CUET UG Applied Mathematics Booster Test 2 - Sampling Methods & Errors -

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CUET UG Applied Mathematics Booster Test 2 - Sampling Methods & Errors

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Applied Mathematics Booster Test 2 - Sampling Methods & Errors

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QUESTION 1 OF 20

Which of the following analytical approaches fundamentally contradicts the core principle of probability sampling?

Answer Review

1 Which of the following analytical approaches fundamentally contradicts the core principle of probability sampling?

A Every unit has a known, non-zero probability of selection

B Selection is based on subjective judgment of the investigator

C Uses objective chance mechanisms (e.g., random number generators)

D Aims to produce unbiased population estimates

💡 Short Explanation

Probability sampling requires non-subjective, chance-based selection mechanics. Subjective selection introduces investigator bias, violating randomization. Every element must have a known, positive inclusion probability.

✅ Correct Answer Explanation

Probability sampling relies on objective chance mechanisms where every element in the population has a known, non-zero chance of being selected. Subjective selection by an investigator removes the objective element of chance and turns the design into non-probability sampling (such as judgment or purposive sampling). Hence, Option B fundamentally contradicts probability sampling principles.

❓ Why Other Options Are Incorrect

Option A → Incorrect because having a known, non-zero selection probability is a defining feature of probability sampling.
Option C → Incorrect because utilizing objective chance tools like random number generators is standard in probability sampling.
Option D → Incorrect because aiming for unbiased population estimates is a primary purpose of probability sampling designs.

🎯 Strategy

Used

Extreme Word Filter

Application:

Identify terms that imply arbitrary or biased human intervention ("subjective judgment") vs standard probability framework terms ("known probability", "objective chance", "unbiased estimates").

Final Logic: Subjective investigator choice completely violates objective randomization rules.

🧠 Memory Trick

Probability = Pure Chance, Non-Probability = Personal Choice.

📚 Class XI – Statistics for Economics – Chapter 2: Collection of Data – Topic: Sampling Methods (Probability vs Non-Probability Sampling)

2 An electoral district has \(N=100,000\) voters. Each voter has a selection probability of \(P=0.01\).
Find the sample size \(n\):

A 100

B 10,000

C 100,000

D 1,000

💡 Short Explanation

Under equal chance probability sampling, selection probability \(P=\frac{n}{N}\). Substitute \(N=100,000\) and \(P=0.01\). Solve for sample size \(n\): \(n=P\times N=0.01\times 100,000=1,000\).

✅ Correct Answer Explanation

In simple random sampling without replacement (or equal probability sampling), the probability \(P\) of any individual being included in the sample of size \(n\) out of population size \(N\) is given by \(P=\frac{n}{N}\). Rearranging to solve for sample size: \(n=P\times N=0.01\times 100,000=1,000\). Thus, Option D is the correct value.

❓ Why Other Options Are Incorrect

Option A → Incorrect because \(100\) corresponds to \(P=0.001\).
Option B → Incorrect because \(10,000\) corresponds to \(P=0.1\).
Option C → Incorrect because \(100,000\) corresponds to taking the entire population (\(P=1.0\)).

🎯 Strategy

Used

Substitution

Application:

Plug the known variables \(N=100,000\) and \(P=0.01\) directly into the inclusion probability formula \(n=P\times N\).

Final Logic: Multiplying population \(N\) by inclusion probability \(P\) yields the exact sample size \(n=1,000\).

🧠 Memory Trick

\(n=P\times N\) (Sample = Probability \(\times\) Population).

📚 Class XI – Statistics for Economics – Chapter 2: Collection of Data – Topic: Simple Random Sampling Properties

3 In a medical study of a rare disease, existing patients refer other patients for participation.
This sampling method is:

A Snowball sampling

B Stratified sampling

C Simple random sampling

D Systematic sampling

💡 Short Explanation

Referrals from existing participants form a growing network chain. Used primarily for hard-to-reach or rare target populations. Non-probability technique dependent on primary subject networks.

✅ Correct Answer Explanation

Snowball sampling (or chain referral sampling) is a non-probability sampling technique where existing study subjects recruit future subjects from among their acquaintances. It is especially useful in medical research involving rare conditions or hidden populations where sample frames are unavailable.

❓ Why Other Options Are Incorrect

Option B → Incorrect because stratified sampling divides population into homogenous subgroups before random selection.
Option C → Incorrect because simple random sampling chooses subjects randomly from a complete sampling frame.
Option D → Incorrect because systematic sampling selects every \(k\)-th unit from an ordered list.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Match the phrase "existing patients refer other patients" with the concept of chain growth (snowballing).

Final Logic: Referrals leading to additional referrals uniquely define snowball sampling.

🧠 Memory Trick

Referral rolls like a growing snowball.

📚 Class XI – Statistics for Economics – Chapter 2: Collection of Data – Topic: Non-Random Sampling Methods

4 Match List I with List II:

List I	List II
1. Voluntary online polls	a. Unbiased probability sampling
2. Selecting easily reachable chips	b. High convenience bias
3. Selecting only top-performing students	c. High selection bias
4. Lottery-based selection	d. High voluntary response bias

A 1–b, 2–c, 3–d, 4–a

B 1–a, 2–b, 3–c, 4–d

C 1–d, 2–b, 3–c, 4–a

D 1–c, 2–d, 3–b, 4–a

💡 Short Explanation

Voluntary online polls rely on self-selection \(\rightarrow\) Voluntary response bias (1–d). Easily reachable items \(\rightarrow\) Convenience bias (2–b). Choosing only top performers targets a specific biased group \(\rightarrow\) Selection bias (3–c). Lottery selection \(\rightarrow\) Unbiased probability sampling (4–a).

✅ Correct Answer Explanation

1–d: Online polls where participants choose whether to respond suffer from voluntary response bias. 2–b: Choosing easily available items (chips) introduces convenience bias. 3–c: Systematically choosing a specific sub-segment like top performers introduces selection bias. 4–a: A lottery system gives every element an equal chance, making it unbiased probability sampling. Matching these gives 1–d, 2–b, 3–c, 4–a, which corresponds to Option C.

❓ Why Other Options Are Incorrect

Option A → Incorrectly links online polls to convenience bias (b) and top-performing students to voluntary response bias (d).
Option B → Incorrectly pairs voluntary online polls with unbiased probability sampling (a).
Option D → Incorrectly pairs voluntary online polls with selection bias (c) and reachable chips with voluntary response bias (d).

🎯 Strategy

Used

Elimination

Application:

Match item 4 (Lottery-based selection) to 'a' (Unbiased probability sampling) first. Only Options A and C end with 4–a. Next, match item 1 (Voluntary online polls) to 'd' (High voluntary response bias), confirming Option C.

Final Logic: Pairing item 1 with 'd' and item 4 with 'a' uniquely isolates Option C.

🧠 Memory Trick

Lottery = Unbiased (4-a); Online Polls = Voluntary (1-d).

📚 Class XI – Statistics for Economics – Chapter 2: Collection of Data – Topic: Types of Sampling and Sampling Biases

5 Given population \(N=10\), sample size \(n=2\):
1. Total samples = 5
2. Total possible simple random samples without replacement = \(\left(\frac{10}{2}\right)=45\)
3. This is non-probability sampling
4. Each element has equal chance of exclusion

A 1 and 3 only

B 2 and 4 only

C 1, 2, and 4 only

D 1, 2, 3, and 4

💡 Short Explanation

Statement 1 is false (\(\left(\frac{10}{2}\right)=45\), not \(5\)). Statement 2 is true (\(\left(\frac{10}{2}\right)=45\)). Statement 3 is false (Simple random sampling is probability sampling). Statement 4 is true (Equal chance of inclusion implies equal chance of exclusion \(1-\frac{n}{N}\)).

✅ Correct Answer Explanation

Simple Random Sampling (SRS) without replacement allows \(\left(\frac{N}{n}\right)=\left(\frac{10}{2}\right)=45\) total possible samples. Thus, Statement 2 is correct, and Statement 1 is incorrect. SRS is a classic probability sampling method, making Statement 3 false. Since every element has an equal probability of inclusion (\(\frac{2}{10}=0.2\)), every element also has an equal probability of exclusion (\(1-0.2=0.8\)), making Statement 4 correct. Hence, statements 2 and 4 are correct (Option B).

❓ Why Other Options Are Incorrect

Option A → Incorrect because statements 1 and 3 are conceptually false.
Option C → Incorrect because statement 1 states total samples = 5, which is mathematically false.
Option D → Incorrect because it includes false statements 1 and 3.

🎯 Strategy

Used

Elimination

Application:

Evaluate Statement 3: "This is non-probability sampling" is clearly false since SRS is a probability method. Eliminate any options containing 3 (Options A, D). Evaluate Statement 1: \(\left(\frac{10}{2}\right)=45\neq 5\), so Statement 1 is false. Eliminate Option C.

Final Logic: Statements 2 and 4 are mathematically and conceptually accurate.

🧠 Memory Trick

Equal inclusion means equal exclusion; combinations set the sample total.

📚 Class XI – Statistics for Economics – Chapter 2: Collection of Data – Topic: Simple Random Sampling Equations

6 Let sample mean vector be \(\hat{x}\). The expected value across all samples \(E\left(\hat{x}\right)\) is:

A \(0\)

B Population mean \(\mu\)

C Sample size \(n\)

D Standard error \(\sigma\)

💡 Short Explanation

Sample mean vector \(\hat{x}\) is an unbiased estimator. \(E\left(\hat{x}\right)=\mu\) holds true under probability sampling. Demonstrates mathematical lack of systematic bias in SRS.

✅ Correct Answer Explanation

In simple random sampling, the sample mean vector \(\hat{x}\) is an unbiased estimator of the true population mean vector \(\mu\). By definition of unbiasedness, the expectation taken across all possible random samples equals the population parameter: \(E\left(\hat{x}\right)=\mu\).

❓ Why Other Options Are Incorrect

Option A → Incorrect because expected value is not zero unless the population mean itself is zero.
Option C → Incorrect because \(n\) represents sample size (a scalar integer count), not the expected mean vector.
Option D → Incorrect because standard error measures the standard deviation/dispersion of the sampling distribution, not its expectation/center.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Expected value of an estimator for a mean parameter vector must match the vector dimensions and metric of the population mean vector (\(\mu\)).

Final Logic: Unbiasedness guarantees \(E\left(\hat{x}\right)=\mu\).

🧠 Memory Trick

Expected value of Sample Mean = Population Mean.

📚 Class XII – Introductory Econometrics / Statistics – Chapter: Sampling Distributions – Topic: Properties of Unbiased Estimators

7 Population size \(N=2,500\), sample size \(n=50\). Find sampling interval \(k\):

A 25

B 100

C 20

D 50

💡 Short Explanation

Sampling interval \(k\) formula: \(k=\frac{N}{n}\). Population \(N=2,500\), Sample \(n=50\). Calculate: \(k=\frac{2500}{50}=50\).

✅ Correct Answer Explanation

In systematic sampling, the sampling interval \(k\) determines the spacing between selected units. It is computed as \(k=\frac{N}{n}\), where \(N\) is total population size and \(n\) is desired sample size. Substituting values: \(k=\frac{2500}{50}=50\) Hence, every 50th item is selected, making Option D correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect calculation resulting from dividing by 100 (\(2500/100=25\)).
Option B → Incorrect calculation resulting from using sample size \(n=25\) (\(2500/25=100\)).
Option C → Incorrect calculation resulting from using \(n=125\).

🎯 Strategy

Used

Substitution

Application:

Directly apply \(N=2500\) and \(n=50\) into the standard interval formula \(k=N/n\).

Final Logic: \(\frac{2500}{50}=50\).

🧠 Memory Trick

Interval \(k=Population/Sample\).

📚 Class XI – Statistics for Economics – Chapter 2: Collection of Data – Topic: Systematic Sampling Formula

8 Assertion (A): In systematic random sampling, if the ordered population contains a hidden periodic pattern aligned with the chosen interval \(k\), the resulting sample becomes biased.
Reason (R): Systematic sampling assumes that the interval \(k\) does not correlate with any inherent cyclical pattern in the population.

A Both A and R are true, and R is the correct explanation of A

B Both A and R are true, but R is not the correct explanation of A

C A is true, R is false

D A is false, R is true

💡 Short Explanation

Hidden periodicity aligned with interval \(k\) causes sampling bias. Occurs because selection repeatedly hits the same recurring peak/trough. R correctly provides the theoretical reason why A happens.

✅ Correct Answer Explanation

Periodicity is a known limitation of systematic sampling. If a population exhibits cyclical behavior at intervals matching \(k\), the sample will consistently pick elements from the exact same position in each cycle (e.g., selecting only weekend sales data). This leads to severe bias. Reason (R) correctly explains that systematic sampling relies on the assumption of non-correlation with cyclical patterns for sample validity. Therefore, both A and R are true, and R correctly explains A.

❓ Why Other Options Are Incorrect

Option B → Incorrect because R directly explains the conceptual mechanism behind Assertion A.
Option C → Incorrect because Reason R is a true fundamental assumption of systematic sampling.
Option D → Incorrect because Assertion A is a well-established statistical fact.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Connect the vulnerability of fixed intervals to recurring cycles; the presence of periodicity breaks the assumption of randomness, directly validating both statement A and explanation R.

Final Logic: Hidden cycles matching interval \(k\) break sample representative validity.

🧠 Memory Trick

Periodicity + Fixed Interval = Systematic Bias.

📚 Class XI – Statistics for Economics – Chapter 2: Collection of Data – Topic: Systematic Sampling Limitations

9 Arrange the following sampling methods from least structured (pure randomness) to most structurally biased:
1. Systematic sampling (fixed interval)
2. Simple random sampling (lottery method)
3. Convenience sampling
4. Stratified random sampling

A 2, 1, 3, 4

B 3, 2, 1, 4

C 4, 3, 2, 1

D 2, 4, 1, 3

💡 Short Explanation

Least structured / Pure randomness: Simple random sampling (2). Structured probability: Stratified random sampling (4). Fixed structural interval: Systematic sampling (1). Most structurally biased: Convenience sampling (3).

✅ Correct Answer Explanation

Ordering from pure randomness (least forced structure) to non-probability/biased structure: 1. Simple Random Sampling (2): Pure unstructured randomness (e.g., lottery). 2. Stratified Random Sampling (4): Structured randomness across pre-defined homogeneous groups. 3. Systematic Sampling (1): Rigid structural spacing (every \(k\)-th item). 4. Convenience Sampling (3): Highly subjective, unrandomized, and prone to maximum structural bias. Thus, the correct sequence is 2, 4, 1, 3 (Option D).

❓ Why Other Options Are Incorrect

Option A → Places convenience sampling before stratified sampling, incorrectly ranking a non-probability method as less structurally biased than a probability design.
Option B → Begins with convenience sampling, reversing the required order from least structured/biased to most biased.
Option C → Starts with stratified sampling and puts simple random sampling near the end.

🎯 Strategy

Used

Option Grouping

Application:

Identify extremes: Simple random sampling (2) must be first (least structured), and convenience sampling (3) must be last (most biased). Only Option D starts with 2 and ends with 3.

Final Logic: SRS (2) is purest randomness; Convenience (3) is most structurally biased.

🧠 Memory Trick

Unstructured Random \(\rightarrow\) Stratified \(\rightarrow\) Fixed Interval \(\rightarrow\) Convenience Bias.

📚 Class XI – Statistics for Economics – Chapter 2: Collection of Data – Topic: Comparison of Sampling Techniques

10 A vaccine company has:
Batch A = 600 doses, Batch B = 400 doses (Total \(N=1,000\)).
A representative sample of \(n=100\) doses is required. Using proportional allocation, find sample size \(n_{A}\) for Batch A:

A 60

B 50

C 40

D 100

💡 Short Explanation

Proportional allocation formula: \(n_{A}=\left(\frac{N_{A}}{N}\right)\times n\). \(N_{A}=600\), \(N=1,000\), \(n=100\). Calculate: \(n_{A}=\left(\frac{600}{1000}\right)\times 100=60\).

✅ Correct Answer Explanation

Under proportional stratified sampling, each stratum's sample size is proportional to its representation in the total population. \(n_{A}=\left(\frac{N_{A}}{N}\right)\times n=\left(\frac{600}{1000}\right)\times 100=60\) Thus, Batch A contributes 60 doses to the sample, making Option A correct.

❓ Why Other Options Are Incorrect

Option B → Incorrect because 50 assumes equal allocation instead of proportional allocation.
Option C → Incorrect because 40 is the proportional sample size for Batch B (\(n_{B}=\frac{400}{1000}\times 100\)).
Option D → Incorrect because 100 is the total sample size across both batches combined.

🎯 Strategy

Used

Substitution

Application:

Substitute population stratum size (\(600\)), total population (\(1,000\)), and desired total sample size (\(100\)) into \(n_{i}=\left(N_{i}/N\right)\times n\).

Final Logic: \(60\%\) of total population resides in Batch A, so Batch A gets \(60\%\) of sample \(=60\).

🧠 Memory Trick

Match stratum proportion to sample proportion (\(60\%\) of \(100=60\)).

📚 Class XI – Statistics for Economics – Chapter 2: Collection of Data – Topic: Stratified Sampling Proportional Allocation

11 In which scenario is using a non-representative sample justified?

A Studying a rare trait in a specific subgroup without generalizing

B Estimating national mean income

C Predicting election outcomes

D Measuring total industrial output

💡 Short Explanation

Non-representative/purposive sampling is acceptable when generalization is not the goal. Useful for exploratory research or rare subgroup case studies. Population-level estimates (income, elections) strictly require representative probability samples.

✅ Correct Answer Explanation

Non-representative sampling methods (such as purposive, quota, or snowball sampling) are justified when the research goal is qualitative, exploratory, or focused entirely on specific subgroups without aiming to generalize findings to a broader population. In studying rare traits in specific subgroups, obtaining a strictly representative probability sample is often impractical or unnecessary.

❓ Why Other Options Are Incorrect

Option B → Incorrect because estimating national mean income requires a representative sample to avoid extreme income distribution bias.
Option C → Incorrect because predicting election outcomes requires an unbiased representative sample of voter demographics.
Option D → Incorrect because measuring total industrial output requires representative cross-sector sampling.

🎯 Strategy

Used

Extreme Word Filter

Application:

Identify the option that limits scope ("without generalizing") versus options requiring broad population generalizations (national mean, election predictions, total output).

Final Logic: Non-generable exploratory studies do not mandate representative sampling frames.

🧠 Memory Trick

No Generalization = Non-Representative Allowed.

📚 Class XI – Statistics for Economics – Chapter 2: Collection of Data – Topic: Uses of Non-Probability Sampling

12 For a 95% confidence interval: What is the combined probability in both rejection tails?

A \(5\%\) or \(0.05\)

B \(2.5\%\) or \(0.025\)

C \(95\%\) or \(0.95\)

D \(1\%\) or \(0.01\)

💡 Short Explanation

Total area under probability distribution curve = \(100\%\) (\(1.0\)). Significance level \(\alpha =1-Confidence Level\). \(\alpha =1-0.95=0.05\) (\(5\%\)) combined across both tails.

✅ Correct Answer Explanation

The total area under a normal probability curve is \(100\%\) (\(1.0\)). A \(95\%\) confidence level leaves a level of significance \(\alpha =1-0.95=0.05\) (\(5\%\)). This \(\alpha =5\%\) represents the total combined probability split between both rejection tails (with \(2.5\%\) in each individual tail). Thus, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrect because \(2.5\%\) (\(0.025\)) represents the probability in each individual tail, not the combined probability in both tails.
Option C → Incorrect because \(95\%\) is the acceptance region / confidence level.
Option D → Incorrect because \(1\%\) corresponds to a \(99\%\) confidence interval.

🎯 Strategy

Used

Substitution

Application:

Use the formula \(\alpha =1-Confidence Level=1-0.95=0.05\) (\(5\%\)).

Final Logic: Combined tail area is \(\alpha =100\%-95\%=5\%\).

🧠 Memory Trick

Confidence (\(95\%\)) + Both Tails (\(5\%\)) = \(100\%\).

📚 Class XII – Introductory Econometrics / Statistics – Chapter: Estimation and Hypothesis Testing – Topic: Confidence Intervals

13 Which variables definitively contribute directly to selection bias during statistical sampling setups?
1. Utilizing an incomplete sampling frame (e.g., omitting unlisted directories)
2. Methodologically over-representing a specific target demographic
3. The natural mathematical variance (sampling error) between \(\hat{x}\) and \(\mu\)
4. Deliberately picking a non-representative, heavily localized sample

A 1, 2, and 4 only

B 1 and 3 only

C 2, 3, and 4 only

D 1, 2, 3, and 4

💡 Short Explanation

Statement 1 causes selection bias (undercoverage from incomplete frame). Statement 2 causes selection bias (systematic over-representation). Statement 3 is standard chance sampling error, not systematic selection bias. Statement 4 causes selection bias (deliberate non-representative choice).

✅ Correct Answer Explanation

Selection bias occurs when certain members of a population are systematically more or less likely to be selected in a sample than others due to human or methodological flaws. Statements 1, 2, and 4 are direct causes of selection bias. Statement 3 describes natural sampling error—the random fluctuation that occurs naturally between sample statistics and population parameters even in perfectly randomized samples—which is distinct from systematic selection bias. Therefore, 1, 2, and 4 only are correct.

❓ Why Other Options Are Incorrect

Option B → Incorrect because statement 3 is random error, not selection bias, and omits valid causes 2 and 4.
Option C → Incorrect because it includes natural mathematical variance (statement 3).
Option D → Incorrect because it includes natural sampling error (statement 3) as selection bias.

🎯 Strategy

Used

Elimination

Application:

Evaluate Statement 3: Natural variance (\(\hat{x}-\mu\)) due to chance is sampling error, not selection bias. Eliminate options containing 3 (B, C, D).

Final Logic: Removing standard sampling error (statement 3) leaves Option A.

🧠 Memory Trick

Bias = Systematic Flaw; Sampling Error = Natural Random Variance.

📚 Class XI – Statistics for Economics – Chapter 2: Collection of Data – Topic: Sampling Errors vs Non-Sampling Biases

14 A professor wants to model the average cognitive score of a large auditorium of 100 students but calculates the sample mean using only the 5 students seated in the front row because their papers are easily accessible. This represents:

A Voluntary response bias

B Convenience sampling bias

C Undercoverage bias

D Systematic sampling error

💡 Short Explanation

Selecting subjects based purely on ease of access = Convenience bias. Front-row students were picked because they were "easily accessible". Lacks randomization or self-selection dynamics.

✅ Correct Answer Explanation

Convenience sampling bias occurs when a researcher chooses elements from a population based on proximity, ease of access, or availability rather than objective randomization. Choosing the 5 closest students in the front row solely because their papers were easily reachable is a classic textbook example of convenience sampling bias.

❓ Why Other Options Are Incorrect

Option A → Incorrect because voluntary response bias requires participants to self-select into the sample willingly (e.g., opting into a poll).
Option C → Incorrect because undercoverage is a frame defect, whereas picking front-row students is a sampling method choice based on ease.
Option D → Incorrect because systematic error follows a rigid mathematical interval \(k\), not proximity convenience.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Match key phrase "easily accessible" with the statistical concept of convenience.

Final Logic: Proximity and accessibility selection defines convenience bias.

🧠 Memory Trick

Easily accessible = Convenience Sampling.

📚 Class XI – Statistics for Economics – Chapter 2: Collection of Data – Topic: Non-Random Sampling Biases

15 Match List I with List II:

List I	List II
1. A survey question is worded in a confusing or emotionally leading manner	a. Undercoverage bias
2. A web poll allows users to vote multiple times	b. Standard sampling error
3. A sample yields x̄ = 50 while μ = 52, despite proper random selection	c. Voluntary response bias
4. A survey is conducted only during working hours, excluding employed individuals	d. Response bias

A 1–b, 2–c, 3–d, 4–a

B 1–a, 2–b, 3–c, 4–d

C 1–c, 2–d, 3–a, 4–b

D 1–d, 2–c, 3–b, 4–a

💡 Short Explanation

Confusing/leading question \(\rightarrow\) Response bias (1–d). Web poll multiple voting \(\rightarrow\) Voluntary response bias (2–c). Difference despite proper random selection \(\rightarrow\) Standard sampling error (3–b). Excluding working individuals during work hours \(\rightarrow\) Undercoverage bias (4–a).

✅ Correct Answer Explanation

1–d: Leading or ambiguous survey questions cause measurement/response bias. 2–c: Open web polls that allow repeated self-selected voting suffer from voluntary response bias. 3–b: Difference between sample statistic \(\hat{x}\) and population parameter \(\mu\) under proper randomization is standard sampling error. 4–a: Omitting employed people by surveying only during work hours leaves out a demographic segment, causing undercoverage bias. Matching gives 1–d, 2–c, 3–b, 4–a (Option D).

❓ Why Other Options Are Incorrect

Option A → Incorrectly matches item 1 (leading question) to standard sampling error (b).
Option B → Incorrectly pairs item 1 to undercoverage bias (a) and item 3 to voluntary response bias (c).
Option C → Incorrectly matches item 1 to voluntary response bias (c).

🎯 Strategy

Used

Elimination

Application:

Match item 1 (leading wording) to 'd' (Response bias) and item 3 (difference despite random selection) to 'b' (Standard sampling error). Option D is the only option with both 1–d and 3–b.

Final Logic: Matching items 1–d and 3–b uniquely leaves Option D.

🧠 Memory Trick

Leading wording = Response Bias; Random chance difference = Sampling Error.

📚 Class XI – Statistics for Economics – Chapter 2: Collection of Data – Topic: Types of Errors and Biases

16 Assertion (A): Undercoverage bias invalidates a dataset when the sample provides disproportionately less representation to certain members of the population.
Reason (R): This occurs when the sampling frame is incomplete or ignores specific demographic groups.

A Both A and R are false

B A is true, R is false

C Both A and R are true, and R is the correct explanation of A

D A is false, R is true

💡 Short Explanation

Undercoverage distorts data representation for target populations. Incomplete sampling frames directly cause undercoverage. Reason R provides the exact causal mechanism for Assertion A.

✅ Correct Answer Explanation

Assertion (A) accurately defines undercoverage bias: it occurs when certain population members are under-represented in a sample, invalidating population-wide inferences. Reason (R) correctly explains why this happens: the sampling frame (the list from which the sample is drawn) is incomplete, omitting specific groups. Thus, both statements are true, and R is the correct explanation of A.

❓ Why Other Options Are Incorrect

Option A → Incorrect because both statements are established statistical principles.
Option B → Incorrect because Reason R is true.
Option D → Incorrect because Assertion A is true.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Connect the definition of undercoverage (A) directly to its structural cause—an incomplete sampling frame (R).

Final Logic: Incomplete sampling frames (R) directly create under-representation (A).

🧠 Memory Trick

Incomplete Frame = Undercoverage Bias.

📚 Class XI – Statistics for Economics – Chapter 2: Collection of Data – Topic: Sampling Frame Errors

17 A population contains values: \(P=\left\{10,20,30,40,50\right\}\).
A random sample of size \(n=3\) yields: \(S=\left\{20,30,40\right\}\).
Compute the sampling error: \(e=\hat{x}-\mu\):

A \(10\)

B \(-5\)

C \(0\)

D \(5\)

💡 Short Explanation

Calculate population mean \(\mu =\frac{10+20+30+40+50}{5}=\frac{150}{5}=30\). Calculate sample mean \(\hat{x}=\frac{20+30+40}{3}=\frac{90}{3}=30\). Sampling error \(e=\hat{x}-\mu =30-30=0\).

✅ Correct Answer Explanation

Sampling error is the difference between sample statistic and true population parameter: \(e=\hat{x}-\mu\). 1. Population Mean (\(\mu\)): \(\mu =\frac{10+20+30+40+50}{5}=\frac{150}{5}=30\) 1. Sample Mean (\(\hat{x}\)): \(\hat{x}=\frac{20+30+40}{3}=\frac{90}{3}=30\) 1. Sampling Error (\(e\)): \(e=\hat{x}-\mu =30-30=0\) Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect value resulting from subtracting single items rather than means.
Option B → Incorrect negative deviation calculation.
Option D → Incorrect positive deviation calculation.

🎯 Strategy

Used

Substitution

Application:

Compute sample and population arithmetic means directly and subtract: \(30-30=0\).

Final Logic: Sample mean equals population mean (\(\hat{x}=\mu =30\)), so error is 0.

🧠 Memory Trick

Error = Sample Mean - Population Mean.

📚 Class XI – Statistics for Economics – Chapter 2: Collection of Data – Topic: Calculation of Sampling Error

18 Which of the following is not a legitimate source of sampling error?

A Different samples estimate population parameters differently

B Faulty sample selection process

C Very small sample size

D Using symbols \(\hat{x}\) and \(\mu\) to denote statistics

💡 Short Explanation

Sampling error arises from sample variation, small samples, or faulty design. Symbol notation is merely a standard mathematical convention. Choice of notation has zero effect on numerical errors or statistical bias.

✅ Correct Answer Explanation

Sampling error arises from intrinsic variability between samples, inadequate sample sizes, or poor sampling frame/selection procedures. Standard statistical notation—such as using \(\hat{x}\) for sample mean and \(\mu\) for population mean—is simply a labeling convention and has no mathematical or practical impact on sampling error. Hence, Option D is NOT a source of sampling error.

❓ Why Other Options Are Incorrect

Option A → Incorrect because sample-to-sample variability is a core source of sampling error.
Option B → Incorrect because faulty sample selection systematically causes sampling error/bias.
Option C → Incorrect because small sample sizes increase sampling variance and error.

🎯 Strategy

Used

Extreme Word Filter

Application:

Identify notation/labeling as a passive representation, which cannot physically introduce mathematical error into data.

Final Logic: Notational symbols are labels and cannot generate sampling error.

🧠 Memory Trick

Labels/Symbols don't create errors; process and sample size do.

📚 Class XI – Statistics for Economics – Chapter 2: Collection of Data – Topic: Sources of Sampling Errors

📄 Passage

The difference between a population parameter and a sample statistic is called sampling error. Even random samples contain such errors because they are not identical to the population. These errors may be positive or negative, and they decrease as sample size increases. According to the Central Limit Theorem, the sampling distribution becomes approximately normal as sample size increases.

Based on this, if \(\mu =50\), which value of \(\hat{x}\) produces a positive sampling error?

A \(\hat{x}=55\)

B \(\hat{x}=45\)

C \(\hat{x}=50\)

D \(\hat{x}=40\)

💡 Short Explanation

Sampling error formula given in passage: \(e=\hat{x}-\mu\). For error to be positive (\(e>0\)), \(\hat{x}\) must be strictly greater than \(\mu\). With \(\mu =50\), \(\hat{x}=55⟹e=55-50=+5\).

✅ Correct Answer Explanation

According to the passage, sampling error is the numerical difference between the sample statistic (\(\hat{x}\)) and the population parameter (\(\mu\)), expressed as \(e=\hat{x}-\mu\). For the error to be positive, \(\hat{x}\) must exceed \(\mu\). Given \(\mu =50\): For Option A: \(\hat{x}=55⟹e=55-50=+5\) (Positive error). Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrect because \(\hat{x}=45⟹e=45-50=-5\) (Negative error).
Option C → Incorrect because \(\hat{x}=50⟹e=50-50=0\) (Zero error).
Option D → Incorrect because \(\hat{x}=40⟹e=40-50=-10\) (Negative error).

🎯 Strategy

Used

Substitution

Application:

Plug each option's \(\hat{x}\) value into \(e=\hat{x}-\mu\):
\(55-50=+5>0\).

Final Logic: \(\hat{x}>\mu\) yields a positive sampling error.

🧠 Memory Trick

Positive Error \(\rightarrow\) Sample Statistic > Population Parameter.

📚 Class XI – Statistics for Economics – Chapter 2: Collection of Data – Topic: Sampling Error Sign and Direction

📄 Passage

Based on the passage, what happens as sample size increases?

A Sampling error increases; distribution becomes skewed

B Sampling error decreases; distribution approaches normal (bell-shaped)

C Sampling error remains constant; distribution becomes uniform

D Sampling error becomes zero; distribution becomes exponential

💡 Short Explanation

Passage explicitly states: "errors... decrease as sample size increases." Passage explicitly states: "sampling distribution becomes approximately normal as sample size increases." Option B directly mirrors both statements from the passage.

✅ Correct Answer Explanation

The provided passage explicitly specifies two effects of increasing sample size: 1. "These errors may be positive or negative, and they decrease as sample size increases." 2. "According to the Central Limit Theorem, the sampling distribution becomes approximately normal as sample size increases." Combining these two statements directly yields Option B.

❓ Why Other Options Are Incorrect

Option A → Incorrect because it contradicts the passage by stating sampling error increases and distribution becomes skewed.
Option C → Incorrect because it claims sampling error remains constant and distribution becomes uniform.
Option D → Incorrect because sampling error approaches zero but does not automatically become absolute zero for finite samples, and the distribution becomes normal, not exponential.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Directly match the two key text statements from the passage regarding sample size increase to the corresponding option text.

Final Logic: Passage explicitly confirms error decreases and distribution becomes normal.

🧠 Memory Trick

Larger Sample = Smaller Error + Normal Curve.

📚 Class XI – Statistics for Economics – Chapter 2: Collection of Data – Topic: Central Limit Theorem and Sample Size Effects

CUET UG Applied Mathematics Booster Test 2 - Sampling Methods & Errors

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