CUET UG Applied Mathematics Booster Test 2

🚀

CUET UG Applied Mathematics Booster Test 2 - t-Test & Applications

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Applied Mathematics Booster Test 2 - t-Test & Applications

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QUESTION 1 OF 20

Match the notations mapping the parameters and statistics Strategy Used when evaluating mean comparisons:

List I	List II
1. μ₀	a. Sample size
2. x̄	b. Target hypothesized population mean
3. S	c. Sample standard deviation
4. n	d. Sample mean

Answer Review

1 Match the notations mapping the parameters and statistics Strategy Used when evaluating mean comparisons:

List I	List II
1. μ₀	a. Sample size
2. x̄	b. Target hypothesized population mean
3. S	c. Sample standard deviation
4. n	d. Sample mean

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

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

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

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

💡 Short Explanation

μ₀ denotes hypothesized population mean x̄ represents sample mean S and n denote sample SD and size

✅ Correct Answer Explanation

The symbols in hypothesis testing have fixed meanings: • μ₀ → hypothesized population mean • x̄ → sample mean • S → sample standard deviation • n → sample size Thus the correct matching becomes: 1-b, 2-d, 3-c, 4-a Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrectly swaps μ₀ and x̄.
Option C → Mislabels S and μ₀.
Option D → Incorrectly maps S and x̄.

🎯 Strategy

Used

Option Grouping

Application:

Match standard statistical notation systematically.

Final Logic:

Standard symbols directly determine the correct mapping.

🧠 Memory Trick

"μ₀ → Population, x̄ → Sample"

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

2 When calculating the sample variance S² to substitute for an unknown population variance, arrange the required algebraic steps in the correct order:
1. Divide by n−1 (degrees of freedom)
2. Calculate the sample mean x̄
3. Sum all squared deviations
4. Subtract the sample mean from each data point xᵢ

A 1, 2, 3, 4

B 4, 3, 2, 1

C 2, 4, 3, 1

D 2, 3, 4, 1

💡 Short Explanation

Mean is computed first Deviations are calculated next Variance ends with division by df

✅ Correct Answer Explanation

The correct sequence for computing sample variance is: Step 1: \(\hat{x}=\frac{\sum x_{i}}{n}\) Step 2: subtract x̄ from each observation. Step 3: square and sum all deviations. Step 4: \(S^{2}=\frac{\sum (x_{i}-\hat{x})^{2}}{n-1}\) \(\sigma\) \(n\) \(\hat{x}\approx -0.01,;s^{2}\approx 1.934\) xbar=-0.01s^2 ~ 1.934 Thus, the correct arrangement is: 2 → 4 → 3 → 1 Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Division cannot occur before calculations.
Option B → Mean must be calculated first.
Option D → Deviations must be computed before summing squares.

🎯 Strategy

Used

Substitution

Application:

Follow the variance formula step-by-step.

Final Logic:

Variance computation always ends with division by n−1.

🧠 Memory Trick

"Mean → Deviate → Square → Divide"

📚 Class XI – Statistics for Economics – Chapter: Measures of Dispersion – Topic: Sample Variance –

3 The absolute magnitude of a computed t-ratio depends on which of the following components from the test statistic formula?
I. The difference between sample mean and population mean (x̄−μ₀)
II. The sample standard deviation (S)
III. The sample size (n)
IV. The significance level (α)

A I and II only

B II and III only

C I, III, and IV only

D I, II, and III only

💡 Short Explanation

t-ratio depends on numerator and denominator Standard error includes S and n α affects decision, not t-value calculation

✅ Correct Answer Explanation

The one-sample t-statistic is: \(t=\frac{\hat{x}-{\mu}_{0}}{S/\sqrt{n}}\) Thus, the magnitude of t depends on: • Difference between x̄ and μ₀ • Sample standard deviation S • Sample size n The significance level α is Strategy Used only for decision-making, not for computing the t-ratio. Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Omits sample size n.
Option B → Omits the numerator difference.
Option C → α does not affect computed t-value.

🎯 Strategy

Used

Substitution

Application:

Inspect the formula directly for included variables.

Final Logic:

Only variables inside the formula affect t magnitude.

🧠 Memory Trick

"t Depends on Mean, SD, and n"

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: t-Statistic Formula –

4 Identify the incorrect statement regarding decision indicators in hypothesis testing:

A If the p-value > α, we reject the null hypothesis

B If |t_calc| > t_crit, we reject H₀

C A two-tailed test splits the rejection regions across both tails

D A wider confidence interval corresponds to a lower significance level

💡 Short Explanation

p-value greater than α implies weak evidence H₀ is rejected only for small p-values Two-tailed tests divide α equally

✅ Correct Answer Explanation

The standard decision rule is: \(p-value<\alpha \Rightarrow Reject H_{0}\) If the p-value is greater than α, we fail to reject H₀. Thus, Option A is incorrect. Options B, C, and D correctly describe hypothesis testing principles.

❓ Why Other Options Are Incorrect

Option B → Correct rejection rule.
Option C → Correct for two-tailed testing.
Option D → Lower α gives wider confidence intervals.

🎯 Strategy

Used

Elimination

Application:

Compare each option with the standard hypothesis-testing rule.

Final Logic:

Large p-values do not justify rejection.

🧠 Memory Trick

"Small p → Reject H₀"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: p-Value Decision Rule –

5 Assertion (A):
In a right-tailed test, the alternative hypothesis is H₁: μ > μ₀.
Reason (R):
The null hypothesis in the same test is H₀: μ ≠ μ₀.

A Both A and R are false

B A is true, R is false

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

D A is false, R is true

💡 Short Explanation

Right-tailed tests use "greater than" Null hypothesis complements H₁ "≠" belongs to two-tailed tests

✅ Correct Answer Explanation

In a right-tailed test: \(H_{1}:\mu >{\mu}_{0}\) This makes Assertion A true. However, the null hypothesis should be: \(H_{0}:\mu \leq {\mu}_{0}\) not μ ≠ μ₀. Therefore, Reason R is false. Thus, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Assertion is true.
Option C → Reason is incorrect.
Option D → Assertion is not false.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Match tail direction with appropriate hypothesis symbols.

Final Logic:

Right-tailed tests require ">" in H₁.

🧠 Memory Trick

"Right Tail → Greater Than"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Right-Tailed Tests –

6 For a significance level α = 0.05 in a two-tailed test, the rejection regions are split equally across both tails.
What is the probability in the right tail rejection region?

A 0.05

B 0.95

C 0.025

D 0.10

💡 Short Explanation

Two-tailed tests divide α equally Each tail gets α/2 0.05 ÷ 2 = 0.025

✅ Correct Answer Explanation

In a two-tailed test: \(\frac{\alpha }{2}=\frac{0.05}{2}=0.025\) Thus, each rejection tail contains probability 0.025. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Represents total α, not one tail.
Option B → Represents confidence level area.
Option D → Incorrect total probability.

🎯 Strategy

Used

Substitution

Application:

Divide α equally across both tails.

Final Logic:

Two tails always receive α/2 each.

🧠 Memory Trick

"Two Tails → Half α"

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Two-Tailed Rejection Regions –

7 If the calculated t-value becomes extremely large (t → ∞), what does the tail probability
\(\int_{t}^{\infty }\,f(x) dx\)
approach?

A 1

B 0.5

C α

D 0

💡 Short Explanation

Extreme t-values move deep into tails Tail area shrinks continuously Probability approaches zero

✅ Correct Answer Explanation

As: \(t\rightarrow \infty\) the area beyond t under the curve becomes negligible. Thus: \(\int_{t}^{\infty }\,f(x) dx\rightarrow 0\) Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Total area under the curve equals 1, not the tail area.
Option B → Tail probability is not constant.
Option C → α is unrelated to the limiting behavior.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Visualize shrinking tail area for extreme t-values.

Final Logic:

Far tails contain almost zero probability.

🧠 Memory Trick

"Extreme Tail → Zero Area"

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Tail Probability –

8 If a computed p-value is 0.034 and the significance level is α = 0.05, what is the correct decision?

A Reject H₀ since 0.034 < 0.05

B Accept H₀ unconditionally

C Fail to reject H₀

D Recalculate the standard error

💡 Short Explanation

Compare p-value with α Smaller p-value means significant evidence Reject H₀ when p < α

✅ Correct Answer Explanation

The decision rule is: \(p-value<\alpha \Rightarrow Reject H_{0}\) Since: \(0.034<0.05\) the result is statistically significant. Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → H₀ is not automatically accepted.
Option C → Contradicts the p-value rule.
Option D → No recalculation is necessary.

🎯 Strategy

Used

Substitution

Application:

Directly compare numerical values.

Final Logic:

Smaller p-value implies rejection.

🧠 Memory Trick

"p Small → Reject All"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: p-Value Comparison –

9 A researcher evaluates significance levels using the vector
\(A=[0.01, 0.05, 0.10]\)
As we move from index 1 to index 3 (increasing α), how does the critical t-value behave?

A Increases indefinitely

B Decreases monotonically

C Remains constant

D Matches the sample standard deviation

💡 Short Explanation

Larger α allows easier rejection Critical boundary shifts toward center Critical t-value decreases

✅ Correct Answer Explanation

As α increases, the rejection region becomes larger. Thus, smaller t-values are sufficient for rejection. Hence, the critical t-value decreases steadily. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Critical values do not grow with α.
Option C → Critical values vary with α.
Option D → Critical t-values are unrelated to sample SD.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Associate larger α with easier rejection criteria.

Final Logic:

Easier rejection means lower critical t-values.

🧠 Memory Trick

"Higher α → Lower Critical t"

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Critical Values and Alpha –

10 A dataset initially has n = 30 observations (df = 29). If 10 more observations are added (n = 40), how does the t-distribution change?

A Becomes infinitely flat and wide

B Becomes a uniform distribution

C Becomes more concentrated and approaches the standard normal distribution

D Becomes heavily skewed

💡 Short Explanation

Larger samples increase df t-distribution narrows with larger df Shape approaches normal distribution

✅ Correct Answer Explanation

As sample size increases: \(df=n-1\) also increases. Larger degrees of freedom reduce tail thickness, causing the t-distribution to approach the standard normal distribution. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → t-distribution becomes narrower, not wider.
Option B → It never becomes uniform.
Option D → Increasing df reduces irregularity, not increases skewness.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Connect larger sample size with normal approximation.

Final Logic:

Larger df makes t-distribution more normal.

🧠 Memory Trick

"More df → More Normal"

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

11 A lighting firm expects a population bulb parameter μ = 2000 hours. However, an extracted sample array of n = 25 units shows x̄ = 1950 hours alongside S = 120.
Calculate the respective empirical t-statistic exactly.

A 2.500

B -1.950

C 4.167

D -2.083

💡 Short Explanation

Use one-sample t-test formula Compute standard error first Negative sign indicates sample mean is lower

✅ Correct Answer Explanation

The one-sample t-statistic is: \(t=\frac{\hat{x}-\mu }{S/\sqrt{n}}\) Substituting values: \(t=\frac{1950-2000}{120/\sqrt{25}}\) Since: \(\sqrt{25}=5\) and: \(120/5=24\) Thus: \(t=\frac{-50}{24}=-2.083\) Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect sign and magnitude.
Option B → Incorrect numerical calculation.
Option C → Wrong computation and positive sign.

🎯 Strategy

Used

Substitution

Application:

Substitute directly into the t-statistic formula.

Final Logic:

Lower sample mean produces a negative t-value.

🧠 Memory Trick

"Lower Mean → Negative t"

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: One-Sample t-Test Calculation –

12 Retaining the result t = -2.083 (with df = 24) from the previous bulb test, if the two-tailed significance threshold for α = 0.05 sits at a critical boundary of ±2.064, what statistical conclusion is confirmed?

A Reject H₀, as -2.083 is more extreme than -2.064

B Fail to reject H₀, as the t-value lies within bounds

C Recalculate the standard error

D Accept H₀, ignoring negative magnitude

💡 Short Explanation

Compare calculated t with critical boundary More extreme values enter rejection region Absolute t-value exceeds critical value

✅ Correct Answer Explanation

For a two-tailed test, rejection occurs if: \(∣t_{calc}∣>t_{crit}\) Here: \(∣-2.083∣=2.083>2.064\) Thus, the calculated value lies in the rejection region. Therefore, H₀ is rejected. Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → The t-value lies outside the acceptance region.
Option C → No recalculation is required.
Option D → Negative sign cannot be ignored in tail comparison.

🎯 Strategy

Used

Elimination

Application:

Compare absolute t-value with the critical value.

Final Logic:

More extreme t-values imply rejection.

🧠 Memory Trick

"|t| Bigger → Reject"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: t-Test Decision Rule – 363

13 When bridging the numeric gap between a sample mean and the assumed population parameter, the Standard Error of the Mean (SEM) provides the scaling factor in the denominator.
SEM is expressed as:

A σ × n

B S/√n

C σ/S

D √n/S

💡 Short Explanation

SEM measures variability of sample means Depends on sample SD and sample size Larger n reduces SEM

✅ Correct Answer Explanation

The Standard Error of the Mean is: \(SEM=\frac{S}{\sqrt{n}}\) It represents the standard deviation of the sampling distribution of the sample mean. This value appears in the denominator of the t-statistic formula. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect multiplication form.
Option C → Does not represent SEM.
Option D → Inverted expression.

🎯 Strategy

Used

Substitution

Application:

Recall the standard SEM formula directly.

Final Logic:

SEM always equals SD divided by √n.

🧠 Memory Trick

"SEM = SD over Root n"

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Standard Error of Mean –

14 If a meteorologist claims rainfall levels are greater than average (right-tailed hypothesis), but the calculated t-value turns out to be negative, what is the correct decision rule?

A Adjust degrees of freedom dynamically

B Reject H₀ immediately

C Fail to reject H₀, as the result contradicts the hypothesized direction

D Modify the alpha level

💡 Short Explanation

Right-tailed tests expect positive t-values Negative t contradicts H₁ direction Evidence does not support the claim

✅ Correct Answer Explanation

In a right-tailed test: \(H_{1}:\mu >{\mu}_{0}\) The rejection region lies in the positive tail. A negative t-value moves in the opposite direction of the claim. Thus, there is insufficient evidence to reject H₀. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Degrees of freedom are unrelated to direction.
Option B → Negative t-values do not support right-tailed rejection.
Option D → Alpha level should not be changed afterward.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Match sign of t-value with test direction.

Final Logic:

Right-tailed tests require positive extreme values.

🧠 Memory Trick

"Right Tail Needs Positive t"

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

15 Two independent clinical groups are analyzed: experimental group A with n₁ = 15 and control group B with n₂ = 20.
Assuming equal variances, what is the maximum possible degrees of freedom?

A 14

B 19

C 35

D 33

💡 Short Explanation

Equal variance independent t-tests use pooled df Formula adds sample sizes Subtract 2 for estimated means

✅ Correct Answer Explanation

For independent samples with equal variances: \(df=n_{1}+n_{2}-2\) Substituting values: \(df=15+20-2=33\) Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Uses only first sample df.
Option B → Uses only second sample df.
Option C → Fails to subtract 2.

🎯 Strategy

Used

Substitution

Application:

Apply pooled df formula directly.

Final Logic:

Equal variance t-tests use n₁ + n₂ − 2.

🧠 Memory Trick

"Two Samples → Add and Minus 2"

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

16 When populations exhibit unequal variances, the standard error for the difference between two sample means is calculated using which formula?

A √((s₁²/n₁) + (s₂²/n₂))

B √(sₚ²(1/n₁ + 1/n₂))

C s₁/√n₁ − s₂/√n₂

D s₁² + s₂²

💡 Short Explanation

Unequal variances avoid pooled SD Variances are handled separately Standard errors combine inside root

✅ Correct Answer Explanation

For unequal variance t-tests, the standard error is: \(SE=\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}\) This formula independently incorporates both sample variances. Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Uses pooled variance, appropriate only for equal variances.
Option C → Standard errors are not subtracted directly.
Option D → Variance sum alone is incomplete.

🎯 Strategy

Used

Option Grouping

Application:

Distinguish pooled and unpooled variance formulas.

Final Logic:

Unequal variance requires separate variance terms.

🧠 Memory Trick

"Unequal → Separate Variances"

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Unequal Variance t-Test –

17 When computing pooled variance sₚ² for two distributions, the resulting value functions as:

A A simple arithmetic mean ignoring sample size

B A weighted average based on degrees of freedom

C A multiplicative scaling factor

D A linear subtraction

💡 Short Explanation

Pooled variance combines information from both samples Larger samples contribute more weight Degrees of freedom determine weighting

✅ Correct Answer Explanation

The pooled variance formula is: \(s_{p}^{2}=\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}\) The weights are the degrees of freedom: \(n_{1}-1andn_{2}-1\) Thus, pooled variance acts as a weighted average. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Sample sizes and df are not ignored.
Option C → Pooling is not multiplicative.
Option D → No subtraction process defines pooling.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Observe weighting terms in the formula.

Final Logic:

Pooling combines variances proportionally.

🧠 Memory Trick

"Pooled = Weighted"

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Pooled Variance –

18 For Country A (s₁ = 38, n₁ = 8) and Country B (s₂ = 12, n₂ = 10), why is the degrees of freedom approximated as 7?

A Because 10 − 3 = 7

B Because 199 − 191 is close numerically

C Because the smaller of n₁ − 1 = 7 and n₂ − 1 = 9 is chosen

D Because pooled df is halved

💡 Short Explanation

Unequal variance approximation uses smaller df Compute df for each sample separately Choose the smaller value

✅ Correct Answer Explanation

For unequal variance approximation: \(df=min⁡(n_{1}-1, n_{2}-1)\) Here: \(n_{1}-1=7,n_{2}-1=9\) The smaller value is 7. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Arbitrary subtraction.
Option B → Mean difference does not determine df.
Option D → df is not halved.

🎯 Strategy

Used

Substitution

Application:

Calculate df values directly for both samples.

Final Logic:

Smaller df is selected in simplified unequal variance cases.

🧠 Memory Trick

"Unequal Variance → Smaller df"

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

📄 Passage

The Central Limit Theorem (CLT) states that sampling distributions approach a normal distribution as sample size increases, regardless of the original population distribution. Typically, \(n\geq 30\) ensures this approximation. Confidence intervals (CI) are constructed using sample statistics, probability levels, and margins of error. A 95% confidence interval indicates that we are 95% confident the true population parameter lies within the calculated interval.

A political survey estimates Candidate A will receive 30% of votes with a 5% margin of error at a 95% confidence level.
What does this imply?

A Candidate A has exactly 30% votes

B There is a 5% chance of complete failure

C Candidate A will receive 90–100% votes

D There is 95% confidence the vote share lies between 25% and 35%

💡 Short Explanation

Confidence interval uses estimate ± margin of error 30% ± 5% gives interval bounds CI expresses confidence in parameter range

✅ Correct Answer Explanation

The confidence interval is: \(30\%\pm 5\%\) Thus: \(25\%\leq p\leq 35\%\) A 95% confidence level means we are 95% confident the true population proportion lies in this interval. Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Sample estimates are not exact.
Option B → Margin of error does not represent failure probability.
Option C → Completely unrelated interval.

🎯 Strategy

Used

Substitution

Application:

Apply estimate ± margin of error.

Final Logic:

Confidence interval equals 30% ± 5%.

🧠 Memory Trick

"CI = Estimate ± Error"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Confidence Interval – 368

📄 Passage

To ensure valid application of the Central Limit Theorem without additional adjustments, what minimum sample size is typically required?

A 30 or more observations

B Exactly 10 observations

C Fewer than 5 observations

D Any random size

💡 Short Explanation

CLT improves with larger samples n ≥ 30 is the common guideline Sampling distribution approaches normality

✅ Correct Answer Explanation

The Central Limit Theorem states that the sampling distribution approaches normality as sample size increases. A widely accepted practical guideline is: \(n\geq 30\) Thus, 30 or more observations are generally sufficient. Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Usually insufficient for reliable CLT approximation.
Option C → Very small samples do not ensure normality.
Option D → Sample size does matter for CLT validity.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Associate CLT with the standard n ≥ 30 guideline.

Final Logic:

Larger samples ensure near-normal sampling distributions.

🧠 Memory Trick

"CLT → Thirty Plus"

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Central Limit Theorem –

CUET UG Applied Mathematics Booster Test 2 - t-Test & Applications

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