SSC CGL Mock – Paper II Statistics Mock Test 03 – Paper II · Statistics | Online Quiz

1.

Information collected by asking third parties – witnesses who know the facts – rather than the persons concerned is gathered by

Correct Answer: B [indirect oral investigation]

Explanation:

Indirect oral investigation collects information from third parties (witnesses) who know the facts.

2.

A fictional distribution has 'less than' cumulative frequencies: below 10 – 4, below 20 – 12, below 30 – 27, below 40 – 41, below 50 – 50. How many observations lie from 20 to less than 40?

Correct Answer: C [29]

Explanation:

cf(40) − cf(20) = 41 − 12 = 29.

3.

Statement-basedConsider the statements:
I. In an exclusive class interval, the upper limit is not counted in that class.
II. The class mark of a class equals (upper limit − lower limit) ÷ 2.
Which statement(s) is/are correct?

Correct Answer: A [Only I]

Explanation:

I is correct. Class mark = (upper + lower) ÷ 2; (upper − lower) is the width, so II is wrong.

4.

The mode of a grouped distribution can be located graphically from its

Correct Answer: A [histogram]

Explanation:

Joining the corners of the tallest bar to the adjoining bars in a histogram locates the mode.

5.

In a pie diagram of a fictional state's budget, the sector for health has an angle of 54° and stands for ₹ 4,050 crore. What is the whole budget?

Correct Answer: D [₹ 27,000 crore]

Explanation:

Total = 4,050 × 360/54 = ₹ 27,000 crore.

6.

Match the followingMatch each diagram with its main use:
P. Ogive Q. Histogram R. Lorenz curve S. Pie diagram
1. inequality of income distribution 2. locating the median 3. locating the mode 4. parts of one whole

Correct Answer: B [P-2, Q-3, R-1, S-4]

Explanation:

Ogive → median; histogram → mode; Lorenz curve → inequality; pie → parts of a whole.

7.

The mid-values of a continuous frequency distribution are 115, 125, 135 and 145. What is the upper limit of the third class?

Correct Answer: D [140]

Explanation:

Width = 10; third class = 135 ± 5 = 130–140.

8.

Marks of a fictional class are shown in a stem-and-leaf display (stem = tens):
3 | 2 5 5 8 4 | 0 1 3 6 6 9 5 | 2 4 7 6 | 1 5
How many students scored 45 or more?

Correct Answer: C [8]

Explanation:

46, 46, 49, 52, 54, 57, 61, 65 → 8 students.

9.

Books issued per day by a fictional library were grouped as 0–10 books: 5 days, 10–20: f days, 20–30: 10 days, 30–40: 7 days, 40–50: 3 days. The mean number of books issued a day is 23. The missing frequency f is

Correct Answer: A [10]

Explanation:

Σfx = 655 + 15f, N = 25 + f; (655 + 15f) = 23(25 + f) gives 8f = 80, f = 10.

10.

Electricity used in a month (units) by 50 homes of a fictional lane: 0–20: 6, 20–40: 14, 40–60: 20, 60–80: 7, 80–100: 3. The seventh decile D₇ is

Correct Answer: B [55]

Explanation:

7N/10 = 35; class 40–60 (cf before 20): 40 + (35 − 20)/20 × 20 = 55.

11.

A fictional company's output rose 21% in the first year and 44% in the second year. The average annual rate of growth is

Correct Answer: D [32%]

Explanation:

GM of growth factors = √(1.21 × 1.44) = 1.1 × 1.2 = 1.32 → 32% a year (not the AM 32.5%).

12.

Group A has 20 persons with mean income ₹ 50 thousand and group B has mean income ₹ 60 thousand. If the combined mean is ₹ 56 thousand, how many persons are in group B?

Correct Answer: C [30]

Explanation:

20(56 − 50) = n(60 − 56) → n = 120/4 = 30.

13.

A car covers 120 km at 40 km/h, then 90 km at 45 km/h and then 60 km at 30 km/h. Its average speed for the whole journey is

Correct Answer: A [38.57 km/h]

Explanation:

Total distance 270 km; time = 3 + 2 + 2 = 7 h; average = 270/7 = 38.57 km/h (weighted HM).

14.

In a moderately positively skewed distribution, the three averages are in the order

Correct Answer: C [Mode < Median < Mean]

Explanation:

Positive skew pulls the mean to the right (high values); the median lies between mode and mean.

15.

In a fictional examination taken by 400 candidates (no ties), Meena is placed 61st from the top. Her percentile rank, taken as the percentage of candidates who scored below her, is

Correct Answer: B [84.75]

Explanation:

Below her: 400 − 61 = 339 candidates → 339/400 × 100 = 84.75.

16.

The data set 4, 7, 7, 9, 12, 12, 15 is

Correct Answer: A [bimodal, with modes 7 and 12]

Explanation:

7 and 12 both occur twice, more than any other value – two modes.

17.

Statement-basedConsider the statements:
I. The median is not affected by a few extremely large values.
II. The arithmetic mean can be found for a distribution with open-end classes without any assumption.
Which statement(s) is/are correct?

Correct Answer: A [Only I]

Explanation:

I is correct. The mean needs mid-values of all classes, so open-end classes need an assumption.

18.

Two positive numbers have arithmetic mean 6.5 and geometric mean 6. The numbers are

Correct Answer: D [4 and 9]

Explanation:

Sum = 13 and product = 36 → 4 and 9 (all options have sum 13, only 4 × 9 = 36).

19.

The mean and SD of 20 observations were found to be 40 and 5. It was later found that 52 had been copied as 32. The correct SD is

Correct Answer: D [5.29]

Explanation:

Σx = 820, Σx² = 34180; mean 41; σ² = 1,709 − 1,681 = 28; σ ≈ 5.29.

20.

A fictional student scored 18 in the first unit test and 30 in the second. The standard deviation of her two scores is

Correct Answer: B [6]

Explanation:

For two numbers σ = |a − b|/2 = 12/2 = 6.

21.

Weekly overtime (hours) of 20 drivers at a fictional bus depot: 0–10: 2, 10–20: 5, 20–30: 6, 30–40: 5, 40–50: 2. The standard deviation of the overtime hours is

Correct Answer: C [11.40]

Explanation:

Mean 25; Σfd² (h = 10) = 2·4 + 5·1 + 0 + 5·1 + 2·4 = 26; σ² = 26/20 × 100 = 130; σ ≈ 11.40.

22.

Two-quantity comparisonTwo fictional brands of tyres are compared.
Quantity I: CV of brand A (mean life 80 thousand km, SD 12 thousand km)
Quantity II: CV of brand B (mean life 60 thousand km, SD 9 thousand km)

Correct Answer: C [Quantity I = Quantity II]

Explanation:

CV(A) = 12/80 × 100 = 15% and CV(B) = 9/60 × 100 = 15% – equal.

23.

Var(x) = 9. If y = 5 − 2x, the variance of y is

Correct Answer: A [36]

Explanation:

Var(a + bx) = b² Var(x) = (−2)² × 9 = 36; the constant 5 has no effect.

24.

Two-quantity comparisonFor the values 3, 7, 8, 12, 20:
Quantity I: mean deviation about the mean
Quantity II: standard deviation

Correct Answer: B [Quantity I < Quantity II]

Explanation:

Mean 10; MD = 24/5 = 4.8; SD = √33.2 ≈ 5.76; MD < SD (SD ≥ MD always).

25.

Statement-basedConsider the statements:
I. The standard deviation does not change when the origin is changed.
II. The coefficient of variation does not change when all values are multiplied by the same positive number.
Which statement(s) is/are correct?

Correct Answer: C [Both I and II]

Explanation:

I: shifting does not change deviations.

26.

For the fictional frequency data x: 2, 4, 6, 8, 10 with f: 1, 4, 6, 4, 1, the mean deviation about the mean is

Correct Answer: C [1.5]

Explanation:

Mean 6; Σf|x − 6| = 4 + 8 + 0 + 8 + 4 = 24; MD = 24/16 = 1.5.

27.

The classes of a fictional distribution run 10–20, 20–30, …, 60–70. The coefficient of range is

Correct Answer: D [0.75]

Explanation:

Range uses the extreme boundaries: (70 − 10)/(70 + 10) = 0.75.

28.

Taking 5 kg as the working origin, the first three moments of the weights of some fictional parcels are 2, 20 and 40. The third moment about the mean is

Correct Answer: B [−64]

Explanation:

μ₃ = μ₃′ − 3μ₂′μ₁′ + 2(μ₁′)³ = 40 − 120 + 16 = −64 (moments about any point convert the same way).

29.

Daily earnings of fictional street vendors have mean ₹ 450, median ₹ 420 and coefficient of variation 20%. The median-based Pearson skewness 3(mean − median)/σ equals

Correct Answer: A [1]

Explanation:

σ = 20% of 450 = 90; Sk = 3 × 30/90 = 1.

30.

For a fictional distribution, Bowley's coefficient of skewness is 0.4, Q₁ = 30 and Q₃ = 60. The median is

Correct Answer: B [39]

Explanation:

0.4 = (60 + 30 − 2Md)/30 → 2Md = 90 − 12 = 78 → Md = 39.

31.

For a fictional distribution, D₁ = 12, median = 36 and D₉ = 68. Kelly's coefficient of skewness (D₉ + D₁ − 2Md)/(D₉ − D₁) is

Correct Answer: C [0.14]

Explanation:

(68 + 12 − 72)/(68 − 12) = 8/56 ≈ 0.14.

32.

The third central moment of x is −5. If y = 2x + 3, the third central moment of y is

Correct Answer: D [−40]

Explanation:

Origin (+3) has no effect; scale 2 enters as 2³: μ₃(y) = 8 × (−5) = −40.

33.

Assertion–ReasonAssertion (A): In a symmetrical distribution, all the odd-order central moments are zero.
Reason (R): In a symmetrical distribution, deviations of the same size but opposite signs occur equally often.

Correct Answer: A [Both A and R are true and R is the correct explanation of A]

Explanation:

Equal and opposite deviations cancel in Σ(x − x̄) odd – R explains A.

34.

Scores in a fictional quiz league give the first four moments about the value 5 as 2, 20, 40 and 400. Kurtosis is measured by β₂ = μ₄/μ₂². Its value is about

Correct Answer: A [2]

Explanation:

μ₂ = 20 − 2² = 16; μ₄ = 400 − 4·40·2 + 6·20·2² − 3·2⁴ = 512; β₂ = 512/256 = 2.

35.

Diagram-basedThe scatter diagram shows six fictional students (point labels give the exact values).

Karl Pearson's correlation coefficient between x and y is about

Correct Answer: D [−0.98]

Explanation:

x̄ = 5.5, ȳ = 7.33; Σdxdy = −42, Σdx² = 37.5, Σdy² = 49.33; r ≈ −0.98.

36.

For fictional data, bᵧₓ = 0.8, bₓᵧ = 0.2 and σₓ = 3. Then σᵧ is

Correct Answer: A [6]

Explanation:

bᵧₓ/bₓᵧ = σᵧ²/σₓ² = 4 → σᵧ = 2σₓ = 6.

37.

Ranks given to seven fictional films by two critics: Critic P – 3, 1, 4, 2, 6, 7, 5; Critic Q – 2, 3, 5, 1, 7, 6, 4. The rank correlation coefficient is

Correct Answer: B [0.82]

Explanation:

Σd² = 10; ρ = 1 − 6 × 10/(7 × 48) = 0.82.

38.

A fictional firm studies sales (X₁) against advertising (X₂) and price discount (X₃): r₁₂ = 0.6, r₁₃ = 0.7, r₂₃ = 0.5. How well X₂ and X₃ together explain X₁ is measured by R₁.₂₃, which is about

Correct Answer: C [0.76]

Explanation:

R² = (0.36 + 0.49 − 2 × 0.6 × 0.7 × 0.5)/(1 − 0.25) = 0.43/0.75 ≈ 0.5733; R ≈ 0.76.

39.

In a fictional 2 × 2 table, (AB) = 36, (Aβ) = 9, (αB) = 16, (αβ) = 64. Yule's coefficient of colligation Y = (√ad − √bc)/(√ad + √bc) is

Correct Answer: D [0.6]

Explanation:

√ad = 48, √bc = 12; Y = 36/60 = 0.6 (Q would be 0.88).

40.

r between price and quantity demanded in a fictional market is −0.9. The percentage of the variation in demand that is explained by price (linearly) is

Correct Answer: B [81%]

Explanation:

Coefficient of determination r² = 0.81 → 81% (r² is never negative).

41.

Assertion–ReasonAssertion (A): If bᵧₓ = 1.6, then bₓᵧ cannot be 0.8.
Reason (R): The product of the two regression coefficients cannot exceed 1.

Correct Answer: A [Both A and R are true and R is the correct explanation of A]

Explanation:

1.6 × 0.8 = 1.28 > 1 would make r² > 1; R gives exactly this reason.

42.

For coded variables u = (x − 100)/5 and v = (y − 50)/2, the regression coefficient of v on u is 1.5. The regression coefficient of y on x is

Correct Answer: C [0.6]

Explanation:

bᵧₓ = (h_y/h_x) × b_vu = (2/5) × 1.5 = 0.6 (origins do not matter).

43.

Heights X and arm spans Y of some fictional athletes have correlation 0.5, Var(X) = 16 and Var(Y) = 25. Their covariance is

Correct Answer: A [10]

Explanation:

Cov = r σₓσᵧ = 0.5 × 4 × 5 = 10.

44.

For fictional used cars, age x (months) and resale value y (₹ ten thousand) have x̄ = 30, ȳ = 45, σₓ = 6, σᵧ = 9 and r = 0.75. Estimate the age of a car whose resale value is y = 57.

Correct Answer: D [36]

Explanation:

Use x on y: bₓᵧ = r σₓ/σᵧ = 0.5; x = 30 + 0.5(57 − 45) = 36.

45.

The two regression lines are at right angles to each other when

Correct Answer: B [r = 0]

Explanation:

With r = 0 the lines are y = ȳ and x = x̄ – perpendicular.

46.

Machines X, Y, Z of a fictional factory make 50%, 30% and 20% of the bolts; 1%, 2% and 3% of their bolts are faulty. A bolt picked at random is faulty. The probability that machine Z made it is

Correct Answer: A [6/17]

Explanation:

Bayes: 0.006/(0.005 + 0.006 + 0.006) = 6/17.

47.

Two fair dice are thrown. Given that the first die shows an even number, the probability that the sum is 10 is

Correct Answer: C [1/9]

Explanation:

Even first die: 18 outcomes; (4,6), (6,4) → 2/18 = 1/9.

48.

P(A) = 0.4, P(B) = 0.5 and P(A ∪ B) = 0.7. The events A and B are

Correct Answer: A [independent but not mutually exclusive]

Explanation:

P(A ∩ B) = 0.4 + 0.5 − 0.7 = 0.2 = P(A)P(B) → independent; it is not 0, so not mutually exclusive.

49.

Three fictional engineers independently try to fix a fault; their chances of success are 1/2, 1/3 and 1/4. The probability that the fault gets fixed is

Correct Answer: D [3/4]

Explanation:

1 − (1/2)(2/3)(3/4) = 1 − 1/4 = 3/4.

50.

Four people are chosen at random. Assuming each is equally likely to be born on any day of the week, the probability that all four were born on different days of the week is

Correct Answer: C [120/343]

Explanation:

7 × 6 × 5 × 4/7⁴ = 840/2401 = 120/343.

51.

The letters of the word STATISTICS are arranged at random. The probability that the arrangement begins with S is

Correct Answer: B [3/10]

Explanation:

By symmetry, the first letter is any of the 10 letters equally likely; 3 of them are S → 3/10.

52.

Raffle tickets of a fictional club are numbered 1 to 100 and one is drawn. A prize goes to multiples of 4 and to multiples of 6. The chance that the drawn ticket wins is

Correct Answer: B [0.33]

Explanation:

25 + 16 − 8 (multiples of 12) = 33 → 0.33.

53.

P(A) = 0.5, P(B) = 0.3 and P(A ∩ B) = 0.1. The probability that exactly one of A and B happens is

Correct Answer: C [0.6]

Explanation:

P(A) + P(B) − 2P(A ∩ B) = 0.6.

54.

A fictional coin is tossed 500 times and shows heads 320 times. Taking P(head) as 320/500 = 0.64 uses the

Correct Answer: A [relative-frequency (empirical) approach]

Explanation:

Probability estimated as the long-run proportion of occurrences is the relative-frequency approach.

55.

A box has 6 good and 4 defective bulbs. Two bulbs are taken out one after the other without replacement. The probability that one is good and the other defective is

Correct Answer: D [8/15]

Explanation:

GD or DG: 2 × (6/10)(4/9) = 48/90 = 8/15.

56.

Each of 10 fictional patients recovers independently with probability 0.3. The most likely number of recoveries (the mode) is

Correct Answer: C [3]

Explanation:

Mode = integer part of (n + 1)p = integer part of 3.3 = 3.

57.

1% of the bulbs made by a fictional plant are defective. In a box of 200 bulbs, using the Poisson approximation and e⁻² = 0.1353, the probability that the box has no defective bulb is

Correct Answer: B [0.1353]

Explanation:

m = np = 200 × 0.01 = 2; P(X = 0) = e⁻² = 0.1353.

58.

Marks in a fictional test are normal with mean 60. 10% of the candidates score above 76. Taking the z-value that cuts off the top 10% as 1.28, the SD of the marks is

Correct Answer: A [12.5]

Explanation:

(76 − 60)/σ = 1.28 → σ = 16/1.28 = 12.5.

59.

A random variable has mean 3 and variance 4. The value of E(X²) is

Correct Answer: D [13]

Explanation:

E(X²) = Var(X) + [E(X)]² = 4 + 9 = 13.

60.

In a fictional game you win ₹ 50 with probability 0.1, win ₹ 10 with probability 0.3 and lose ₹ 8 otherwise. Your expected gain per game is

Correct Answer: C [₹ 3.2]

Explanation:

0.1 × 50 + 0.3 × 10 − 0.6 × 8 = 5 + 3 − 4.8 = ₹ 3.2.

61.

The life (years) of a fictional battery is exponential with λ = 0.2 per year. A battery has already lasted 2 years. Using e^−0.6 = 0.5488, the probability that it lasts beyond 5 years is

Correct Answer: A [0.5488]

Explanation:

Memoryless: P(X > 5 | X > 2) = P(X > 3) = e −0.2 × 3 = 0.5488.

62.

For customers of a fictional bank, X = 1 if the customer has a savings account and Y = 1 if she has a loan (0 otherwise), with P(0, 0) = 0.2, P(0, 1) = 0.2, P(1, 0) = 0.1, P(1, 1) = 0.5. The covariance of X and Y is

Correct Answer: B [0.08]

Explanation:

E(X) = 0.6, E(Y) = 0.7, E(XY) = 0.5; Cov = 0.5 − 0.42 = 0.08.

63.

For a binomial distribution with n = 16, what is the largest possible variance (over all p)?

Correct Answer: D [4]

Explanation:

npq is largest at p = q = 1/2: 16 × 1/4 = 4.

64.

The number of misprints per page in a fictional book is Poisson with variance 3. Using e⁻³ = 0.0498, the probability that a page has no misprint is

Correct Answer: B [0.0498]

Explanation:

Mean = variance = 3; P(X = 0) = e⁻³ = 0.0498.

65.

Scores on a fictional aptitude scale are normal with mean 50 and SD 10. Using Q₁ = μ − 0.6745σ, the lower quartile is

Correct Answer: A [43.255]

Explanation:

Q₁ = 50 − 0.6745 × 10 = 43.255.

66.

The cumulative distribution function of a fictional random variable is F(x) = x²/9 for 0 ≤ x ≤ 3. P(1 < X ≤ 2) is

Correct Answer: C [1/3]

Explanation:

F(2) − F(1) = 4/9 − 1/9 = 1/3.

67.

A random sample of 100 is drawn without replacement from 500 fictional shops; σ = 20. Using the finite population correction √((N − n)/(N − 1)), the standard error of the mean is about

Correct Answer: D [1.79]

Explanation:

(20/√100) × √(400/499) = 2 × 0.8953 ≈ 1.79.

68.

A fictional district has 400 large farms (SD of yield 10) and 600 small farms (SD 5). A sample of 70 farms is allocated by Neyman (optimum) allocation. The number of large farms in the sample is

Correct Answer: D [40]

Explanation:

nₕ ∝ NₕSₕ: 4000 : 3000 → n₁ = 70 × 4/7 = 40.

69.

Some villages of a fictional block are chosen at random and every household in the chosen villages is surveyed. This design is

Correct Answer: A [cluster sampling]

Explanation:

Whole groups (clusters) are selected and all their units studied – cluster sampling.

70.

A fictional poll wants to estimate the proportion of voters favouring a proposal within ±5 percentage points with 95% confidence (z = 1.96), using p = 0.5 as the safest guess. The minimum sample size is

Correct Answer: C [385]

Explanation:

n = z²pq/E² = 3.8416 × 0.25/0.0025 = 384.16 → 385.

71.

Fill in the blankIn simple random sampling, the mean of the sampling distribution of the sample mean x̄ equals

Correct Answer: B [the population mean μ]

Explanation:

E(x̄) = μ: the sample mean is an unbiased estimator of the population mean.

72.

In multistage sampling, the units chosen at the very first stage are called

Correct Answer: B [primary sampling units]

Explanation:

First-stage units (e.g. districts) are primary sampling units; the last-stage units (e.g.

73.

A study of a hidden group starts with a few known members, each of whom names others to be interviewed. This method is

Correct Answer: C [snowball sampling]

Explanation:

Each respondent leads to more respondents – the sample grows like a snowball.

74.

Six fictional workers' outputs were measured before and after a training; the differences (after − before) are 2, 4, −1, 3, 2, 2. Using the paired t-test (two-sided, 5%, t₀.₀₂₅,₅ = 2.571), the result is

Correct Answer: A [t ≈ 2.93; the training made a difference]

Explanation:

d̄ = 2, s_d² = 2.8; t = 2/√(2.8/6) ≈ 2.93 > 2.571 → reject H₀.

75.

Yields (kg per plot) of two fictional varieties: A – 12, 14, 15, 11, 13; B – 16, 15, 18, 14, 17. The pooled two-sample t statistic (df = 8) is about

Correct Answer: D [3]

Explanation:

x̄_A = 13, x̄_B = 16; s_p² = 2.5; t = 3/√(2.5 × 2/5) ≈ 3 (> 2.306, significant at 5%).

76.

A fictional die rolled 120 times shows faces 1–6 with frequencies 16, 22, 18, 27, 20, 17. With χ²₀.₀₅,₅ = 11.07, the goodness-of-fit test gives

Correct Answer: B [χ² = 4.1; the die may be taken as fair]

Explanation:

Expected 20 each; χ² = (16 + 4 + 4 + 49 + 0 + 9)/20 = 4.1 < 11.07 → fair.

77.

T₁ and T₂ are unbiased estimators of θ with variances 4 and 5. The efficiency of T₂ relative to T₁ is

Correct Answer: A [0.8]

Explanation:

Efficiency = Var(T₁)/Var(T₂) = 4/5 = 0.8: T₁ is more efficient.

78.

Other things equal, if the sample size is made 4 times as large, the width of a confidence interval for the mean (σ known)

Correct Answer: C [becomes half]

Explanation:

Width ∝ 1/√n; √4 = 2 → half the width.

79.

Times (minutes) between arrivals at a fictional counter are exponential with density λe^−λx. The observed times are 3, 8, 2, 7, 5. The maximum likelihood estimate of λ is

Correct Answer: D [0.2]

Explanation:

Answer: 0.2. λ̂ = 1/x̄ = 1/5 = 0.2.

80.

Waiting times at a fictional bus stop are taken as uniform on (0, θ) minutes; the average of the observed waits is 7.5 minutes. Equating θ/2 to this average, θ is estimated as

Correct Answer: C [15]

Explanation:

E(X) = θ/2; equate to x̄: θ̂ = 2x̄ = 15.

81.

A Type II error means

Correct Answer: B [accepting H₀ when H₁ is in fact true]

Explanation:

Answer: accepting H₀ when H₁ is in fact true. β = P(accept H₀ | H₀ false).

82.

A sample of 10 fictional bolts has s² = 18 mm². To test H₀: σ² = 12 against σ² > 12 at 5% (χ²₀.₀₅,₉ = 16.919), the conclusion is

Correct Answer: A [χ² = 13.5; do not reject H₀]

Explanation:

χ² = (n − 1)s²/σ₀² = 9 × 18/12 = 13.5 < 16.919.

83.

For a random sample from a Bernoulli (0–1) population with parameter p, a sufficient statistic for p is

Correct Answer: D [the total number of successes Σxᵢ]

Explanation:

The likelihood p Σx (1 − p) n − Σx depends on the data only through Σxᵢ.

84.

Yields of three fictional wheat varieties (rows) under three fertilisers (columns), one plot each: row 1 – 6, 8, 10; row 2 – 9, 12, 12; row 3 – 12, 13, 17. In the two-way ANOVA, the F ratio for varieties is

Correct Answer: D [27]

Explanation:

SS rows = 54, SS columns = 24, SS total = 82, SSE = 4; F = (54/2)/(4/4) = 27.

85.

Every observation in a fictional one-way ANOVA is reduced by 50 to simplify the arithmetic. The F ratio

Correct Answer: C [remains unchanged]

Explanation:

Subtracting a constant leaves all sums of squares, hence F, unchanged.

86.

A fictional experiment has 30 observations in 6 groups. The total degrees of freedom in its one-way ANOVA table are

Correct Answer: B [29]

Explanation:

Total df = N − 1 = 29 (= 5 between + 24 within).

87.

A fictional one-way ANOVA with 4 diets and 20 animals gives F = 4.2. With F₀.₀₅(3, 16) = 3.24, the conclusion is

Correct Answer: A [the diets differ significantly at 5%]

Explanation:

4.2 > 3.24 → reject H₀ of equal mean gains.

88.

Three fictional brands of tyre, 5 tyres each, give total distances (thousand km) 30, 45 and 45. The between-brands sum of squares is

Correct Answer: C [30]

Explanation:

CF = 120²/15 = 960; Σ T²/n = (900 + 2025 + 2025)/5 = 990; SSB = 990 − 960 = 30.

89.

The trend of a fictional series is y = 120 + 6x with origin 2018 (x in years). Shifted to origin 2022, the equation becomes

Correct Answer: B [y = 144 + 6x]

Explanation:

New a = 120 + 6 × (2022 − 2018) = 144; slope unchanged.

90.

The annual trend of a fictional firm's sales is Y = 240 + 36x (Y = annual total, x in years, origin at the middle of a year). The equivalent monthly trend (y = monthly sales, x in months, same origin) is

Correct Answer: A [y = 20 + 0.25x]

Explanation:

Divide a by 12 (monthly level) and b by 12 × 12 = 144 (monthly level, monthly step): 20 and 0.25.

91.

In a fictional quarter, actual sales were 132 and the trend value 120. The ratio-to-trend percentage for that quarter is

Correct Answer: D [110]

Explanation:

Answer: 110. 132/120 × 100 = 110.

92.

Using the additive model, a fictional month has trend 500, seasonal effect −20, cyclical effect 15 and irregular effect 5. The observed value is

Correct Answer: D [500]

Explanation:

Y = T + S + C + I = 500 − 20 + 15 + 5 = 500.

93.

To remove a regular cycle of 5 years from a yearly series by moving averages, the period of the moving average should be

Correct Answer: B [5 years]

Explanation:

A moving average whose period equals the cycle length wipes out that cycle.

94.

For a fictional product, the average sales of quarters 1–4 over several years are 45, 60, 50, 45. By the method of simple averages, the seasonal index of quarter 2 is

Correct Answer: A [120]

Explanation:

Grand average 50; index = 60/50 × 100 = 120.

95.

Fictional data:
Item A: base price ₹ 10, base quantity 20, current price ₹ 12, current quantity 30
Item B: base price ₹ 8, base quantity 10, current price ₹ 10, current quantity 10
Item C: base price ₹ 5, base quantity 40, current price ₹ 6, current quantity 50
The Marshall–Edgeworth price index (weights q₀ + q₁) is about

Correct Answer: C [120.72]

Explanation:

Σp₁(q₀ + q₁)/Σp₀(q₀ + q₁) × 100 = 1340/1110 × 100 ≈ 120.72.

96.

Laspeyres' and Paasche's indices for a fictional basket are 132.4 and 118.8. The Dorbish–Bowley index is

Correct Answer: B [125.6]

Explanation:

(L + P)/2 = (132.4 + 118.8)/2 = 125.6.

97.

An old fictional index (base 2010 = 100) reads 160 in 2018. A new series starts with 2018 = 100 and reads 125 in 2021. Spliced to the old base, the 2021 figure is

Correct Answer: D [200]

Explanation:

Answer: 200. 125 × 160/100 = 200.

98.

For a fictional basket, base prices are ₹ 10, 5, 20, base quantities 20, 30, 10 and current quantities 25, 33, 12. Laspeyres' QUANTITY index is

Correct Answer: C [119.09]

Explanation:

Σq₁p₀/Σq₀p₀ × 100 = 655/550 × 100 = 119.09.

99.

The factor reversal test requires that (P = price index, Q = quantity index, V = value ratio)

Correct Answer: A [P₀₁ × Q₀₁ = Σp₁q₁/Σp₀q₀]

Explanation:

Price index × quantity index must equal the value ratio; Fisher's index satisfies it.

100.

Link relatives of a fictional price series for years 1, 2 and 3 are 110, 120, 90 (year 0 = 100). The chain index for year 3 is

Correct Answer: D [118.8]

Explanation:

100 × 1.10 × 1.20 × 0.90 = 118.8.

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