Parametric Hypothesis Tests — Part 3
Tests for Mean and Variance of Normal Populations. Mean differences.
Lisbon Accounting and Business School – Polytechnic University of Lisbon
2025-04-21
Recap from Last Week
In Lecture 6, we established:
- The general framework: \(H_0\) vs. \(H_1\), test statistic, \(\alpha\), critical region
- Type I and Type II errors, power
- The \(p\)-value: smallest \(\alpha\) at which we reject \(H_0\)
- The relationship between CIs and two-tailed tests
Today: We apply the framework systematically to every test case in the syllabus.
Road Map for Today
| Section | Topic |
|---|---|
| 3.3 | Tests for \(\mu\) (Normal, \(\sigma\) known and unknown) |
| Tests for \(\sigma^2\) (Normal population) | |
| 3.4 | Tests for \(\mu_1 - \mu_2\) (two Normal populations) |
| 3.5 | Large-sample (asymptotic) tests: mean and proportion |
3.3 — Tests for Parameters of Normal Populations
Overview
We consider a random sample \((X_1, X_2, \ldots, X_n)\) from a Normal population \(X \sim N(\mu, \sigma)\).
We will cover four cases:
- Case 1: Test for \(\mu\) with \(\sigma\) known
- Case 2: Test for \(\mu\) with \(\sigma\) unknown (small sample)
- Case 2’: Test for \(\mu\) with \(\sigma\) unknown and \(n > 30\)
- Case 3: Test for \(\sigma^2\) (or \(\sigma\))
Case 1: Test for \(\mu\), \(\sigma\) Known
Test for \(\mu\) — Normal population, \(\sigma\) known
Test statistic: \[Z_0 = \frac{\bar{X} - \mu_0}{\sigma / \sqrt{n}} \underset{\text{under } H_0}{\sim} N(0, 1)\]
Critical regions (at significance level \(\alpha\)):
| \(H_1\) | Rejection region |
|---|---|
| \(\mu \neq \mu_0\) | \(\|z_{obs}\| \geq z_{\alpha/2}\) |
| \(\mu > \mu_0\) | \(z_{obs} \geq z_{\alpha}\) |
| \(\mu < \mu_0\) | \(z_{obs} \leq -z_{\alpha}\) |
Recall: \(z_{\alpha}\) is the value such that \(P(Z > z_{\alpha}) = \alpha\), from the Standard Normal table.
Example 1: Test for \(\mu\), \(\sigma\) known
The nominal capacity of a soft drink bottle is 330 mL. The filling process has \(\sigma = 5\) mL. A random sample of \(n = 40\) bottles has \(\bar{x} = 328.2\) mL. At \(\alpha = 0.05\), test whether the mean capacity differs from 330 mL.
Example 1: Solution
Hypotheses: \(H_0: \mu = 330\) vs. \(H_1: \mu \neq 330\) (two-tailed)
Test statistic: \(Z_0 = \frac{\bar{X} - 330}{5/\sqrt{40}} \sim N(0,1)\)
Observed value: \[z_{obs} = \frac{328.2 - 330}{5/\sqrt{40}} = \frac{-1.8}{0.7906} \approx -2.28\]
Rejection region (two-tailed, \(\alpha = 0.05\)): \[|z_{obs}| \geq z_{0.025} = 1.96\]
Since \(|-2.28| = 2.28 > 1.96\), we reject \(H_0\).
\(p\)-value: \(p = 2 \times P(Z \geq 2.28) = 2 \times (1 - 0.9887) = 2 \times 0.0113 = 0.0226\)
Case 2: Test for \(\mu\), \(\sigma\) Unknown
Test for \(\mu\) — Normal population, \(\sigma\) unknown
Test statistic: \[T_0 = \frac{\bar{X} - \mu_0}{S' / \sqrt{n}} \underset{\text{under } H_0}{\sim} t_{(n-1)}\]
Critical regions (at significance level \(\alpha\)):
| \(H_1\) | Rejection region |
|---|---|
| \(\mu \neq \mu_0\) | \(\|t_{obs}\| \geq t_{\alpha/2,(n-1)}\) |
| \(\mu > \mu_0\) | \(t_{obs} \geq t_{\alpha,(n-1)}\) |
| \(\mu < \mu_0\) | \(t_{obs} \leq -t_{\alpha,(n-1)}\) |
Recall: \(S'\) is the corrected sample standard deviation, and \(t_{\alpha,(\nu)}\) is from Table 7 (\(t\)-Student).
Example 2: Test for \(\mu\), \(\sigma\) unknown
An engineer suspects that the mean weight of packages produced by a machine exceeds the nominal value of 100 g. A random sample of \(n = 20\) packages from a Normal population gives \(\bar{x} = 101\) g and \(s' = 3\) g.
Test at \(\alpha = 0.05\).
Example 2: Solution
Hypotheses: \(H_0: \mu \leq 100\) vs. \(H_1: \mu > 100\) (right-tailed)
Test statistic: \(T_0 = \frac{\bar{X} - 100}{S'/\sqrt{20}} \sim t_{(19)}\)
Observed value: \[t_{obs} = \frac{101 - 100}{3/\sqrt{20}} = \frac{1}{0.6708} \approx 1.49\]
Rejection region (right-tailed, \(\alpha = 0.05\)): \[t_{obs} \geq t_{0.05,(19)} = 1.729\]
(Table 7)
Since \(1.49 < 1.729\), we do not reject \(H_0\).
Conclusion: At 5%, there is insufficient evidence that the mean weight exceeds 100 g.
Case 2’: Test for \(\mu\), \(\sigma\) Unknown, \(n > 30\)
Test for \(\mu\) — Normal population, \(\sigma\) unknown, large sample (\(n > 30\))
When \(n > 30\), \(S'\) approximates \(\sigma\) well and the \(t\)-distribution approaches the Standard Normal:
\[Z_0 = \frac{\bar{X} - \mu_0}{S' / \sqrt{n}} \underset{\text{under } H_0}{\dot{\sim}} N(0, 1)\]
The critical regions are the same as in Case 1, using \(z_{\alpha}\) and \(z_{\alpha/2}\) from the Normal table.
This is the same approach as in Case 1, but replacing \(\sigma\) by \(s'\). The approximation is justified by the CLT and by the fact that \(t_{(\nu)} \to N(0,1)\) as \(\nu \to \infty\).
Case 3: Test for \(\sigma^2\) (Variance)
Test for \(\sigma^2\) — Normal population
Test statistic: \[Q_0 = \frac{(n-1)S'^2}{\sigma_0^2} \underset{\text{under } H_0}{\sim} \chi^2_{(n-1)}\]
Critical regions (at significance level \(\alpha\)):
| \(H_1\) | Rejection region |
|---|---|
| \(\sigma^2 \neq \sigma_0^2\) | \(q_{obs} \leq q_{1-\alpha/2}\) or \(q_{obs} \geq q_{\alpha/2}\) |
| \(\sigma^2 > \sigma_0^2\) | \(q_{obs} \geq q_{\alpha}\) |
| \(\sigma^2 < \sigma_0^2\) | \(q_{obs} \leq q_{1-\alpha}\) |
Here \(q_{\alpha}\) denotes the value from the \(\chi^2\) table (Table 6) such that \(P(Q > q_{\alpha}) = \alpha\).
Important: The \(\chi^2\) Distribution is Not Symmetric
Because the \(\chi^2\) distribution is right-skewed, the two-tailed rejection region is not symmetric around the center. The lower critical value is \(q_{1-\alpha/2}\) and the upper is \(q_{\alpha/2}\).
Example 3: Test for \(\sigma^2\)
Consider the test: \(H_0: \sigma^2 \geq 10\) vs. \(H_1: \sigma^2 < 10\), with \(n = 20\) from a Normal population and \(s'^2 = 9\) (i.e., \(s' = 3\)).
Test at \(\alpha = 0.05\).
Example 3: Solution
Test statistic: \[Q_0 = \frac{(n-1)S'^2}{\sigma_0^2} \underset{\text{under } H_0}{\sim} \chi^2_{(19)}\]
Observed value: \[q_{obs} = \frac{(20-1) \times 9}{10} = \frac{171}{10} = 17.1\]
Rejection region (left-tailed, \(\alpha = 0.05\)):
We need \(q_{1-\alpha} = q_{0.95}\) from the \(\chi^2\) table with 19 d.f.
From Table 6: \(q_{0.95,(19)} = 10.117\).
Rejection region: \(\left[0\,;\, 10.117\right]\).
Since \(q_{obs} = 17.1 > 10.117\), we do not reject \(H_0\).
Summary of Cases — Section 3.3
| Parameter | \(\sigma\) | Test Stat. | Distribution | Table |
|---|---|---|---|---|
| \(\mu\) | known | \(Z_0 = \frac{\bar{X}-\mu_0}{\sigma/\sqrt{n}}\) | \(N(0,1)\) | Normal |
| \(\mu\) | unknown, \(n\leq 30\) | \(T_0 = \frac{\bar{X}-\mu_0}{S'/\sqrt{n}}\) | \(t_{(n-1)}\) | Table 7 |
| \(\mu\) | unknown, \(n > 30\) | \(Z_0 = \frac{\bar{X}-\mu_0}{S'/\sqrt{n}}\) | \(\dot\sim N(0,1)\) | Normal |
| \(\sigma^2\) | — | \(Q_0 = \frac{(n-1)S'^2}{\sigma_0^2}\) | \(\chi^2_{(n-1)}\) | Table 6 |
This mirrors the cases from interval estimation — the pivot variables are the same.
Practice Exercise A
A quality inspector measures the temperature of a chemical process. Historical data show \(\sigma = 2.5°\)C. A random sample of \(n = 16\) measurements gives \(\bar{x} = 98.3°\)C.
Test \(H_0: \mu = 100\) vs. \(H_1: \mu \neq 100\) at \(\alpha = 0.01\).
Compute the \(p\)-value.
Practice Exercise A: Solution
a) \(Z_0 = \frac{\bar{X} - 100}{2.5/\sqrt{16}} \sim N(0,1)\)
\[z_{obs} = \frac{98.3 - 100}{2.5/4} = \frac{-1.7}{0.625} = -2.72\]
Rejection region (two-tailed, \(\alpha = 0.01\)): \(|z_{obs}| \geq z_{0.005} = 2.576\)
Since \(|-2.72| = 2.72 > 2.576\), we reject \(H_0\).
b) \(p = 2 \times P(Z \geq 2.72) = 2 \times (1 - 0.9967) = 2 \times 0.0033 = 0.0066\)
Conclusion: At 1%, there is evidence that the process temperature differs from 100°C.
Practice Exercise B
From a Normal population, a sample of \(n = 15\) observations gives \(s' = 5.04\).
Test \(H_0: \sigma^2 = 16\) vs. \(H_1: \sigma^2 \neq 16\) at \(\alpha = 0.10\).
Practice Exercise B: Solution
Test statistic: \[Q_0 = \frac{(n-1)S'^2}{\sigma_0^2} \sim \chi^2_{(14)}\]
\[q_{obs} = \frac{14 \times (5.04)^2}{16} = \frac{14 \times 25.4016}{16} = \frac{355.62}{16} \approx 22.23\]
Rejection region (two-tailed, \(\alpha = 0.10\)):
From Table 6, \(\nu = 14\): \(q_{0.95,(14)} = 6.571\) and \(q_{0.05,(14)} = 23.685\)
Reject if \(q_{obs} \leq 6.571\) or \(q_{obs} \geq 23.685\).
Since \(6.571 < 22.23 < 23.685\), we do not reject \(H_0\).
Conclusion: At 10%, there is no evidence that \(\sigma^2 \neq 16\).
Quick Recap Before the Break
What we covered (Section 3.3)
- Case 1: Test for \(\mu\) with \(\sigma\) known → \(Z_0 \sim N(0,1)\)
- Case 2: Test for \(\mu\) with \(\sigma\) unknown → \(T_0 \sim t_{(n-1)}\)
- Case 2’: Test for \(\mu\) with \(\sigma\) unknown, \(n>30\) → \(Z_0 \dot\sim N(0,1)\)
- Case 3: Test for \(\sigma^2\) → \(Q_0 \sim \chi^2_{(n-1)}\)
3.4 — Tests for the Difference of Means
Setup
We have two independent Normal populations:
\[X_1 \sim N(\mu_1, \sigma_1) \qquad X_2 \sim N(\mu_2, \sigma_2)\]
Two independent random samples:
- From population 1: \(n_1\) observations → \(\bar{X}_1\), \(S'^2_1\)
- From population 2: \(n_2\) observations → \(\bar{X}_2\), \(S'^2_2\)
Goal: Test hypotheses about \(\mu_1 - \mu_2\) (usually testing if \(\mu_1 - \mu_2 = 0\), i.e., if the two means are equal).
The General Hypotheses
| \(H_0\) | \(H_1\) | Type |
|---|---|---|
| \(\mu_1 - \mu_2 = d_0\) | \(\mu_1 - \mu_2 \neq d_0\) | Two-tailed |
| \(\mu_1 - \mu_2 \leq d_0\) | \(\mu_1 - \mu_2 > d_0\) | Right-tailed |
| \(\mu_1 - \mu_2 \geq d_0\) | \(\mu_1 - \mu_2 < d_0\) | Left-tailed |
Often \(d_0 = 0\), i.e., we test whether the two population means are equal.
Case A: Both \(\sigma_1\) and \(\sigma_2\) Known
Test for \(\mu_1 - \mu_2\) — variances known
Test statistic: \[Z_0 = \frac{(\bar{X}_1 - \bar{X}_2) - d_0}{\sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2}}} \underset{\text{under } H_0}{\sim} N(0, 1)\]
Critical regions use \(z_{\alpha}\) and \(z_{\alpha/2}\) from the Normal table, as in the one-sample case.
Case B: Variances Unknown but Equal (\(\sigma_1^2 = \sigma_2^2\))
Test for \(\mu_1 - \mu_2\) — variances unknown, assumed equal
We estimate the common variance using the pooled estimator:
\[S_p'^2 = \frac{(n_1 - 1)S'^2_1 + (n_2 - 1)S'^2_2}{n_1 + n_2 - 2}\]
Test statistic: \[T_0 = \frac{(\bar{X}_1 - \bar{X}_2) - d_0}{S'_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}} \underset{\text{under } H_0}{\sim} t_{(n_1 + n_2 - 2)}\]
The degrees of freedom are \(\nu = n_1 + n_2 - 2\). Critical values from Table 7.
When Can We Assume Equal Variances?
In practice, the assumption \(\sigma_1^2 = \sigma_2^2\) can be tested using an \(F\)-test (not in our syllabus), or it may be given in the problem statement.
Practical rule: If the ratio of the larger sample variance to the smaller is less than about 2–3, the equal-variance assumption is often reasonable.
When the problem says “assume equal variances” or “assume \(\sigma_1 = \sigma_2\)”, use the pooled estimator.
Example 4: Difference of Means (Equal Variances)
Two production lines manufacture the same component. To compare average lengths, independent random samples are taken:
- Line 1: \(n_1 = 12\), \(\bar{x}_1 = 50.3\) mm, \(s'_1 = 1.2\) mm
- Line 2: \(n_2 = 15\), \(\bar{x}_2 = 49.7\) mm, \(s'_2 = 1.4\) mm
Assuming equal population variances and Normal populations, test whether the mean lengths differ at \(\alpha = 0.05\).
Example 4: Solution — Step 1
Hypotheses: \(H_0: \mu_1 - \mu_2 = 0\) vs. \(H_1: \mu_1 - \mu_2 \neq 0\) (two-tailed)
Pooled variance: \[s_p'^2 = \frac{(12-1)(1.2)^2 + (15-1)(1.4)^2}{12 + 15 - 2} = \frac{11 \times 1.44 + 14 \times 1.96}{25}\]
\[= \frac{15.84 + 27.44}{25} = \frac{43.28}{25} = 1.7312\]
\[s'_p = \sqrt{1.7312} \approx 1.3157\]
Example 4: Solution — Step 2
Test statistic: \(T_0 \sim t_{(25)}\)
\[t_{obs} = \frac{(50.3 - 49.7) - 0}{1.3157 \sqrt{\frac{1}{12} + \frac{1}{15}}} = \frac{0.6}{1.3157 \times \sqrt{0.0833 + 0.0667}}\]
\[= \frac{0.6}{1.3157 \times \sqrt{0.15}} = \frac{0.6}{1.3157 \times 0.3873} = \frac{0.6}{0.5095} \approx 1.18\]
Example 4: Solution — Step 3
Rejection region (two-tailed, \(\alpha = 0.05\), \(\nu = 25\)):
From Table 7: \(t_{0.025,(25)} = 2.060\)
Reject if \(|t_{obs}| \geq 2.060\).
Since \(|1.18| = 1.18 < 2.060\), we do not reject \(H_0\).
Conclusion: At 5%, there is no evidence of a difference in mean lengths between the two production lines.
Case C: Variances Unknown, Large Samples
Test for \(\mu_1 - \mu_2\) — variances unknown, \(n_1 > 30\) and \(n_2 > 30\)
When both samples are large, the CLT applies and we can use:
\[Z_0 = \frac{(\bar{X}_1 - \bar{X}_2) - d_0}{\sqrt{\dfrac{S'^2_1}{n_1} + \dfrac{S'^2_2}{n_2}}} \underset{\text{under } H_0}{\dot{\sim}} N(0, 1)\]
No assumption about equal variances is needed.
Example 5: Difference of Means (Large Samples)
A company wants to compare the average monthly sales (in thousands of euros) of two regions:
- Region A: \(n_1 = 50\), \(\bar{x}_1 = 87.3\), \(s'_1 = 12.5\)
- Region B: \(n_2 = 45\), \(\bar{x}_2 = 82.1\), \(s'_2 = 14.8\)
Test \(H_0: \mu_1 \leq \mu_2\) vs. \(H_1: \mu_1 > \mu_2\) at \(\alpha = 0.05\).
Example 5: Solution
Hypotheses: \(H_0: \mu_1 - \mu_2 \leq 0\) vs. \(H_1: \mu_1 - \mu_2 > 0\) (right-tailed)
\[z_{obs} = \frac{(87.3 - 82.1) - 0}{\sqrt{\frac{(12.5)^2}{50} + \frac{(14.8)^2}{45}}} = \frac{5.2}{\sqrt{\frac{156.25}{50} + \frac{219.04}{45}}}\]
\[= \frac{5.2}{\sqrt{3.125 + 4.868}} = \frac{5.2}{\sqrt{7.993}} = \frac{5.2}{2.827} \approx 1.84\]
Rejection region (right-tailed, \(\alpha = 0.05\)): \(z_{obs} \geq z_{0.05} = 1.645\)
Since \(1.84 > 1.645\), we reject \(H_0\).
\(p\)-value: \(p = P(Z > 1.84) = 1 - \Phi(1.84) = 1 - 0.9671 = 0.0329\)
Summary: Tests for \(\mu_1 - \mu_2\)
| Case | Conditions | Test Statistic | Distribution |
|---|---|---|---|
| A | \(\sigma_1, \sigma_2\) known | \(Z_0\) with \(\sigma_1, \sigma_2\) | \(N(0,1)\) |
| B | \(\sigma_1, \sigma_2\) unknown, equal | \(T_0\) with \(S'_p\) | \(t_{(n_1+n_2-2)}\) |
| C | Large samples (\(n_1, n_2 > 30\)) | \(Z_0\) with \(S'_1, S'_2\) | \(\dot\sim N(0,1)\) |
3.5 — Large-Sample Tests (Asymptotic Normality)
Motivation
So far, we assumed Normal populations (except for Case 2’ and Case C, where the CLT helped).
What if we do not know the population distribution, but have a large sample (\(n > 30\))?
By the Central Limit Theorem, the sampling distribution of \(\bar{X}\) is approximately Normal, regardless of the population distribution.
This allows us to perform tests even without the Normality assumption.
Test for \(\mu\) — Large Sample, Non-Normal Population
Test for \(\mu\) — large sample, any distribution
If \(n > 30\), by the CLT:
\[Z_0 = \frac{\bar{X} - \mu_0}{S' / \sqrt{n}} \underset{\text{under } H_0}{\dot{\sim}} N(0, 1)\]
This is the same formula as Case 2’ (Normal population, \(\sigma\) unknown, \(n > 30\)), but now the Normality assumption on the population is relaxed.
The critical regions and \(p\)-value calculations use the Standard Normal table.
Example 6: Large-Sample Test for \(\mu\)
A logistics company records the delivery time for packages. The distribution is unknown (likely right-skewed). A random sample of \(n = 100\) deliveries gives \(\bar{x} = 3.2\) days and \(s' = 0.8\) days.
Test \(H_0: \mu \leq 3\) vs. \(H_1: \mu > 3\) at \(\alpha = 0.05\).
Example 6: Solution
Even though the population is not Normal, \(n = 100 > 30\), so the CLT applies.
\[z_{obs} = \frac{3.2 - 3}{0.8/\sqrt{100}} = \frac{0.2}{0.08} = 2.50\]
Rejection region (right-tailed, \(\alpha = 0.05\)): \(z_{obs} \geq 1.645\)
Since \(2.50 > 1.645\), we reject \(H_0\).
\(p\)-value: \(p = P(Z > 2.50) = 1 - \Phi(2.50) = 1 - 0.9938 = 0.0062\)
Strong evidence that the mean delivery time exceeds 3 days.
Test for a Proportion \(p\) — Bernoulli Population
Test for \(p\) — large sample from a Bernoulli population
Let \(X \sim \text{Bernoulli}(p)\). The sample proportion is \(\hat{p} = X/n\), where \(X = \sum_{i=1}^n X_i\).
For large \(n\) (typically \(np_0 \geq 5\) and \(n(1-p_0) \geq 5\)), by the CLT:
\[Z_0 = \frac{\hat{p} - p_0}{\sqrt{\dfrac{p_0(1 - p_0)}{n}}} \underset{\text{under } H_0}{\dot{\sim}} N(0, 1)\]
Note: Under \(H_0\), we use \(p_0\) (not \(\hat{p}\)) in the denominator, because we assume \(H_0\) is true when computing the test statistic.
Example 7: Test for a Proportion
A company claims that at most 5% of its products are defective. In a random sample of \(n = 200\) items, 15 are found to be defective.
Test the company’s claim at \(\alpha = 0.05\).
Example 7: Solution
Hypotheses: \(H_0: p \leq 0.05\) vs. \(H_1: p > 0.05\) (right-tailed)
\(\hat{p} = 15/200 = 0.075\), \(p_0 = 0.05\)
Checking conditions: \(np_0 = 200 \times 0.05 = 10 \geq 5\) ✓ and \(n(1-p_0) = 190 \geq 5\) ✓
\[z_{obs} = \frac{0.075 - 0.05}{\sqrt{\frac{0.05 \times 0.95}{200}}} = \frac{0.025}{\sqrt{\frac{0.0475}{200}}} = \frac{0.025}{\sqrt{0.0002375}}\]
\[= \frac{0.025}{0.01541} \approx 1.62\]
Example 7: Solution (cont.)
Rejection region (right-tailed, \(\alpha = 0.05\)): \(z_{obs} \geq 1.645\)
Since \(1.62 < 1.645\), we do not reject \(H_0\).
\(p\)-value: \(p = P(Z > 1.62) = 1 - \Phi(1.62) = 1 - 0.9474 = 0.0526\)
Conclusion: At 5%, there is marginal but insufficient evidence to reject the company’s claim. The \(p\)-value (\(0.0526\)) is just barely above \(\alpha\).
This is a “borderline” case. At \(\alpha = 0.10\) we would reject \(H_0\).
Example 8: Test for Proportion (Two-tailed)
In a survey of \(n = 400\) voters, 224 favor a certain policy. Test whether the true proportion differs from 50% at \(\alpha = 0.01\).
Example 8: Solution
\(H_0: p = 0.50\) vs. \(H_1: p \neq 0.50\) (two-tailed)
\(\hat{p} = 224/400 = 0.56\)
\[z_{obs} = \frac{0.56 - 0.50}{\sqrt{\frac{0.50 \times 0.50}{400}}} = \frac{0.06}{\sqrt{0.000625}} = \frac{0.06}{0.025} = 2.40\]
Rejection region (two-tailed, \(\alpha = 0.01\)): \(|z_{obs}| \geq z_{0.005} = 2.576\)
Since \(|2.40| = 2.40 < 2.576\), we do not reject \(H_0\) at 1%.
\(p\)-value: \(p = 2 \times P(Z > 2.40) = 2 \times 0.0082 = 0.0164\)
At \(\alpha = 0.05\) we would reject. At \(\alpha = 0.01\) we do not.
Comprehensive Summary
All Test Cases — Complete Reference
| # | Parameter | Conditions | Test Statistic | Distribution |
|---|---|---|---|---|
| 1 | \(\mu\) | Normal, \(\sigma\) known | \(\frac{\bar{X}-\mu_0}{\sigma/\sqrt{n}}\) | \(N(0,1)\) |
| 2 | \(\mu\) | Normal, \(\sigma\) unknown | \(\frac{\bar{X}-\mu_0}{S'/\sqrt{n}}\) | \(t_{(n-1)}\) |
| 2’ | \(\mu\) | Normal, \(\sigma\) unknown, \(n>30\) | \(\frac{\bar{X}-\mu_0}{S'/\sqrt{n}}\) | \(\dot\sim N(0,1)\) |
| 3 | \(\sigma^2\) | Normal | \(\frac{(n-1)S'^2}{\sigma_0^2}\) | \(\chi^2_{(n-1)}\) |
| 4a | \(\mu_1-\mu_2\) | Normal, \(\sigma_1,\sigma_2\) known | \(\frac{(\bar{X}_1-\bar{X}_2)-d_0}{\sqrt{\sigma_1^2/n_1+\sigma_2^2/n_2}}\) | \(N(0,1)\) |
| 4b | \(\mu_1-\mu_2\) | Normal, \(\sigma_1=\sigma_2\) unknown | \(\frac{(\bar{X}_1-\bar{X}_2)-d_0}{S'_p\sqrt{1/n_1+1/n_2}}\) | \(t_{(n_1+n_2-2)}\) |
| 4c | \(\mu_1-\mu_2\) | Large samples | \(\frac{(\bar{X}_1-\bar{X}_2)-d_0}{\sqrt{S'^2_1/n_1+S'^2_2/n_2}}\) | \(\dot\sim N(0,1)\) |
| 5 | \(\mu\) | Any distrib., \(n > 30\) | \(\frac{\bar{X}-\mu_0}{S'/\sqrt{n}}\) | \(\dot\sim N(0,1)\) |
| 6 | \(p\) | Bernoulli, large \(n\) | \(\frac{\hat{p}-p_0}{\sqrt{p_0(1-p_0)/n}}\) | \(\dot\sim N(0,1)\) |
Critical Regions — Quick Reference
For a test statistic \(T\) with observed value \(t_{obs}\):
| \(H_1\) | Rejection region | \(p\)-value |
|---|---|---|
| \(\theta \neq \theta_0\) | \(\|t_{obs}\| \geq c_{\alpha/2}\) | \(2 \times P(T \geq \|t_{obs}\|)\) |
| \(\theta > \theta_0\) | \(t_{obs} \geq c_{\alpha}\) | \(P(T \geq t_{obs})\) |
| \(\theta < \theta_0\) | \(t_{obs} \leq -c_{\alpha}\) (or \(\leq c_{1-\alpha}\) for \(\chi^2\)) | \(P(T \leq t_{obs})\) |
\(c_{\alpha}\) is the critical value at level \(\alpha\) from the appropriate table (\(z_{\alpha}\), \(t_{\alpha,(\nu)}\), or \(q_{\alpha,(\nu)}\)).
The General Procedure — Revisited
Step-by-step (for any case)
- Identify the parameter, population(s), and what is known/unknown
- State \(H_0\) and \(H_1\)
- Choose \(\alpha\)
- Select the appropriate test statistic (from the summary table)
- Determine the critical region (using the correct table and type of test)
- Compute \(t_{obs}\) (and the \(p\)-value, if requested)
- Decide: reject or do not reject \(H_0\)
- Conclude in context
Additional Practice
Exercise C
In a random sample of \(n = 250\) consumers surveyed, 75 stated they prefer brand A.
Test \(H_0: p = 0.35\) vs. \(H_1: p \neq 0.35\) at \(\alpha = 0.05\).
Compute the \(p\)-value.
Exercise C: Solution
\(\hat{p} = 75/250 = 0.30\), \(p_0 = 0.35\)
\[z_{obs} = \frac{0.30 - 0.35}{\sqrt{\frac{0.35 \times 0.65}{250}}} = \frac{-0.05}{\sqrt{0.00091}} = \frac{-0.05}{0.03017} \approx -1.66\]
Rejection region (two-tailed, \(\alpha = 0.05\)): \(|z_{obs}| \geq 1.96\)
Since \(|-1.66| = 1.66 < 1.96\), we do not reject \(H_0\).
\(p\)-value: \(p = 2 \times P(Z \geq 1.66) = 2 \times (1 - 0.9515) = 2 \times 0.0485 = 0.0970\)
No evidence at 5% that \(p \neq 0.35\).
Exercise D
Two factories produce steel rods. Quality control takes samples:
- Factory 1: \(n_1 = 10\), \(\bar{x}_1 = 502.1\) mm, \(s'_1 = 4.5\) mm
- Factory 2: \(n_2 = 12\), \(\bar{x}_2 = 498.3\) mm, \(s'_2 = 5.1\) mm
Assume Normal populations with equal variances. Test whether Factory 1 produces longer rods at \(\alpha = 0.05\).
Exercise D: Solution
\(H_0: \mu_1 - \mu_2 \leq 0\) vs. \(H_1: \mu_1 - \mu_2 > 0\) (right-tailed)
Pooled variance: \[s'^2_p = \frac{9 \times (4.5)^2 + 11 \times (5.1)^2}{10 + 12 - 2} = \frac{9 \times 20.25 + 11 \times 26.01}{20} = \frac{182.25 + 286.11}{20} = \frac{468.36}{20} = 23.418\]
\[s'_p = \sqrt{23.418} \approx 4.839\]
\[t_{obs} = \frac{(502.1 - 498.3) - 0}{4.839\sqrt{\frac{1}{10} + \frac{1}{12}}} = \frac{3.8}{4.839 \times \sqrt{0.1833}} = \frac{3.8}{4.839 \times 0.4282} = \frac{3.8}{2.072} \approx 1.83\]
Exercise D: Solution (cont.)
Rejection region (right-tailed, \(\alpha = 0.05\), \(\nu = 20\)):
From Table 7: \(t_{0.05,(20)} = 1.725\)
Since \(1.83 > 1.725\), we reject \(H_0\).
\(p\)-value: From Table 7, \(t_{0.05,(20)} = 1.725\) and \(t_{0.025,(20)} = 2.086\).
Since \(1.725 < 1.83 < 2.086\): \(0.025 < p < 0.05\).
Conclusion: At 5%, there is evidence that Factory 1 produces longer rods on average than Factory 2.
Summary of Today’s Lecture
Key takeaways — Lecture 7
- Section 3.3: Tests for \(\mu\) (\(\sigma\) known/unknown) and \(\sigma^2\) of Normal populations use the same pivot variables as confidence intervals (\(Z\), \(T\), \(Q\))
- Section 3.4: Tests for \(\mu_1 - \mu_2\) require identifying whether variances are known, unknown-but-equal (pooled), or whether samples are large
- Section 3.5: Large-sample tests (\(n > 30\)) work for any distribution thanks to the CLT; tests for proportions use \(\hat{p}\) and \(p_0\)
- In all cases, the procedure is the same 8-step process
- The \(p\)-value provides the strength of evidence and can be used alongside (or instead of) the critical region approach
What’s Next
The second midterm covers Topics 2.2, 3, and 4.
Next: Topic 4 — The Linear Regression Model.
Disclaimer
These slides are a free adaptation of the course material for Estatística II by Prof. Teresa Ferreira and Prof. Sandra Custódio from the Lisbon Accounting and Business School — Polytechnic University of Lisbon.
Primary reference: Newbold, P., Carlson, W. & Thorne, B. — Statistics for Business and Economics, Global Edition.
Statistics II — Parametric Hypothesis Tests