Learning Objectives

By the end of this chapter, you will be able to:

When to Use t-Test

Use t-test when:

  1. Testing claims about population means
  2. Sample size is small (n < 30)
  3. Population is approximately normal
  4. σ is unknown (always for small samples)
flowchart TD
    A["Testing mean?"]
    B{"Is σ known?"}
    C{"Is n ≥ 30?"}
    D["Z-test"]
    E{"Is population<br/>normal?"}
    F["t-test"]
    G["Non-parametric test"]

    A --> B
    B -->|Yes| D
    B -->|No| C
    C -->|Yes| D
    C -->|No| E
    E -->|Yes/Approx| F
    E -->|No| G

The t-Distribution

Properties

flowchart LR
    A["df = 5: Heavy tails"]
    B["df = 15: Moderate tails"]
    C["df = 30: Near normal"]
    D["df = ∞: Normal"]

    A --> B --> C --> D

One-Sample t-Test

Test Statistic

\[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\]

Degrees of freedom: df = n - 1

Step-by-Step Example 1: One-Sample t-Test

Problem: A government claims average processing time is 30 minutes. A sample of 16 cases shows:

Test at α = 0.05 if processing time exceeds 30 minutes.

Solution:

Step 1: State hypotheses

Step 2: Check conditions

Step 3: Calculate test statistic \(t = \frac{34 - 30}{8/\sqrt{16}} = \frac{4}{2} = 2.00\)

Step 4: Find critical value

Step 5: Decision

Step 6: Conclusion At the 0.05 level of significance, there is sufficient evidence to conclude that average processing time exceeds 30 minutes.

t-Table Reference (Selected Values)

df α = 0.10 α = 0.05 α = 0.025 α = 0.01 α = 0.005
5 1.476 2.015 2.571 3.365 4.032
10 1.372 1.812 2.228 2.764 3.169
15 1.341 1.753 2.131 2.602 2.947
20 1.325 1.725 2.086 2.528 2.845
25 1.316 1.708 2.060 2.485 2.787
30 1.310 1.697 2.042 2.457 2.750

Note: For two-tailed test at α = 0.05, use α/2 = 0.025 column.

Independent Samples t-Test

Used to compare means from two independent groups when at least one sample is small.

Assumptions

  1. Both samples are random and independent
  2. Both populations are approximately normal
  3. Population variances are equal (for pooled t-test)

Pooled Variance

When we assume equal population variances:

\[s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}\]

Pooled Standard Error

\[SE = s_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}\]

Test Statistic

\[t = \frac{\bar{x}_1 - \bar{x}_2}{s_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\]

Degrees of freedom: df = n₁ + n₂ - 2

Step-by-Step Example 2: Independent Samples t-Test

Problem: Compare productivity of two teams:

  Team A Team B
n 12 10
Mean 85 78
SD 8 10

Test at α = 0.05 if Team A has higher productivity.

Solution:

Step 1: State hypotheses

Step 2: Calculate pooled variance \(s_p^2 = \frac{(12-1)(8)^2 + (10-1)(10)^2}{12 + 10 - 2}\) \(= \frac{11(64) + 9(100)}{20} = \frac{704 + 900}{20} = \frac{1604}{20} = 80.2\)

\[s_p = \sqrt{80.2} = 8.96\]

Step 3: Calculate standard error \(SE = 8.96\sqrt{\frac{1}{12} + \frac{1}{10}} = 8.96\sqrt{0.0833 + 0.1} = 8.96\sqrt{0.1833}\) \(= 8.96 \times 0.428 = 3.84\)

Step 4: Calculate test statistic \(t = \frac{85 - 78}{3.84} = \frac{7}{3.84} = 1.82\)

Step 5: Find critical value

Step 6: Decision

Step 7: Conclusion At the 0.05 level of significance, there is sufficient evidence to conclude that Team A has higher productivity than Team B.

Step-by-Step Example 3: Two-Tailed t-Test

Problem: Compare exam scores of two teaching methods:

  Method 1 Method 2
n 15 18
Mean 72 78
SD 10 12

Test at α = 0.05 if means differ.

Solution:

Step 1: State hypotheses

Step 2: Calculate pooled variance \(s_p^2 = \frac{(14)(100) + (17)(144)}{31} = \frac{1400 + 2448}{31} = \frac{3848}{31} = 124.13\)

\[s_p = 11.14\]

Step 3: Calculate standard error \(SE = 11.14\sqrt{\frac{1}{15} + \frac{1}{18}} = 11.14\sqrt{0.0667 + 0.0556}\) \(= 11.14\sqrt{0.1222} = 11.14 \times 0.350 = 3.90\)

Step 4: Calculate test statistic \(t = \frac{72 - 78}{3.90} = \frac{-6}{3.90} = -1.54\)

Step 5: Find critical value

Step 6: Decision

Step 7: Conclusion At the 0.05 level of significance, there is insufficient evidence to conclude that the teaching methods produce different mean scores.

Confidence Interval for Difference of Means

\[(\bar{x}_1 - \bar{x}_2) \pm t^* \times SE\]

Example 4: 95% CI

Using Example 2 data:

\(95\% \text{ CI} = 7 \pm 2.086 \times 3.84 = 7 \pm 8.01\) \(= (-1.01, 15.01)\)

Since 0 is in the interval, at 95% confidence the difference is not significant (for two-tailed test).

Unequal Variances (Welch’s t-Test)

When population variances are NOT equal, use Welch’s approximation:

Test Statistic

\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]

Degrees of Freedom (Welch-Satterthwaite)

\[df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{(s_1^2/n_1)^2}{n_1-1} + \frac{(s_2^2/n_2)^2}{n_2-1}}\]

(This is complex - often calculated by software)

Summary: Choosing the Right Test

Condition Test to Use
σ known, any n Z-test
σ unknown, n ≥ 30 Z-test (with s)
σ unknown, n < 30, normal pop t-test
Equal variances assumed Pooled t-test
Unequal variances Welch’s t-test

Practice Problems

Problem 1

A sample of 20 employees has mean productivity 95 with s = 15. Test at α = 0.05 if mean differs from 100.

Problem 2

Compare two groups:

Test at α = 0.05 if means differ.

Problem 3

For Problem 2, construct a 95% CI for the difference in means.

Problem 4

Test scores:

Test if morning class has higher scores at α = 0.01.

Problem 5

A sample of 25 gives $\bar{x}$ = 50, s = 8. Find the 99% CI for the population mean.

Summary

Component Formula
One-sample t $t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$, df = n-1
Pooled variance $s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}$
Two-sample t $t = \frac{\bar{x}_1 - \bar{x}_2}{s_p\sqrt{1/n_1 + 1/n_2}}$, df = n₁+n₂-2
Decision Compare t to critical value from t-table

Next Topic

In the next chapter, we will study Paired t-Test - testing for differences when samples are dependent (matched pairs).