CUET UG Applied Mathematics Booster Test 2 - Hypothesis & Testing Basics -

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CUET UG Applied Mathematics Booster Test 2 - Hypothesis & Testing Basics

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Applied Mathematics Booster Test 2 - Hypothesis & Testing Basics

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QUESTION 1 OF 20

Which of the following accurately describe the concept of a statistical hypothesis?
1. It is a tentative, declarative statement.
2. It relates two or more variables.
3. It is always proven absolutely true by the sample.
4. It assumes values for population parameters to enable decision making.

Answer Review

1 Which of the following accurately describe the concept of a statistical hypothesis?
1. It is a tentative, declarative statement.
2. It relates two or more variables.
3. It is always proven absolutely true by the sample.
4. It assumes values for population parameters to enable decision making.

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

Hypotheses are tentative statements They may relate variables or parameters Samples do not absolutely prove hypotheses

✅ Correct Answer Explanation

A statistical hypothesis is a tentative and declarative statementStrategy Used for statistical decision-making. Statement 1 is correct because hypotheses are provisional assumptions. Statement 2 is correct because hypotheses often describe relationships between variables or parameters. Statement 4 is correct because hypotheses assume values for population parameters to guide inference. Statement 3 is incorrect because sample evidence never proves a hypothesis with absolute certainty. Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Statement 3 is incorrect because statistical proof is probabilistic, not absolute.
Option C → Includes the incorrect statement that hypotheses are always proven true.
Option D → Statement 3 makes the entire option incorrect.

🎯 Strategy

Used

Elimination

Application:

Remove options containing the extreme word "always."

Final Logic:

Statistical hypotheses are testable assumptions, not absolute truths.

🧠 Memory Trick

"Hypothesis = Tentative Test"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Meaning of Hypothesis –

2 Match the hypothesis testing steps in List I with their purposes in List II:

List I	List II
1. State H₀ and H₁	a. To establish thresholds for rejection
2. Find critical value	b. To calculate the t-ratio from data
3. Compute test value	c. To define the core assumptions
4. Make decision	d. To accept or reject H₀

A 1-c, 2-a, 3-b, 4-d

B 1-a, 2-b, 3-c, 4-d

C 1-d, 2-c, 3-b, 4-a

D 1-b, 2-d, 3-a, 4-c

💡 Short Explanation

Hypotheses define assumptions Critical values set rejection limits Decisions follow test computation

✅ Correct Answer Explanation

State H₀ and H₁ → defines assumptions → 1–c Find critical value → establishes rejection threshold → 2–a Compute test value → calculates t-ratio/statistic → 3–b Make decision → accept or reject H₀ → 4–d Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrectly swaps assumptions and computation roles.
Option C → Decision-making does not define assumptions.
Option D → Test values are not equivalent to rejection thresholds.

🎯 Strategy

Used

Option Grouping

Application:

Match each hypothesis-testing step with its standard statistical purpose.

Final Logic:

Testing sequence aligns correctly only in Option A.

🧠 Memory Trick

"State → Threshold → Compute → Decide"

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

3 Assertion (A):
The null hypothesis explicitly states that the two groups under study are the same.
Reason (R):
If the calculated absolute t-ratio is smaller than the critical value, we reject the null hypothesis.

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

H₀ assumes no difference Small t-values do not reject H₀ Rejection occurs only for extreme values

✅ Correct Answer Explanation

Assertion A is true because the null hypothesis generally assumes that groups are statistically equal or show no difference. Reason R is false because when: \(∣t_{calculated}∣<∣t_{critical}∣\) the null hypothesis is not rejected. Rejection occurs only when the calculated value exceeds the critical value. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Assertion A is true.
Option C → Reason incorrectly describes the rejection rule.
Option D → Assertion correctly describes H₀.

🎯 Strategy

Used

Substitution

Application:

Apply the standard t-test decision rule.

Final Logic:

Small t-values support retaining H₀.

🧠 Memory Trick

"Small t → Stay with H₀"

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

4 When testing a new baking process against a conventional one with average life μ₀, the statement "The new method is better than the conventional method" is biased.
How is this transformed into a formal alternative hypothesis if we suspect the new method yields longer life?

A H₁: μ = μ₀

B H₁: μ > μ₀

C H₁: μ < μ₀

D H₁: μ ≤ μ₀

💡 Short Explanation

"Better" implies greater value Longer life means larger mean Directional claims use one-tailed hypotheses

✅ Correct Answer Explanation

If the new method is suspected to produce longer life, the alternative hypothesis must indicate an increase. Thus: \(H_{1}:\mu >{\mu}_{0}\) Option B is correct because it represents a right-tailed alternative hypothesis.

❓ Why Other Options Are Incorrect

Option A → Equality belongs to H₀.
Option C → Represents shorter life instead of longer life.
Option D → Includes equality and opposite direction.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Interpret "better" and "longer life" as an increase.

Final Logic:

Longer average life implies μ > μ₀.

🧠 Memory Trick

"Better → Greater"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: One-Tailed Alternative Hypothesis –

5 If the null hypothesis is H₀: μ ≤ 15, what must be the corresponding alternative hypothesis H₁?

A =

B ≤

C >

D ≠

💡 Short Explanation

H₁ complements H₀ Opposite directional signs areStrategy Used Equality stays in H₀

✅ Correct Answer Explanation

The null hypothesis is: \(H_{0}:\mu \leq 15\) The alternative hypothesis must represent the complementary opposite: \(H_{1}:\mu >15\) Thus, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Equality alone is incomplete for H₁.
Option B → Repeats the null hypothesis sign.
Option D → Represents a two-tailed hypothesis instead of the complementary right-tail.

🎯 Strategy

Used

Option Grouping

Application:

Match null and alternative hypotheses as complementary pairs.

Final Logic:

Opposite of "≤" is ">".

🧠 Memory Trick

"≤ flips to >"

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

6 Which of the following represents a strictly complementary and valid pair of null (H₀) and alternative (H₁) hypotheses?

A H₀: < and H₁: >

B H₀: ≠ and H₁: =

C H₀: ≥ and H₁: <

D H₀: = and H₁: ≤

💡 Short Explanation

H₀ usually contains equality H₁ contains strict inequality Complementary signs avoid overlap

✅ Correct Answer Explanation

Valid null and alternative hypotheses must: • be complementary • avoid overlap • include equality in H₀ Option C satisfies these conditions: \(H_{0}:\geq ; H_{1}:<\) Thus, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → H₀ should generally include equality.
Option B → Equality should belong to H₀, not H₁.
Option D → H₁ cannot include equality if H₀ already does.

🎯 Strategy

Used

Option Grouping

Application:

Identify complementary hypothesis signs systematically.

Final Logic:

H₀ keeps equality; H₁ uses strict opposite sign.

🧠 Memory Trick

"H₀ keeps ="

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Valid Hypothesis Pairs –

7 Arrange the logical progression for framing correct statistical hypotheses:
1. Identify possible bias (e.g., assuming a new method is inherently superior).
2. Adopt a neutral attitude as the safest approach.
3. Define H₀ as "no difference" (μ = μ₀).
4. Set H₁ representing the existence of a difference.

A 4, 3, 2, 1

B 1, 4, 2, 3

C 2, 1, 3, 4

D 1, 2, 3, 4

💡 Short Explanation

Bias must be recognized first Neutrality guides hypothesis framing H₀ precedes H₁

✅ Correct Answer Explanation

Correct statistical framing follows this order: 1. Identify bias 2. Adopt neutrality 3. Define H₀ as no difference 4. Define H₁ as difference exists Thus, the correct sequence is: 1 → 2 → 3 → 4 Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Begins with H₁ before neutrality.
Option B → Sets H₁ before neutral H₀.
Option C → Ignores initial identification of bias.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Follow the logical flow of objective hypothesis construction.

Final Logic:

Bias identification precedes formal hypothesis writing.

🧠 Memory Trick

"Bias → Neutral → H₀ → H₁"

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

8 Why are preferential statements (e.g., "Method A is inferior to Method B") inappropriate for a null hypothesis?

A They perfectly reflect the sample standard deviation

B They ensure zero standard error

C They imply infinite degrees of freedom

D They introduce bias instead of maintaining neutrality

💡 Short Explanation

H₀ must remain neutral Preferential language creates bias Objective testing avoids assumptions of superiority

✅ Correct Answer Explanation

Null hypotheses are framed neutrally to avoid researcher bias. Preferential statements such as "inferior" or "better" assume conclusions before testing. Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Sample standard deviation is unrelated to wording bias.
Option B → No statement guarantees zero SEM.
Option C → Degrees of freedom are unrelated to preference wording.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Identify subjective wording that violates neutrality.

Final Logic:

H₀ must avoid preferential assumptions.

🧠 Memory Trick

"Neutral H₀ = Fair Test"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Avoiding Bias in Hypotheses –

9 A financial analyst tests the mean equated monthly installment (EMI) of a sample of 100 borrowers. The historical population mean EMI is ₹5000.
If the analyst suspects the mean has changed due to interest rate fluctuations, what are the appropriate hypotheses?

A H₀: μ > 5000; H₁: μ ≤ 5000

B H₀: μ = 5000; H₁: μ ≠ 5000

C H₀: μ < 5000; H₁: μ ≥ 5000

D H₀: μ = 5000; H₁: μ = 5000

💡 Short Explanation

"Changed" implies two-tailed testing H₀ assumes no change H₁ tests for any difference

✅ Correct Answer Explanation

The analyst suspects the mean EMI has changed, without specifying increase or decrease. Thus, the alternative hypothesis is: \(H_{1}:\mu \neq 5000\) The corresponding null hypothesis is: \(H_{0}:\mu =5000\) Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Represents a one-tailed hypothesis.
Option C → Also represents a directional one-tailed test.
Option D → H₀ and H₁ cannot both be identical.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

The word "changed" indicates a two-tailed alternative.

Final Logic:

Change without direction implies μ ≠ 5000.

🧠 Memory Trick

"Changed → ≠"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Two-Tailed Hypothesis –

10 In a clinical trial predicting viral infection rates, if the assumed probability of infection is under 25%, what are the correct hypotheses?

A H₀: p = 0.25; H₁: p ≠ 0.25

B H₀: p = 0.25; H₁: p < 0.25

C H₀: p ≤ 0.25; H₁: p > 0.25

D H₀: p < 0.25; H₁: p ≥ 0.25

💡 Short Explanation

"Under 25%" implies less-than alternative H₀ keeps equality Left-tailed test applies here

✅ Correct Answer Explanation

The assumption "under 25%" corresponds to the alternative hypothesis: \(H_{1}:p<0.25\) The null hypothesis therefore becomes: \(H_{0}:p=0.25\) Thus, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Represents a two-tailed hypothesis.
Option C → Reverses the intended direction.
Option D → Incorrectly places the claim in H₀.

🎯 Strategy

Used

Option Grouping

Application:

Match directional wording with correct null–alternative structure.

Final Logic:

"Under" corresponds to a left-tailed hypothesis.

🧠 Memory Trick

"Under → <"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Left-Tailed Hypothesis –

11 Assertion (A):
If the calculated t-value exceeds the positive critical t-value in a right-tailed test, the null hypothesis is rejected.
Reason (R):
A calculated t-value larger than the critical value indicates that the observed difference is statistically significant and not due to random error.

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

Right-tailed tests reject H₀ for large positive t-values Critical values define rejection boundaries Large t-values suggest statistical significance

✅ Correct Answer Explanation

In a right-tailed t-test, the null hypothesis is rejected when: \(t_{calculated}>t_{critical}\) Thus, Assertion A is true. Reason R is also true because a calculated t-value exceeding the critical value indicates that the observed difference is unlikely to have occurred due to random sampling fluctuation alone. The reason correctly explains why the null hypothesis is rejected. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Both statements are actually true.
Option B → Reason R correctly explains statistical significance.
Option D → Assertion A is also true.

🎯 Strategy

Used

Substitution

Application:

Apply the standard right-tailed rejection rule.

Final Logic:

Exceeding the critical value implies statistical significance.

🧠 Memory Trick

"Large t → Reject H₀"

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Right-Tailed Hypothesis Testing –

12 Regarding the decision-making process in hypothesis testing, which of the following statements is incorrect?

A Rejecting the null hypothesis implies sufficient evidence in favor of the alternative hypothesis.

B Not rejecting the null hypothesis indicates insufficient evidence to support a difference.

C Rejecting the null hypothesis proves with complete certainty that the sample mean equals the true population parameter.

D A larger absolute value of the t-statistic increases the likelihood of rejecting the null hypothesis.

💡 Short Explanation

Hypothesis testing is probabilistic Statistical conclusions are never absolutely certain Large t-values increase rejection probability

✅ Correct Answer Explanation

Option C is incorrect because statistical hypothesis testing never proves anything with complete certainty. Rejecting H₀ only indicates sufficient evidence against it based on sample data. Option A is correct because rejection of H₀ supports H₁. Option B is correct because failure to reject H₀ means evidence is insufficient. Option D is correct because larger absolute t-values are more likely to fall in rejection regions. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Correctly describes evidence in favor of H₁.
Option B → Correct interpretation of "do not reject H₀."
Option D → Larger |t| values increase the chance of rejection.

🎯 Strategy

Used

Extreme Word Filter

Application:

The phrase "complete certainty" signals statistical inaccuracy.

Final Logic:

Statistical inference is based on probability, not absolute proof.

🧠 Memory Trick

"Statistics ≠ Absolute Proof"

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

13 Given the formula for Standard Error of the Mean (SEM):
\(SEM=\frac{\sigma }{\sqrt{N}}\)
match the variables in List I with their meanings in List II:

List I	List II
1. σM	a. Standard deviation of the original distribution
2. σ	b. Sample size
3. N	c. Denominator in the formula
4. √N	d. Standard Error of the Mean

A 1-a, 2-b, 3-c, 4-d

B 1-c, 2-d, 3-a, 4-b

C 1-b, 2-c, 3-d, 4-a

D 1-d, 2-a, 3-b, 4-c

💡 Short Explanation

σ_M denotes SEM σ denotes population standard deviation √N appears in the denominator

✅ Correct Answer Explanation

σ_M represents the Standard Error of the Mean → 1–d σ represents the standard deviation of the original distribution → 2–a N denotes sample size → 3–b √N forms the denominator in the SEM formula → 4–c Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrectly swaps SEM and standard deviation meanings.
Option B → Misidentifies sample size and SEM notation.
Option C → Incorrectly assigns denominator and sample size.

🎯 Strategy

Used

Option Grouping

Application:

Match symbols directly with standard statistical definitions.

Final Logic:

Statistical notation aligns correctly only in Option D.

🧠 Memory Trick

"σ = Spread, N = Number"

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

14 The Standard Error of the Mean (SEM) measures the standard deviation of which distribution?

A The original population distribution

B A single random sample

C The distribution of degrees of freedom

D The sampling distribution of the sample mean

💡 Short Explanation

SEM measures variability of sample means It relates to repeated sampling SEM belongs to sampling distributions

✅ Correct Answer Explanation

SEM measures the spread or standard deviation of the sampling distribution of the sample mean. Mathematically: \(SEM=\frac{\sigma }{\sqrt{N}}\) Option D is correct because SEM quantifies how sample means vary across repeated samples.

❓ Why Other Options Are Incorrect

Option A → Population spread is measured by σ, not SEM.
Option B → SEM does not describe one specific sample alone.
Option C → Degrees of freedom are unrelated to distribution spread.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Connect SEM with repeated sampling behavior.

Final Logic:

SEM belongs to the sampling distribution of means.

🧠 Memory Trick

"SEM = Spread of Means"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Sampling Distribution and SEM –

15 Assertion (A):
A large sample size (N) always results in a drastically increased Standard Error of the Mean (SEM).
Reason (R):
SEM is directly proportional to the sample size N.

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

Larger samples reduce SEM SEM is inversely related to √N Increased N improves estimation accuracy

✅ Correct Answer Explanation

SEM is calculated using: \(SEM=\frac{\sigma }{\sqrt{N}}\) As sample size N increases, the denominator becomes larger, causing SEM to decrease. Therefore, Assertion A is false. Reason R is also false because SEM is inversely proportional to √N, not directly proportional to N. Thus, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Assertion A is incorrect.
Option C → Both the assertion and reason are mathematically wrong.
Option D → Reason R is also false.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Analyze how increasing N affects SEM mathematically.

Final Logic:

Larger N reduces SEM rather than increasing it.

🧠 Memory Trick

"Big N → Small SEM"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Behaviour of SEM –

16 A small sample size combined with high variance affects hypothesis testing in which of the following ways?
1. It increases the Standard Error of the Mean (SEM).
2. It makes estimation of the population mean less accurate.
3. It reduces the degrees of freedom compared to a larger sample.
4. It automatically leads to rejection of the null hypothesis.

A 1, 2, and 3 only

B 1 and 4 only

C 2, 3, and 4 only

D 1, 2, 3, and 4

💡 Short Explanation

Small N increases SEM High variance reduces estimation accuracy Null hypothesis rejection is not automatic

✅ Correct Answer Explanation

Statement 1 is correct because: \(SEM=\frac{\sigma }{\sqrt{N}}\) Small N and high variance increase SEM. Statement 2 is correct because larger SEM reduces estimation precision. Statement 3 is correct because smaller samples produce smaller degrees of freedom. Statement 4 is incorrect because rejection depends on statistical evidence, not merely sample size or variance. Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Statement 4 is incorrect.
Option C → Automatic rejection never occurs purely from sample characteristics.
Option D → Includes the incorrect Statement 4.

🎯 Strategy

Used

Elimination

Application:

Remove options containing the extreme statement "automatically."

Final Logic:

Sample characteristics affect SEM and accuracy, not guaranteed rejection.

🧠 Memory Trick

"Small N + Large σ = Weak Estimate"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: SEM and Sample Size Effects –

17 If an experiment involves assigning 3 distinct tasks to 3 available time slots, the number of values that are free to vary before the final assignment is determined represents:

A Standard deviation

B Degrees of freedom

C Critical t-value

D Statistical significance

💡 Short Explanation

Degrees of freedom represent free choices Final value becomes constrained automatically df measures independent information

✅ Correct Answer Explanation

Degrees of freedom refer to the number of independent values free to vary before constraints determine the remaining values. In the task-assignment example, once several assignments are fixed, the final assignment becomes predetermined. Thus, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Standard deviation measures dispersion.
Option C → Critical t-values are decision thresholds.
Option D → Statistical significance concerns evidence strength.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Match the idea of "free to vary" with degrees of freedom.

Final Logic:

Independent assignable values define df.

🧠 Memory Trick

"df = Free Choices"

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

18 For a two-independent sample t-test with unequal variances, suppose sample sizes are:
n₁ = 8, n₂ = 10
If the degrees of freedom are taken as the smaller of (n₁ − 1) and (n₂ − 1), what is the value used

A 18

B 9

C 7

D 16

💡 Short Explanation

Compute n₁ − 1 and n₂ − 1 Choose the smaller value Smaller df isStrategy Used here

✅ Correct Answer Explanation

Calculate: \(n_{1}-1=8-1=7\) and \(n_{2}-1=10-1=9\) The smaller value is 7. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrectly adds sample sizes.
Option B → Represents the larger df value.
Option D → Does not follow the stated rule.

🎯 Strategy

Used

Substitution

Application:

Substitute sample sizes directly into the df expressions.

Final Logic:

Minimum of 7 and 9 is 7.

🧠 Memory Trick

"Take the Smaller df"

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Degrees of Freedom in Two-Sample Tests –

19 Arrange the following sample sizes in descending order of their resulting degrees of freedom for a one-sample t-test:
1. N = 15
2. N = 30
3. N = 5
4. N = 50

A 1, 2, 3, 4

B 4, 3, 2, 1

C 3, 1, 2, 4

D 4, 2, 1, 3

💡 Short Explanation

df = N − 1 Larger sample size gives larger df Arrange from largest N to smallest N

✅ Correct Answer Explanation

For a one-sample t-test: \(df=N-1\) Thus: • N = 50 → df = 49 • N = 30 → df = 29 • N = 15 → df = 14 • N = 5 → df = 4 Descending order: 50 → 30 → 15 → 5 Hence: 4 → 2 → 1 → 3 Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Not arranged in descending order.
Option B → Places smaller sample sizes incorrectly.
Option C → Starts with the smallest sample size.

🎯 Strategy

Used

Substitution

Application:

Compute df values directly from sample sizes.

Final Logic:

Higher sample size produces higher df.

🧠 Memory Trick

"Higher N → Higher df"

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Relation Between Sample Size and df –

20 Assertion (A):
A lower degree of freedom provides more statistical power to reject a false null hypothesis.
Reason (R):
Lower degrees of freedom indicate an infinitely large sample size, eliminating sampling error.

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

Higher df generally improves statistical power Lower df usually means smaller samples Infinite samples reduce sampling error, not lower df

✅ Correct Answer Explanation

Assertion A is false because higher—not lower—degrees of freedom generally increase statistical power. Higher df usually arise from larger sample sizes, improving estimation reliability. Reason R is also false because lower df correspond to smaller sample sizes, not infinitely large samples. Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Assertion A is incorrect.
Option C → Both statements are statistically incorrect.
Option D → Reason R is also false.

🎯 Strategy

Used

Extreme Word Filter

Application:

The phrase "infinitely large sample size" signals incorrect reasoning.

Final Logic:

Larger df improve power; lower df do not.

🧠 Memory Trick

"Higher df = Higher Power"

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

CUET UG Applied Mathematics Booster Test 2 - Hypothesis & Testing Basics

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