaller = 30/10 = 3 with (7, 10) df; 3 < 3.14, so the variances do not differ significantly.

7.  
Two-quantity comparison
Quantity I: t₀.₀₂₅ for 10 degrees of freedom
Quantity II: z₀.₀₂₅

Correct Answer: B [Quantity I > Quantity II]

Explanation:

2.228 > 1.960 – t is wider for small samples.

8.  
Assertion–Reason
A: At the same α, a right-tailed z-test rejects H₀ for z = 1.8 while a two-tailed test does not.
R: The one-tailed 5% critical value 1.645 is smaller than the two-tailed value 1.96.

Correct Answer: A [Both A and R are true and R explains A]

Explanation:

Answer: Both A and R are true and R explains A. 1.645 < 1.8 < 1.96.

9.  
Trick SpotterFor a 2 × 2 contingency table, the fastest correct way to get χ² is:

Correct Answer: C [N(ad − bc)² ÷ product of the four marginal totals]

Explanation:

In a 2 × 2 table all four |O − E| equal |ad − bc|/N, so Σ(O − E)²/E collapses to the one-line formula exactly.

10.  
Spot My Mistake“x̄ = 60, σ = 12, n = 36. The 99% interval is 60 ± 1.96 × 2 = 56.08 to 63.92.” The correct 99% interval is:

Correct Answer: A [54.848 to 65.152]

Explanation:

99% needs z = 2.576: 60 ± 2.576 × 2.

11.  
Reverse questionσ = 10 and a 95% confidence interval for μ is 46.08 to 53.92. The sample size was:

Correct Answer: D [25]

Explanation:

Half-width 3.92 = 1.96 × 10/√n → √n = 5 → n = 25.

12.  
Lifetimes (in hundreds of hours) of four lamps are 2, 3, 5 and 6, assumed exponential with rate λ. The maximum likelihood estimate of λ is:

Correct Answer: B [1/4]

Explanation:

The likelihood λⁿe −λΣx is largest at λ̂ = n/Σx = 1/x̄ = 4/16 = 1/4.

13.  
Nine students are tested before and after coaching. The differences have mean 2 and SD 3 (divisor n − 1). Using t₀.₀₂₅(8) = 2.306, the paired t-test gives:

Correct Answer: A [t = 2; not significant]

Explanation:

t = d̄/(s_d/√n) = 2/(3/3) = 2 < 2.306.

14.  
Town A: 50 of 200 voters favour a plan; town B: 30 of 200. Testing H₀: P₁ = P₂ with the pooled proportion, z equals:

Correct Answer: D [2.5 – significant at 5% but not at 1%]

Explanation:

Pooled p = 80/400 = 0.2; SE = √(0.2 × 0.8 × 2/200) = 0.04; z = 0.10/0.04 = 2.5.

15.  
Odd one outWhich of these is the odd one out among methods of point estimation?

Correct Answer: B [Analysis of Type II error]

Explanation:

The other three are methods of estimation; Type II error belongs to testing.

➤ 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 ₹299 →
Score: 0 Answered: 0 / 0