CUET UG Applied Mathematics Booster Test 2 - Parameter & Statistical Concepts -

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CUET UG Applied Mathematics Booster Test 2 - Parameter & Statistical Concepts

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Applied Mathematics Booster Test 2 - Parameter & Statistical Concepts

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QUESTION 1 OF 20

Suppose an institute analyzes the effect of a new COVID-19 vaccine. If it were practically possible to perform clinical trials that include the entire population without exception, the mean effectiveness calculated would be termed as:

Answer Review

1 Suppose an institute analyzes the effect of a new COVID-19 vaccine. If it were practically possible to perform clinical trials that include the entire population without exception, the mean effectiveness calculated would be termed as:

A Sample mean (x ˉ )

B Standard error estimate

C Population parameter (μ)

D Degree of freedom

💡 Short Explanation

Parameter describes an entire population μ denotes population mean Entire population data gives a parameter

✅ Correct Answer Explanation

A population parameter is a numerical measure calculated using data from the entire population. Since the question states that clinical trials include the entire population without exception, the calculated mean effectiveness becomes a population parameter represented by μ. Option C is correct because μ specifically denotes the population mean in statistics. Option A is incorrect because a sample mean is calculated only from a subset of the population. Option B is incorrect because standard error measures variability of sample means, not population effectiveness. Option D is incorrect because degree of freedom relates to statistical testing procedures, not population averages.

❓ Why Other Options Are Incorrect

Option A → Sample mean is based on sample data, not the whole population.
Option B → Standard error estimates variability in sampling distributions.
Option D → Degree of freedom is Strategy Used in inferential calculations like t-tests.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

The phrase "entire population without exception" directly indicates a population parameter.

Final Logic:

Whole population measurement = Population parameter (μ).

🧠 Memory Trick

"μ means whole universe"

📚 Class XI – Statistics for Economics – Chapter: Introduction to Statistics – Topic: Population and Parameter – 352

2 In inferential statistics, if μ and σ represent parameters of a standard normal population, they denote:

A Population mean and population standard deviation

B Sample mean and sample variance

C Statistic mean and statistic error

D Standard error and confidence interval

💡 Short Explanation

μ represents population mean σ represents population standard deviation Both are population parameters

✅ Correct Answer Explanation

In inferential statistics, Greek symbols are generally Strategy Used for population parameters. μ represents the population mean, while σ represents the population standard deviation. Therefore, Option A is correct. Option B is incorrect because sample statistics are generally represented using x ˉ and s. Option C is incorrect because "statistic error" is not a standard notation concept. Option D is incorrect because σ does not represent confidence intervals.

❓ Why Other Options Are Incorrect

Option B → Sample measures use sample notation, not Greek population symbols.
Option C → Statistical error terminology is incorrectly Strategy Used here.
Option D → Standard error and confidence interval are separate inferential concepts.

🎯 Strategy

Used

Option Grouping

Application:

Recognize standard statistical notation pairs.

Final Logic:

μ and σ are standard symbols for population mean and standard deviation.

🧠 Memory Trick

"Greek letters = Population"

📚 Class XI – Statistics for Economics – Chapter: Introduction to Statistics – Topic: Statistical Symbols and Parameters –

3 Which statement incorrectly describes a statistic?

A It is calculated from a portion of the population

B It is a constant value unaffected by the sample chosen

C The sample mean (x ˉ )is a statistic

D It is Strategy Used to estimate a population parameter

💡 Short Explanation

Statistics vary from sample to sample Sample mean is a statistic Statistics estimate parameters

✅ Correct Answer Explanation

A statistic is a numerical value computed from a sample and may change when different samples are selected. Option B is incorrect because a statistic is not constant; it depends on the sample chosen. Option A is correct because statistics are computed from subsets of the population. Option C is correct because the sample mean x ˉ is a commonly Strategy Used statistic. Option D is correct because statistics are Strategy Used to estimate unknown population parameters.

❓ Why Other Options Are Incorrect

Option A → Correctly defines a statistic as sample-based.
Option C → Sample mean is one of the most common statistics.
Option D → Inferential statistics uses sample statistics to estimate parameters.

🎯 Strategy

Used

Elimination

Application:

Identify the statement contradicting the variable nature of statistics.

Final Logic:

Statistics change with samples, so they are not constant.

🧠 Memory Trick

"Statistic shifts with sample"

📚 Class XI – Statistics for Economics – Chapter: Collection of Data – Topic: Sample Statistics – 354

4 Match List I with List II:

List I	List II
1. Sample Variance	a. σ²
2. Sample Mean	b. σ
3. Population Variance	c. S²
4. Population Standard Deviation	d. x̄

A 1–a, 2–d, 3–c, 4–b

B 1–b, 2–a, 3–c, 4–d

C 1–c, 2–b, 3–a, 4–d

D 1–c, 2–d, 3–a, 4–b

💡 Short Explanation

S² denotes sample variance x ˉ denotes sample mean σ² and σ are population measures

✅ Correct Answer Explanation

Sample variance is represented by S², so 1–c is correct. Sample mean is represented by x ˉ, so 2–d is correct. Population variance is denoted by σ², so 3–a is correct. Population standard deviation is represented by σ, so 4–b is correct. Thus, Option D correctly matches all statistical notations.

❓ Why Other Options Are Incorrect

Option A → Incorrectly swaps sample and population variance symbols.
Option B → Sample mean and variance notations are mismatched.
Option C → Population standard deviation is incorrectly matched with x ˉ.

🎯 Strategy

Used

Option Grouping

Application:

Pair standard statistical symbols systematically.

Final Logic:

Only Option D correctly aligns all notation pairs.

🧠 Memory Trick

"S² = Sample Square"

📚 Class XI – Statistics for Economics – Chapter: Measures of Dispersion – Topic: Statistical Notation –

5 Assertion (A): Sampling error can be derived by subtracting a population parameter from a sample statistic.
Reason (R):
"Sampling Error" = x ˉ - μ

A Both A and R are false

B A is true, R is false

C A is false, R is true

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

💡 Short Explanation

Sampling error compares statistic and parameter x ˉ − μ measures deviation Formula explains the assertion directly

✅ Correct Answer Explanation

Sampling error refers to the difference between a sample statistic and the corresponding population parameter. The formula: \(Sampling Error=\hat{x}-\mu\) shows this mathematically. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

❓ Why Other Options Are Incorrect

Option A → Both statements are actually true.
Option B → The reason is not false; it correctly defines sampling error.
Option C → The assertion is true because sampling error compares sample and population values.

🎯 Strategy

Used

Substitution

Application:

Use the given formula directly to verify the assertion.

Final Logic:

Sampling error equals sample statistic minus population parameter.

🧠 Memory Trick

"Error = Sample − Population"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Sampling Error –

6 Given rainfall data (in inches) for 5 years:
8, 5, 7, 5, 6
Compute the sample mean:
x ˉ = (8+5+7+5+6)/5 = 31/5 = 6.2
Which statement correctly distinguishes this from a parameter?

A 6.0; Parameter uses the same 5 years

B 6.4; Parameter comes from subset data

C 6.2; Parameter is computed from entire population data

D 31.0; Both are identical

💡 Short Explanation

Sample mean here equals 6.2 Parameters use entire population data Sample statistics use subset observations

✅ Correct Answer Explanation

The sample mean is correctly calculated as: \(\hat{x}=\frac{8+5+7+5+6}{5}=6.2\) A parameter differs because it is computed using the entire population rather than a sample subset. Therefore, Option C is correct. Option A is incorrect because the parameter is not based on the same sample necessarily. Option B is incorrect because parameters are not derived from subsets. Option D is incorrect because 31 is the sum, not the mean.

❓ Why Other Options Are Incorrect

Option A → Parameter refers to full population data, not just these 5 observations.
Option B → Parameters are not computed from subset data.
Option D → 31 is total rainfall, not the average.

🎯 Strategy

Used

Substitution

Application:

Compute the sample mean directly using the given data.

Final Logic:

6.2 is the sample mean, while parameters use whole population data.

🧠 Memory Trick

"Sample = Part, Parameter = Whole"

📚 Class XI – Statistics for Economics – Chapter: Measures of Central Tendency – Topic: Arithmetic Mean and Parameter – 347

📄 Passage

"In order to make decisions, assumptions about the population are made, called hypotheses. The null hypothesis \(\left(H_{0}\right)\)states that there is no difference between a parameter and a specified value, while the alternative hypothesis \(\left(H_{1}\right)\)states that a difference exists."

If a researcher rejects H₀, what conclusion is drawn?

A There is sufficient evidence supporting H₁

B H₀ is proven true

C Parameter equals the value exactly

D No conclusion can be made

💡 Short Explanation

Rejecting H₀ supports alternative hypothesis Statistical evidence favors H₁ Null hypothesis is not accepted

✅ Correct Answer Explanation

In hypothesis testing, rejecting the null hypothesis H₀ means that sufficient statistical evidence exists in favor of the alternative hypothesis H₁. Therefore, Option A is correct. Option B is incorrect because rejecting H₀ does not prove it true. Option C is incorrect because exact equality is not concluded from testing. Option D is incorrect because a statistical conclusion is indeed made.

❓ Why Other Options Are Incorrect

Option B → Rejecting H₀ cannot prove it true.
Option C → Statistical inference does not establish exact equality.
Option D → Rejection leads to inferential conclusions.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Link rejection of H₀ with acceptance/support of H₁.

Final Logic:

Rejecting null hypothesis supports the alternative.

🧠 Memory Trick

"Reject H₀ → Support H₁"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Hypothesis Testing – 359

📄 Passage

Arrange the steps in sampling for hypothesis testing:
1. Determine sample size
2. Choose sampling method
3. Identify target population
4. Select sampling frame
5. Collect data

A 1, 2, 3, 4, 5

B 3, 4, 2, 1, 5

C 3, 2, 4, 1, 5

D 4, 3, 2, 1, 5

💡 Short Explanation

Population identification comes first Sampling frame precedes sampling method Data collection is final step

✅ Correct Answer Explanation

The correct logical order in hypothesis-testing sampling is: • Identify target population • Select sampling frame • Choose sampling method • Determine sample size • Collect data Thus, the sequence becomes: 3 → 4 → 2 → 1 → 5 Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Begins with sample size before identifying population.
Option C → Sampling frame should generally be selected before method choice.
Option D → Sampling frame cannot precede target population identification.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Follow the natural research methodology sequence.

Final Logic:

Population identification logically precedes sampling operations.

🧠 Memory Trick

"Population → Frame → Method → Size → Data"

📚 Class XI – Statistics for Economics – Chapter: Collection of Data – Topic: Sampling Procedure –

9 Which statements analytically define the concept of statistical significance?
I. It is a measure of the reliability of findings
II. It indicates confidence that the finding is mathematically real
III. It implies the null hypothesis has definitively been proven true

A I only

B III only

C I and II only

D I, II, and III

💡 Short Explanation

Statistical significance measures reliability It supports meaningful findings It never proves H₀ true absolutely

✅ Correct Answer Explanation

Statement I is correct because statistical significance reflects reliability of observed findings. Statement II is also correct because significant results suggest findings are unlikely due to chance. Statement III is incorrect because hypothesis testing never definitively proves the null hypothesis true. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Ignores the correctness of Statement II.
Option B → Statement III is incorrect.
Option D → Statistical significance cannot definitively prove H₀ true.

🎯 Strategy

Used

Extreme Word Filter

Application:

The word "definitively" makes Statement III statistically invalid.

Final Logic:

Statistical significance supports evidence, not absolute proof.

🧠 Memory Trick

"Significant ≠ Absolute Proof"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Statistical Significance –

10 In a one-sample t-test with df = 4, a researcher obtains:
t_calc = -3.10, t_critical = -2.776 (α = 0.05)
What is the correct analytical decision?

A Do not reject H₀ because -3.10 > -2.776

B Reject H₁ because the obtained value is less extreme

C Accept H₀ since the test value is invalid

D Reject H₀ because the obtained value is more extreme than the critical value

💡 Short Explanation

Compare calculated and critical t-values More extreme values fall in rejection region −3.10 is beyond −2.776

✅ Correct Answer Explanation

In a left-tailed t-test, if the calculated t-value is more negative than the critical value, it falls in the rejection region. Since: \(-3.10<-2.776\) the obtained statistic is more extreme than the critical value. Therefore, the null hypothesis H₀ must be rejected. Option D is correct.

❓ Why Other Options Are Incorrect

Option A → −3.10 is actually less than −2.776, not greater.
Option B → Statistical testing concerns rejection of H₀, not H₁ directly.
Option C → The test value is valid and interpretable.

🎯 Strategy

Used

Substitution

Application:

Directly compare numerical t-values with rejection criteria.

Final Logic:

More extreme calculated t-value leads to rejection of H₀.

🧠 Memory Trick

"Beyond Critical → Reject"

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Hypothesis Testing and t-Test –

11 The standard deviation of the sampling distribution of the mean is called:

A Standard Error of the Mean (σ_M )

B Confidence margin of error

C Pooled variance

D Population variance

💡 Short Explanation

Sampling distributions have variability This variability is measured by SEM SEM represents spread of sample means

✅ Correct Answer Explanation

The standard deviation of the sampling distribution of the sample mean is known as the Standard Error of the Mean (SEM). It measures how much sample means are expected to fluctuate around the true population mean. The formula is: \({\sigma}_{M}=\frac{\sigma }{\sqrt{n}}\) Option A is correct because σ_M specifically denotes the standard error of the mean. Option B is incorrect because margin of error is Strategy Used in confidence intervals, not as the standard deviation of sampling distributions. Option C is incorrect because pooled variance combines variances from multiple samples. Option D is incorrect because population variance measures spread in the population, not variability of sample means.

❓ Why Other Options Are Incorrect

Option B → Margin of error is related to interval estimation, not directly to sampling distribution spread.
Option C → Pooled variance is Strategy Used in comparative statistical tests.
Option D → Population variance measures variability among population observations.

🎯 Strategy

Used

Option Grouping

Application:

Identify the option directly associated with sampling distribution variability.

Final Logic:

Standard deviation of sample means = Standard Error of the Mean.

🧠 Memory Trick

"SEM = Spread of Means"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Standard Error of Mean –

12 Assertion (A): The standard error of the mean (SEM) measures how much the sample mean is expected to vary from the population mean.
Reason (R): SEM depends entirely on having a very large standard deviation to increase accuracy.

A A is true, R is false

B Both A and R are false

C A is false, R is true

D Both A and R are true, and R explains A

💡 Short Explanation

SEM measures variability of sample means Smaller SEM improves accuracy Large standard deviation reduces precision

✅ Correct Answer Explanation

Assertion A is true because the Standard Error of the Mean indicates how much the sample mean is expected to vary from the population mean across repeated samples. Reason R is false because a very large standard deviation actually increases variability and decreases accuracy. SEM is given by: \(SEM=\frac{\sigma }{\sqrt{n}}\) A larger σ increases SEM, reducing precision. Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Assertion A is true.
Option C → Reason R is incorrect because larger standard deviation worsens estimation precision.
Option D → The reason does not correctly explain the assertion.

🎯 Strategy

Used

Elimination

Application:

Check whether larger variability truly increases accuracy.

Final Logic:

Larger standard deviation increases SEM, not accuracy.

🧠 Memory Trick

"Large σ → Large Error"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Standard Error –

13 If samples are drawn from a highly skewed population, the sampling distribution of the mean becomes approximately normal when:

A The population becomes normal first

B The sample size is sufficiently large (n ≥ 30)

C The sample standard deviation equals the population mean

D Only non-probability sampling is Strategy Used

💡 Short Explanation

Central Limit Theorem applies here Large samples produce near-normal distributions Parent population shape becomes less important

✅ Correct Answer Explanation

According to the Central Limit Theorem (CLT), even if the population is highly skewed, the sampling distribution of the sample mean becomes approximately normal when the sample size is sufficiently large, commonly n ≥ 30. Therefore, Option B is correct. Option A is incorrect because the population itself need not become normal. Option C is incorrect because equality between standard deviation and mean has no role in CLT. Option D is incorrect because CLT is unrelated to non-probability sampling conditions.

❓ Why Other Options Are Incorrect

Option A → CLT works regardless of original population shape.
Option C → No such requirement exists in sampling theory.
Option D → Non-probability sampling does not define CLT behavior.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

The question directly refers to CLT behavior for skewed populations.

Final Logic:

Large sample size leads to approximate normality.

🧠 Memory Trick

"n ≥ 30 → Near Normal"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Central Limit Theorem –

14 For a one-sample t-test with sample size N = 35, compute the degrees of freedom:
df = N - 1

A 35

B 36

C 34

D 30

💡 Short Explanation

Degrees of freedom for one-sample t-test is N − 1 Substitute N = 35 df = 34

✅ Correct Answer Explanation

For a one-sample t-test, degrees of freedom are calculated using: \(df=N-1\) Substituting N = 35: \(df=35-1=34\) Therefore, Option C is correct. Options A, B, and D do not satisfy the formula.

❓ Why Other Options Are Incorrect

Option A → Uses N directly without subtracting 1.
Option B → Incorrect addition instead of subtraction.
Option D → Does not follow the formula.

🎯 Strategy

Used

Substitution

Application:

Directly substitute the sample size into the df formula.

Final Logic:

35 − 1 = 34.

🧠 Memory Trick

"t-test df = One Less"

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Degrees of Freedom –

15 Assertion (A): As sample size increases, the sample mean approaches the population mean.
Reason (R): This is a direct consequence of the Central Limit Theorem.

A Both A and R are false

B A is true, R is false

C A is false, R is true

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

💡 Short Explanation

Larger samples improve estimation Sample means stabilize near population mean CLT explains this convergence behavior

✅ Correct Answer Explanation

Assertion A is true because as sample size increases, the sample mean becomes a better estimate of the population mean. Reason R is also true because the Central Limit Theorem explains how sampling distributions behave with larger samples. Thus, both statements are true, and the reason correctly explains the assertion. Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Both statements are actually true.
Option B → Reason is not false.
Option C → Assertion is correct because larger samples improve approximation.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Connect increasing sample size with CLT implications.

Final Logic:

CLT explains why sample means approach the population mean.

🧠 Memory Trick

"Large Sample → True Mean"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Central Limit Theorem –

16 Which statement is incorrect regarding the Central Limit Theorem?

A The sampling distribution approaches normality regardless of population shape

B The population must be normally distributed for CLT to hold

C Larger sample sizes improve estimation accuracy

D CLT generally holds when n ≥ 30

💡 Short Explanation

CLT works for many population shapes Large samples create near-normal distributions Population normality is not mandatory

✅ Correct Answer Explanation

The Central Limit Theorem states that the sampling distribution of the sample mean approaches normality for sufficiently large sample sizes regardless of the original population shape. Therefore, Option B is incorrect because population normality is not required. Option A is correct because CLT applies broadly across distributions. Option C is correct because larger samples reduce sampling variability. Option D is correct because n ≥ 30 is a commonly accepted guideline.

❓ Why Other Options Are Incorrect

Option A → Correctly states the core implication of CLT.
Option C → Larger samples improve estimate reliability.
Option D → n ≥ 30 is the standard practical benchmark.

🎯 Strategy

Used

Extreme Word Filter

Application:

The phrase "must be" makes Option B overly restrictive.

Final Logic:

CLT does not require a normally distributed population.

🧠 Memory Trick

"CLT ignores original shape"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Central Limit Theorem –

17 Suppose a 95% confidence interval for a population mean is calculated as:
(100, 300)
What does this interval estimation imply?

A Exactly 95% of all population values lie between 100 and 300

B The sample mean will always lie between 100 and 300 in 95% of samples

C If many such intervals are constructed, the true population parameter will lie within the interval 95% of the time

D The standard error is permanently bounded between 100 and 300

💡 Short Explanation

Confidence intervals relate to repeated sampling 95% confidence refers to interval reliability It does not describe population percentages directly

✅ Correct Answer Explanation

A 95% confidence interval means that if repeated samples are taken and intervals are repeatedly constructed, approximately 95% of those intervals would contain the true population parameter. Therefore, Option C is correct. Option A is incorrect because confidence intervals do not describe the percentage of population values. Option B is incorrect because sample means are not guaranteed to remain within fixed bounds. Option D is incorrect because standard error is unrelated to fixed interval boundaries.

❓ Why Other Options Are Incorrect

Option A → Confidence intervals concern parameters, not raw population observations.
Option B → Sample means can vary across samples.
Option D → Standard error is a variability measure, not a fixed interval.

🎯 Strategy

Used

Elimination

Application:

Remove options confusing confidence intervals with population distribution or standard error.

Final Logic:

Confidence level describes long-run interval success rate.

🧠 Memory Trick

"95% CI = 95% Reliable Intervals"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Confidence Interval –

18 For a right-tailed t-test with sample size:
n = 17
At significance level α = 0.05, the degrees of freedom Strategy Used to find the critical t-value is:
df = n - 1

A 16

B 17

C 18

D 0.05

💡 Short Explanation

Degrees of freedom formula is n − 1 Substitute n = 17 df = 16

✅ Correct Answer Explanation

For a t-test, degrees of freedom are calculated as: \(df=n-1\) Substituting n = 17: \(df=17-1=16\) Therefore, Option A is correct. Other options do not satisfy the formula.

❓ Why Other Options Are Incorrect

Option B → Uses sample size directly without subtracting 1.
Option C → Incorrect arithmetic.
Option D → 0.05 is significance level, not degrees of freedom.

🎯 Strategy

Used

Substitution

Application:

Apply the df formula directly.

Final Logic:

17 − 1 = 16.

🧠 Memory Trick

"df = One Less than n"

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: t-Distribution and Degrees of Freedom –

19 Match List I with List II:

List I	List II
1. Small Standard Error (SEM)	a. Inaccurate population estimation
2. Large Standard Error (SEM)	b. Number of values free to vary
3. Margin of Error	c. Accurate estimation of population mean
4. Degrees of Freedom	d. Describes the accuracy of a sampling method

A 1–b, 2–a, 3–c, 4–d

B 1–a, 2–c, 3–b, 4–d

C 1–b, 2–c, 3–a, 4–d

D 1–c, 2–a, 3–d, 4–b

💡 Short Explanation

Small SEM improves estimation accuracy Large SEM reduces reliability Degrees of freedom describe independent variation

✅ Correct Answer Explanation

A small Standard Error of the Mean indicates accurate estimation of the population mean, so 1–c is correct. A large SEM indicates inaccurate population estimation, so 2–a is correct. Margin of Error describes the accuracy of a sampling method, so 3–d is correct. Degrees of freedom represent the number of values free to vary, so 4–b is correct. Thus, Option D correctly matches all concepts.

❓ Why Other Options Are Incorrect

Option A → Incorrectly matches SEM and margin-of-error concepts.
Option B → Small SEM does not indicate inaccurate estimation.
Option C → Degrees of freedom are incorrectly paired.

🎯 Strategy

Used

Option Grouping

Application:

Match standard inferential statistics concepts carefully.

Final Logic:

Only Option D correctly pairs all definitions.

🧠 Memory Trick

"Small SEM = Strong Estimate"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Standard Error and Margin of Error –

20 Which condition provides the highest confidence that a sample mean accurately estimates the population mean?

A Small N and large standard deviation

B Large N and small standard deviation

C Small N and small standard deviation

D Large N and large standard deviation

💡 Short Explanation

Large samples improve reliability Small variability increases precision Together they minimize sampling error

✅ Correct Answer Explanation

Accurate estimation of the population mean depends on minimizing sampling error. The Standard Error of the Mean is: \(SEM=\frac{\sigma }{\sqrt{n}}\) A larger sample size n reduces SEM, while a smaller standard deviation σ also reduces SEM. Therefore, the best condition for accurate estimation is: • Large N • Small standard deviation Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Small sample size and large variability produce very high sampling error.
Option C → Small N still reduces reliability despite low variability.
Option D → Large variability weakens estimation precision.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Analyze how SEM changes with numerator and denominator.

Final Logic:

Small σ and large n minimize estimation error.

🧠 Memory Trick

"Big Sample, Small Spread"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Standard Error and Accuracy –

CUET UG Applied Mathematics Booster Test 2 - Parameter & Statistical Concepts

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