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CUET UG Applied Mathematics Booster Test 1 - 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 1 - t-Test & Applications

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QUESTION 1 OF 20

For a standard t-distribution centered at zero, evaluate the expected value (mean):
\(E(X)=\int_{-\infty }^{\infty }\,x⋅f(x) dx\)

Answer Review

1 For a standard t-distribution centered at zero, evaluate the expected value (mean):
\(E(X)=\int_{-\infty }^{\infty }\,x⋅f(x) dx\)

A 1

B α

C Degrees of freedom

D 0

💡 Short Explanation

Standard t-distribution is symmetric about zero Positive and negative areas balance Mean of symmetric distribution equals zero

✅ Correct Answer Explanation

The standard t-distribution is centered symmetrically around zero. Because the density function satisfies symmetry: \(f(-x)=f(x)\) the positive and negative contributions in the expected value cancel each other. Thus: \(E(X)=0\) Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Total probability equals 1, not the mean.
Option B → α represents significance level, unrelated to expected value.
Option C → Degrees of freedom are parameters, not the mean.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Use symmetry properties of probability distributions.

Final Logic:

Symmetric distributions centered at zero have mean zero.

🧠 Memory Trick

"Centered Curve → Mean Zero"

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Properties of t-Distribution –

2 When the population variance is unknown and estimated using the sample standard deviation S, how does the t-distribution compare to the standard normal (Z) distribution in terms of tail behavior?

A The tails are thicker, allocating more probability to extreme values

B The tails are thinner, concentrating probability near the mean

C The distributions are identical in shape

D The probability becomes zero beyond t = 3

💡 Short Explanation

t-distribution accounts for uncertainty in variance estimation More probability exists in tails Heavy tails reduce with larger sample sizes

✅ Correct Answer Explanation

The t-distribution has heavier (thicker) tails than the standard normal distribution because the population variance is unknown and estimated from sample data. This extra uncertainty increases the probability of observing extreme values. Therefore, Option A is correct. As sample size increases, the t-distribution gradually approaches the standard normal distribution.

❓ Why Other Options Are Incorrect

Option B → t-distribution tails are heavier, not thinner.
Option C → Shapes become similar only for very large samples.
Option D → Probabilities never suddenly become zero at t = 3.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Connect "unknown variance" with "extra uncertainty."

Final Logic:

More uncertainty produces thicker tails.

🧠 Memory Trick

"Unknown σ → Heavy Tails"

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: t-Distribution Characteristics –

3 As the absolute value of the calculated t-ratio increases (moves away from 0), what happens to the tail area (p-value) beyond this value?

A The area increases exponentially

B The area decreases

C The area remains constant

D The area becomes exactly 1

💡 Short Explanation

Larger t-values move deeper into tails Tail probability shrinks Smaller p-values imply stronger evidence

✅ Correct Answer Explanation

The p-value represents the probability of observing a value as extreme as the calculated t-statistic. As: \(∣t∣↑\) the tail area beyond that value becomes smaller. Thus, the p-value decreases. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Tail probabilities decrease rather than increase.
Option C → p-values vary depending on t-values.
Option D → Tail probability can never equal the entire probability area.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Associate extreme t-values with shrinking tail probabilities.

Final Logic:

Bigger |t| means smaller p-value.

🧠 Memory Trick

"Extreme t → Tiny p"

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Interpretation of p-Value –

4 Consider a decision vector D = [H₀, H₁].
If a two-tailed test yields a calculated t-value within the interval
\(\left(-t_{crit}, t_{crit}\right)\)
which hypothesis is supported?

A Both equally

B H₁

C H₀ (Fail to reject)

D Neither

💡 Short Explanation

Values inside critical limits are non-significant No sufficient evidence against H₀ Null hypothesis is retained

✅ Correct Answer Explanation

In a two-tailed test, the rejection regions lie outside the critical values. If: \(-t_{crit}<t_{calc}<t_{crit}\) then the calculated t-value lies in the acceptance region. Thus, we fail to reject H₀. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Statistical decisions do not support both simultaneously.
Option B → H₁ is supported only in rejection regions.
Option D → One hypothesis must guide the decision.

🎯 Strategy

Used

Elimination

Application:

Identify whether the t-value lies inside or outside rejection regions.

Final Logic:

Inside boundaries implies fail to reject H₀.

🧠 Memory Trick

"Inside Limits → Keep H₀"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Acceptance and Rejection Regions –

5 A 3-day moving average gives x̄ = 1.2, tested against μ = 0 with
\(H_{1}:\mu \neq 0\)
What type of test is this?

A Left-tailed

B Right-tailed

C Single critical point

D Two-tailed

💡 Short Explanation

"Not equal to" implies two directions Both tails are tested Rejection regions occur on both sides

✅ Correct Answer Explanation

The alternative hypothesis uses: \(\mu \neq 0\) The symbol "≠" means deviations in both positive and negative directions are important. Therefore, the hypothesis test is two-tailed. Thus, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Left-tailed uses "<".
Option B → Right-tailed uses ">".
Option C → Two critical points exist in two-tailed tests.

🎯 Strategy

Used

Symbol Recognition

Application:

Identify the inequality sign Strategy Used in H₁.

Final Logic:

"≠" always indicates a two-tailed test.

🧠 Memory Trick

"≠ Means Two Tails"

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

6 Given:
n = 23 ⇒ df = 22, α = 0.01 (left-tailed),
and the one-tailed critical value t = 2.508, determine the correct critical t-value.

A -2.508

B +2.508

C -2.819

D +1.717

💡 Short Explanation

Left-tailed tests use negative critical values Magnitude remains same Sign changes according to tail direction

✅ Correct Answer Explanation

For a left-tailed t-test, the rejection region lies in the negative tail. Thus, if the table gives: \(t=2.508\) the critical value becomes: \(t=-2.508\) Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Positive critical value belongs to right-tailed tests.
Option C → Incorrect numerical value.
Option D → Incorrect table value.

🎯 Strategy

Used

Substitution

Application:

Apply sign convention for left-tailed tests.

Final Logic:

Left tail always uses negative critical values.

🧠 Memory Trick

"Left Tail → Negative t"

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: One-Tailed Critical Values –

7 Match the components of the formula
\(t=\frac{\hat{x}-\mu }{S/\sqrt{n}}\)
with their meanings:

List I	List II
1. x̄	a. Population mean
2. μ	b. Standard Error
3. S/√n	c. Sample size
4. n	d. Sample mean

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

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

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

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

💡 Short Explanation

x̄ denotes sample mean μ denotes population mean S/√n is the standard error

✅ Correct Answer Explanation

The components represent: • x̄ → sample mean • μ → population mean • S/√n → standard error • n → sample size Thus: 1-d, 2-a, 3-b, 4-c Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrectly swaps x̄ and μ.
Option C → Mislabels standard error and sample size.
Option D → Incorrect mapping of all core symbols.

🎯 Strategy

Used

Option Grouping

Application:

Match standard statistical notation systematically.

Final Logic:

Statistical symbols have fixed meanings.

🧠 Memory Trick

"x̄ = sample, μ = population"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Components of t-Formula –

8 Assertion (A):
If the calculated value is t = -3.10 and the critical value is t = -2.776, the null hypothesis is rejected.
Reason (R):
The calculated t-value lies further into the rejection region (tail) than the critical value.

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

Calculated value is more extreme Rejection region lies beyond critical value Evidence opposes H₀

✅ Correct Answer Explanation

Since: \(-3.10<-2.776\) the calculated t-value lies deeper in the rejection region. Thus, H₀ is rejected. Reason R correctly explains why the rejection occurs. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Both statements are statistically correct.
Option B → Reason correctly explains the assertion.
Option D → Assertion is also true.

🎯 Strategy

Used

Substitution

Application:

Compare the numerical locations of t-values on the number line.

Final Logic:

More extreme tail value implies rejection.

🧠 Memory Trick

"Further Tail → Reject"

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

9 Arrange the following significance levels in increasing order of the probability of a Type I error (from strictest to least strict):
1. 0.10
2. 0.05
3. 0.01
4. 0.20

A 1, 2, 3, 4

B 3, 2, 1, 4

C 2, 1, 3, 4

D 4, 1, 2, 3

💡 Short Explanation

Smaller α means stricter testing Larger α increases Type I error risk Arrange from smallest to largest α

✅ Correct Answer Explanation

Type I error probability equals α. Increasing order: \(0.01<0.05<0.10<0.20\) Thus, the correct arrangement is: 3 → 2 → 1 → 4 Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Begins with larger α instead of smallest.
Option C → Incorrect middle ordering.
Option D → Completely reversed strictness order.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Treat α directly as probability magnitude.

Final Logic:

Smaller α means stricter hypothesis testing.

🧠 Memory Trick

"Small α = Strict Test"

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

10 Identify the incorrect statement regarding degrees of freedom (df) in t-tests.

A A lower degree of freedom provides more statistical power to reject a false null hypothesis

B For a one-sample t-test, df = n − 1

C For two independent samples with equal variances, df = n₁ + n₂ − 2

D It represents the number of values free to vary during estimation

💡 Short Explanation

Higher df improves statistical power Lower df increases uncertainty df reflects free information values

✅ Correct Answer Explanation

Lower degrees of freedom produce wider t-distributions with heavier tails, reducing statistical power. Higher df generally improves estimation precision and increases the ability to reject false null hypotheses. Thus, Option A is incorrect. Options B, C, and D correctly define degrees of freedom in t-tests.

❓ Why Other Options Are Incorrect

Option B → Correct formula for one-sample t-tests.
Option C → Correct pooled df formula for equal variances.
Option D → Standard definition of degrees of freedom.

🎯 Strategy

Used

Extreme Word Filter

Application:

Detect the misleading implication that "lower df" improves power.

Final Logic:

Larger samples and larger df generally improve power.

🧠 Memory Trick

"Higher df → Higher Power"

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

11 Which of the following effectively differentiate a sample from a population in statistical inference?
I. A sample statistic utilizes English alphabets (x̄, S).
II. A population parameter utilizes Greek alphabets (μ, σ).
III. The sample size is practically always smaller than the entire population's size.
IV. A population acts as an estimation for the smaller sample bounds.

A I and IV only

B I, II, and III only

C II, III, and IV only

D I, II, III, and IV

💡 Short Explanation

Sample statistics use English symbols Population parameters use Greek symbols Samples are subsets of populations

✅ Correct Answer Explanation

Statement I is correct because sample statistics are commonly represented using English alphabets like: \(\hat{x}, S\) Statement II is correct because population parameters use Greek symbols such as: \(\mu , \sigma\) Statement III is also correct because a sample is generally smaller than the population. Statement IV is incorrect because samples estimate populations, not the reverse. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Omits correct Statements II and III.
Option C → Includes incorrect Statement IV.
Option D → Statement IV is conceptually wrong.

🎯 Strategy

Used

Elimination

Application:

Remove options containing the incorrect Statement IV.

Final Logic:

Samples estimate populations, not populations estimating samples.

🧠 Memory Trick

"Greek = Population, English = Sample"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Sample vs Population –

12 Using the one-sample t-test formula
\(t=\frac{\hat{x}-{\mu}_{0}}{S/\sqrt{n}}\)
if the sample mean exactly matches the expected population mean (x̄ = μ₀), what is the calculated t-value, and where does it map on the graphical curve?

A t = 1, far right tail

B t = -1, extreme left tail

C t = 0, exactly at the apex/center of the curve

D t = α, on the boundary of the rejection zone

💡 Short Explanation

Equal means produce zero numerator t-value becomes zero Zero lies at the center of the distribution

✅ Correct Answer Explanation

If: \(\hat{x}={\mu}_{0}\) then: \(\hat{x}-{\mu}_{0}=0\) Thus: \(t=0\) The value t = 0 lies exactly at the center (peak/apex) of the symmetric t-distribution curve. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → t = 1 is not obtained when means are equal.
Option B → Negative t-values arise when sample mean is smaller.
Option D → α is a probability level, not a t-statistic.

🎯 Strategy

Used

Substitution

Application:

Substitute x̄ = μ₀ directly into the formula.

Final Logic:

Zero numerator gives t = 0.

🧠 Memory Trick

"Equal Means → Zero t"

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

13 An inspector tests 60 fertilizer bags claimed to have a mass of 50 kg. He determines that the sample average is 49.6 kg.
Which numeric values represent the comparison parameter μ₀ and the sample mean x̄, respectively?

A 49.6 and 60

B 50 and 60

C 49.6 and 50

D 50 and 49.6

💡 Short Explanation

Claimed value acts as population parameter Sample average acts as sample mean n = 60 is sample size only

✅ Correct Answer Explanation

The claimed population mean is: \({\mu}_{0}=50\) The calculated sample average is: \(\hat{x}=49.6\) Thus, μ₀ = 50 and x̄ = 49.6. Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Reverses the meanings of μ₀ and x̄.
Option B → 60 is the sample size, not the sample mean.
Option C → Incorrect ordering of values.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Distinguish between claimed population value and observed sample average.

Final Logic:

Claimed mean = μ₀, observed average = x̄.

🧠 Memory Trick

"Claim = μ₀, Observation = x̄"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: One-Sample Mean Testing –

14 If a researcher conducts a hypothesis test and the conclusion states that we fail to reject the null hypothesis, what does this statistically imply?

A There is not enough evidence generated to support the alternative hypothesis claim.

B The sample mean obtained is fundamentally miscalculated.

C The population variance must be forced to equal zero.

D The standard error is infinite.

💡 Short Explanation

Failure to reject H₀ does not prove H₀ true Evidence is insufficient for H₁ Statistical support for difference is lacking

✅ Correct Answer Explanation

Failing to reject the null hypothesis means that the sample evidence is not strong enough to support the alternative hypothesis. It does not mean H₀ is absolutely true; it simply means insufficient evidence exists against it. Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Failure to reject H₀ does not imply calculation error.
Option C → Population variance is unrelated to the decision statement.
Option D → Standard error does not become infinite.

🎯 Strategy

Used

Elimination

Application:

Remove extreme or mathematically impossible interpretations.

Final Logic:

Fail to reject H₀ means weak evidence against H₀.

🧠 Memory Trick

"No Reject ≠ Proven True"

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

15 Country A reports an average farm size of 191 acres (n₁ = 8, s₁ = 38).
Country B reports an average farm size of 199 acres (n₂ = 10, s₂ = 12).
For statistical analysis, these two datasets denote:

A Paired relational samples

B Two independent groups

C A single sample compared against a parameter

D A completely unified parameter body

💡 Short Explanation

Two separate countries form separate samples Observations are unrelated Independent sample comparison applies

✅ Correct Answer Explanation

The farm-size data come from two different countries. The observations in one group do not depend on observations in the other group. Thus, the datasets represent two independent groups. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Paired samples require linked observations.
Option C → Two groups are compared, not one sample against a parameter.
Option D → The datasets remain separate groups.

🎯 Strategy

Used

Odd One Out

Application:

Identify whether observations are related or independent.

Final Logic:

Different countries imply independent groups.

🧠 Memory Trick

"Separate Groups → Independent Test"

📚 Class XII – Mathematics Part II – Chapter: Probability – Topic: Independent Samples –

16 Utilizing the data representing Country A and Country B (191 acres and 199 acres), what is the calculated numerator of the independent t-statistic representing the mean difference?

A 191 + 199 = 390

B 38 − 12 = 26

C 191 − 199 = -8

D 10 − 8 = 2

💡 Short Explanation

Independent t-tests compare means Numerator equals mean difference Subtract one sample mean from another

✅ Correct Answer Explanation

The numerator of the independent t-statistic is: \(\hat{x}_{1}-\hat{x}_{2}\) Substituting values: \(191-199=-8\) Thus, the numerator equals -8. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Means are not added in the t-statistic numerator.
Option B → Standard deviations are unrelated to numerator calculation.
Option D → Sample size difference is irrelevant here.

🎯 Strategy

Used

Substitution

Application:

Directly substitute group means into the numerator formula.

Final Logic:

Mean difference equals 191 − 199.

🧠 Memory Trick

"t Numerator = Mean Difference"

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

17 In the pooled standard deviation formula
\(s_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}}\)
what does the denominator n₁ + n₂ − 2 physically represent?

A The absolute difference of means

B The critical alpha level of pooling

C The combined total sample size

D The total pooled degrees of freedom

💡 Short Explanation

Pooled variance combines two samples Degrees of freedom are added together Minus 2 accounts for estimated means

✅ Correct Answer Explanation

In pooled variance estimation: \(df=n_{1}+n_{2}-2\) This quantity represents the combined pooled degrees of freedom from both samples. The subtraction of 2 occurs because two sample means are estimated. Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Mean difference is unrelated to the denominator.
Option B → Alpha level is unrelated to pooling.
Option C → Total sample size would be n₁ + n₂ only.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Associate pooled variance formulas with pooled df formulas.

Final Logic:

Denominator equals pooled degrees of freedom.

🧠 Memory Trick

"n₁+n₂−2 = pooled df"

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

18 When solving for unequal variances in a two-sample t-test, the formula adapts its denominator.
What is the standard error component Strategy Used for the difference between two sample means without pooling?

A √((s₁²/n₁) + (s₂²/n₂))

B s₁/√n₁ − s₂/√n₂

C s₁² + s₂²

D sₚ/√(n₁ + n₂)

💡 Short Explanation

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

✅ Correct Answer Explanation

For unequal variance t-tests, the standard error for mean difference is: \(SE=\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}\) This formula independently accounts for both sample variances without pooling. Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Standard errors are combined using addition under a square root.
Option C → Variances alone do not produce standard error.
Option D → Uses pooled SD, inappropriate for unequal variances.

🎯 Strategy

Used

Option Grouping

Application:

Identify the formula specifically Strategy Used in unequal variance testing.

Final Logic:

Separate variances remain separate inside the square root.

🧠 Memory Trick

"Unequal Variance → Separate Variance Terms"

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

📄 Passage

The t-test for two independent groups compares experimental and control datasets to determine whether a statistically significant difference exists between them. The null hypothesis \(\left(H_{0}\right)\)assumes no difference between the population means \(\left({\mu}_{1}\ =\ {\mu}_{2}\right)\). The alternative hypothesis \(\left(H_{1}\right)\)states that the population means are not equal \(\left({\mu}_{1}\ \neq \ {\mu}_{2}\right)\). The test procedure varies depending on whether equal or unequal variances are assumed.

A nationwide study concludes that high school boys and girls have statistically significant differences in standardized Mathematics scores.
Based on the passage, which hypothesis must have been rejected to reach this conclusion?

A The alternative hypothesis

B The null hypothesis

C Both hypotheses simultaneously

D Neither hypothesis

💡 Short Explanation

Significant difference supports H₁ Rejecting H₀ allows acceptance of H₁ Null hypothesis assumes no difference

✅ Correct Answer Explanation

The null hypothesis states: \(H_{0}:{\mu}_{1}={\mu}_{2}\) A statistically significant difference implies sufficient evidence against H₀. Thus, H₀ must be rejected to support the conclusion of unequal means. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → H₁ is supported, not rejected.
Option C → Both hypotheses cannot be rejected simultaneously.
Option D → A significant result requires rejecting H₀.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Link "significant difference" directly with rejection of H₀.

Final Logic:

Significant differences contradict the null hypothesis.

🧠 Memory Trick

"Difference Found → Reject H₀"

📚 Class XI – Statistics for Economics – Chapter: Statistical Inference – Topic: Null Hypothesis Rejection –

📄 Passage

An R&D department conducts a two-sample t-test comparing the average lifetime metrics of a legacy shoe and a newly developed shoe model.
If the test result leads to accepting the null hypothesis (H₀), what is the appropriate business decision?

A Discontinue the legacy shoe immediately

B Market the new shoe globally as vastly superior

C Conclude that the new product is not statistically proven to last longer

D Decrease the pooled standard deviation to artificially force a difference

💡 Short Explanation

Accepting H₀ means insufficient evidence of difference Superiority is not statistically established Decisions should follow evidence

✅ Correct Answer Explanation

If H₀ is accepted (or fail to reject H₀), the test does not provide sufficient evidence that the new shoe lasts longer than the legacy model. Thus, the company cannot scientifically claim superiority. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → No evidence supports discontinuation.
Option B → Statistical superiority was not demonstrated.
Option D → Manipulating statistics is scientifically invalid.

🎯 Strategy

Used

Elimination

Application:

Remove exaggerated or unethical conclusions unsupported by evidence.

Final Logic:

Accepting H₀ means insufficient proof of improvement.

🧠 Memory Trick

"No Reject → No Proof"

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

CUET UG Applied Mathematics Booster Test 1 - t-Test & Applications

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