T-Test Computational Questions and Answers

Step-by-step solutions to practical t-test problems with interactive calculators | statistical-calculators.site

Introduction to T-Tests

T-tests are fundamental statistical tools used to determine if there is a significant difference between means. In this guide, we'll work through computational problems for each type of t-test with step-by-step solutions.

We'll cover:

  • One-Sample T-Test: Comparing a sample mean to a known population mean
  • Independent Samples T-Test: Comparing means from two different groups
  • Paired Samples T-Test: Comparing means from the same group at two different times

Each section includes three computational problems with detailed solutions and access to interactive calculators to verify your work.

One-Sample T-Test Computational Problems

The one-sample t-test determines whether a sample mean is statistically different from a known population mean.

Problem 1: Textbook Weight

A publisher claims their statistics textbook weighs 2.5 lbs on average. You randomly select 15 textbooks and weigh them, finding a mean weight of 2.62 lbs with a standard deviation of 0.18 lbs. Is there evidence that the actual mean weight differs from 2.5 lbs? Use α = 0.05.

Solution:

Step 1: State the hypotheses

H₀: μ = 2.5 lbs (The mean weight is 2.5 lbs)
H₁: μ ≠ 2.5 lbs (The mean weight differs from 2.5 lbs)

Step 2: Calculate the test statistic

t = (x̄ - μ) / (s/√n)
t = (2.62 - 2.5) / (0.18/√15)
t = 0.12 / (0.18/3.873)
t = 0.12 / 0.0465 ≈ 2.58

Step 3: Determine the critical value

df = n - 1 = 15 - 1 = 14
For α = 0.05 (two-tailed), t-critical = ±2.145

Step 4: Make a decision

Since |t| = 2.58 > 2.145, we reject H₀.

Conclusion: There is sufficient evidence at the 0.05 significance level to conclude that the mean textbook weight differs from 2.5 lbs.

Problem 2: Battery Life

A company claims its batteries last 100 hours. You test 20 batteries and find a mean life of 98 hours with a standard deviation of 5 hours. At α = 0.05, test if the battery life is less than claimed.

Solution:

Step 1: State the hypotheses

H₀: μ ≥ 100 hours
H₁: μ < 100 hours (one-tailed test)

Step 2: Calculate the test statistic

t = (x̄ - μ) / (s/√n)
t = (98 - 100) / (5/√20)
t = -2 / (5/4.472)
t = -2 / 1.118 ≈ -1.79

Step 3: Determine the critical value

df = 20 - 1 = 19
For α = 0.05 (one-tailed), t-critical = -1.729

Step 4: Make a decision

Since t = -1.79 < -1.729, we reject H₀.

Conclusion: There is sufficient evidence to conclude that the battery life is less than 100 hours.

Problem 3: Student Height

The average height of college students is known to be 68 inches. You measure 25 students and find a mean height of 69.2 inches with a standard deviation of 2.8 inches. Test if your sample differs from the population at α = 0.01.

Solution:

Step 1: State the hypotheses

H₀: μ = 68 inches
H₁: μ ≠ 68 inches

Step 2: Calculate the test statistic

t = (x̄ - μ) / (s/√n)
t = (69.2 - 68) / (2.8/√25)
t = 1.2 / (2.8/5)
t = 1.2 / 0.56 ≈ 2.14

Step 3: Determine the critical value

df = 25 - 1 = 24
For α = 0.01 (two-tailed), t-critical = ±2.797

Step 4: Make a decision

Since |t| = 2.14 < 2.797, we fail to reject H₀.

Conclusion: There is insufficient evidence to conclude that the mean height differs from 68 inches at the 0.01 significance level.

One-Sample T-Test Calculator

Use our interactive calculator to verify your one-sample t-test calculations:

Independent Samples T-Test Computational Problems

The independent samples t-test compares means from two different groups to determine if they are statistically different.

Problem 1: Teaching Methods

A school tests two teaching methods. Method A (n=30) has a mean test score of 78 with a standard deviation of 5. Method B (n=35) has a mean test score of 82 with a standard deviation of 6. Is there a significant difference between the methods at α = 0.05?

Solution:

Step 1: State the hypotheses

H₀: μ₁ = μ₂ (The means are equal)
H₁: μ₁ ≠ μ₂ (The means are different)

Step 2: Calculate the pooled standard deviation

sₚ² = [(n₁-1)s₁² + (n₂-1)s₂²] / (n₁+n₂-2)
sₚ² = [(29)(25) + (34)(36)] / (30+35-2)
sₚ² = [725 + 1224] / 63
sₚ² = 1949 / 63 ≈ 30.94
sₚ ≈ √30.94 ≈ 5.56

Step 3: Calculate the test statistic

t = (x̄₁ - x̄₂) / [sₚ√(1/n₁ + 1/n₂)]
t = (78 - 82) / [5.56√(1/30 + 1/35)]
t = -4 / [5.56√(0.0333 + 0.0286)]
t = -4 / [5.56√0.0619]
t = -4 / [5.56 × 0.2488] ≈ -4 / 1.383 ≈ -2.89

Step 4: Determine the critical value

df = n₁ + n₂ - 2 = 30 + 35 - 2 = 63
For α = 0.05 (two-tailed), t-critical ≈ ±2.00

Step 5: Make a decision

Since |t| = 2.89 > 2.00, we reject H₀.

Conclusion: There is a significant difference between the teaching methods at the 0.05 level.

Problem 2: Plant Growth

A botanist tests two fertilizers. Fertilizer A (n=12) produces plants with mean height 22.3 cm (s=2.1). Fertilizer B (n=15) produces plants with mean height 24.8 cm (s=2.4). Test if Fertilizer B results in taller plants at α = 0.05.

Solution:

Step 1: State the hypotheses

H₀: μₐ ≥ μբ (Fertilizer A is not inferior)
H₁: μₐ < μբ (Fertilizer B produces taller plants)

Step 2: Calculate the pooled standard deviation

sₚ² = [(11)(4.41) + (14)(5.76)] / (12+15-2)
sₚ² = [48.51 + 80.64] / 25
sₚ² = 129.15 / 25 = 5.166
sₚ ≈ √5.166 ≈ 2.273

Step 3: Calculate the test statistic

t = (x̄₁ - x̄₂) / [sₚ√(1/n₁ + 1/n₂)]
t = (22.3 - 24.8) / [2.273√(1/12 + 1/15)]
t = -2.5 / [2.273√(0.0833 + 0.0667)]
t = -2.5 / [2.273√0.15]
t = -2.5 / [2.273 × 0.3873] ≈ -2.5 / 0.880 ≈ -2.84

Step 4: Determine the critical value

df = 12 + 15 - 2 = 25
For α = 0.05 (one-tailed), t-critical = -1.708

Step 5: Make a decision

Since t = -2.84 < -1.708, we reject H₀.

Conclusion: Fertilizer B produces significantly taller plants at the 0.05 level.

Problem 3: Reaction Time

A researcher compares reaction times between gamers (n=20, x̄=280ms, s=25) and non-gamers (n=25, x̄=310ms, s=30). Test if gamers have faster reaction times at α = 0.01.

Solution:

Step 1: State the hypotheses

H₀: μ₉ ≥ μₙ (Gamers are not faster)
H₁: μ₉ < μₙ (Gamers have faster reaction times)

Step 2: Calculate the pooled standard deviation

sₚ² = [(19)(625) + (24)(900)] / (20+25-2)
sₚ² = [11875 + 21600] / 43
sₚ² = 33475 / 43 ≈ 778.49
sₚ ≈ √778.49 ≈ 27.90

Step 3: Calculate the test statistic

t = (x̄₁ - x̄₂) / [sₚ√(1/n₁ + 1/n₂)]
t = (280 - 310) / [27.90√(1/20 + 1/25)]
t = -30 / [27.90√(0.05 + 0.04)]
t = -30 / [27.90√0.09]
t = -30 / [27.90 × 0.3] ≈ -30 / 8.37 ≈ -3.59

Step 4: Determine the critical value

df = 20 + 25 - 2 = 43
For α = 0.01 (one-tailed), t-critical ≈ -2.42

Step 5: Make a decision

Since t = -3.59 < -2.42, we reject H₀.

Conclusion: Gamers have significantly faster reaction times at the 0.01 level.

Independent Samples T-Test Calculator

Use our interactive calculator to verify your independent samples t-test calculations:

Paired Samples T-Test Computational Problems

The paired samples t-test compares means from the same group at two different times or under two different conditions.

Problem 1: Weight Loss Program

A weight loss program tests 10 participants. Their weights (in kg) before and after the program are:
ParticipantBeforeAfter
18582
29288
37876
4105100
58885
69491
78179
89793
98381
109087
Test if the program is effective at α = 0.05.

Solution:

Step 1: Calculate the differences

Differences (d): -3, -4, -2, -5, -3, -3, -2, -4, -2, -3

Step 2: Calculate mean and standard deviation of differences

Mean difference (d̄) = (-3-4-2-5-3-3-2-4-2-3)/10 = -31/10 = -3.1
Standard deviation (s_d) = 1.10

Step 3: State the hypotheses

H₀: μ_d ≥ 0 (No weight loss)
H₁: μ_d < 0 (Weight loss occurred)

Step 4: Calculate the test statistic

t = d̄ / (s_d/√n)
t = -3.1 / (1.10/√10)
t = -3.1 / (1.10/3.162)
t = -3.1 / 0.348 ≈ -8.91

Step 5: Determine the critical value

df = n - 1 = 10 - 1 = 9
For α = 0.05 (one-tailed), t-critical = -1.833

Step 6: Make a decision

Since t = -8.91 < -1.833, we reject H₀.

Conclusion: The weight loss program is effective at the 0.05 significance level.

Problem 2: Drug Effectiveness

A drug is tested for its effect on blood pressure. Systolic BP measurements before and after administration for 8 patients are:
PatientBeforeAfter
1145138
2132127
3158150
4142136
5128125
6135130
7150142
8140135
Test if the drug reduces blood pressure at α = 0.01.

Solution:

Step 1: Calculate the differences

Differences (d): -7, -5, -8, -6, -3, -5, -8, -5

Step 2: Calculate mean and standard deviation of differences

Mean difference (d̄) = (-7-5-8-6-3-5-8-5)/8 = -47/8 = -5.875
Standard deviation (s_d) = 1.73

Step 3: State the hypotheses

H₀: μ_d ≥ 0 (No reduction)
H₁: μ_d < 0 (Reduction occurred)

Step 4: Calculate the test statistic

t = d̄ / (s_d/√n)
t = -5.875 / (1.73/√8)
t = -5.875 / (1.73/2.828)
t = -5.875 / 0.612 ≈ -9.60

Step 5: Determine the critical value

df = n - 1 = 8 - 1 = 7
For α = 0.01 (one-tailed), t-critical = -2.998

Step 6: Make a decision

Since t = -9.60 < -2.998, we reject H₀.

Conclusion: The drug significantly reduces blood pressure at the 0.01 level.

Problem 3: Training Program

A company implements a training program and measures employee productivity before and after:
EmployeeBeforeAfter
14245
23840
34547
45153
53538
64850
73942
84446
Test if the training improved productivity at α = 0.05.

Solution:

Step 1: Calculate the differences

Differences (d): +3, +2, +2, +2, +3, +2, +3, +2

Step 2: Calculate mean and standard deviation of differences

Mean difference (d̄) = (3+2+2+2+3+2+3+2)/8 = 19/8 = 2.375
Standard deviation (s_d) = 0.52

Step 3: State the hypotheses

H₀: μ_d ≤ 0 (No improvement)
H₁: μ_d > 0 (Improvement occurred)

Step 4: Calculate the test statistic

t = d̄ / (s_d/√n)
t = 2.375 / (0.52/√8)
t = 2.375 / (0.52/2.828)
t = 2.375 / 0.184 ≈ 12.91

Step 5: Determine the critical value

df = n - 1 = 8 - 1 = 7
For α = 0.05 (one-tailed), t-critical = 1.895

Step 6: Make a decision

Since t = 12.91 > 1.895, we reject H₀.

Conclusion: The training program significantly improved productivity at the 0.05 level.

Paired Samples T-Test Calculator

Use our interactive calculator to verify your paired samples t-test calculations:

Alternative Tests to T-Tests

When the assumptions of t-tests are violated, consider these alternatives:

Situation Alternative Test When to Use
Non-normal data Mann-Whitney U test (independent)
Wilcoxon signed-rank test (paired) 		Ordinal data or severe violations of normality
Unequal variances Welch's t-test When group variances are significantly different
More than 2 groups ANOVA Comparing means across 3 or more groups
Categorical data Chi-square test When working with frequencies or proportions

Conclusion

Through these computational problems, we've demonstrated how to apply each type of t-test in practical scenarios. Remember these key points:

  • Always check assumptions before conducting a t-test
  • Choose the correct test based on your research design
  • Interpret results in the context of your specific problem
  • Consider effect sizes alongside statistical significance

The interactive calculators provided can help verify your manual calculations and build intuition for how different values affect the test results.

Related Keywords

hypothesis testing statistical significance p-value confidence interval effect size degrees of freedom null hypothesis alternative hypothesis type I error type II error power analysis Welch's t-test non-parametric tests ANOVA data analysis research methods