Recall: Statistical Inference

So far in this course, we have covered:

• Sampling distributions (Topic 1): how sample statistics behave
• Point estimation (Topic 2.1): finding a single “best guess” for a parameter
• Interval estimation (Topic 2.2): constructing a range of plausible values

Why Hypothesis Testing?

In practice, decisions must be made under uncertainty:

• A pharmaceutical company claims a new drug lowers cholesterol by more than 20 mg/dL. Does the clinical trial data support this?
• An engineer suspects that a machine is overfilling packages beyond the nominal 100 g. Is there statistical evidence?
• A bank manager believes the average processing time for loan applications has decreased. Can we confirm this?

From Estimation to Testing

The Structure of a Hypothesis Test

Every hypothesis test involves the same fundamental elements:

1. Null hypothesis (\(H_0\)): the claim we assume to be true until evidence says otherwise
2. Alternative hypothesis (\(H_1\)): the claim we are trying to find evidence for
3. Test statistic: a function of the sample data (computed under \(H_0\))
4. Significance level (\(\alpha\)): the probability of rejecting \(H_0\) when it is actually true
5. Critical (rejection) region: the set of values of the test statistic that lead to rejection of \(H_0\)

The Null Hypothesis (\(H_0\))

The Alternative Hypothesis (\(H_1\))

How to Formulate \(H_0\) and \(H_1\)

A common source of confusion: which claim goes in \(H_0\) and which in \(H_1\)?

Example: Formulating Hypotheses

Scenario: A food company states that each package contains, on average, 500 g of product. A consumer protection agency suspects the company is underfilling.

Example: Formulating Hypotheses (cont.)

Scenario: An engineer suspects that a machine is overfilling packages beyond the nominal 100 g.

The Test Statistic

The Significance Level (\(\alpha\))

The Critical (Rejection) Region

Critical Regions: Visual Summary

The shaded areas represent the rejection region. If the observed test statistic falls in the shaded area, we reject \(H_0\).

The General Test Procedure

Two Types of Mistakes

When we make a decision based on sample data, we can make two kinds of errors:

Type I Error

Type II Error

The Power of a Test

The Trade-off Between \(\alpha\) and \(\beta\)

For a fixed sample size \(n\):

• If we decrease \(\alpha\) (more conservative) \(\Rightarrow\) \(\beta\) increases (harder to detect a real effect)
• If we increase \(\alpha\) (more liberal) \(\Rightarrow\) \(\beta\) decreases (easier to detect, but more false alarms)

Analogy: Fire Alarm

Think of a hypothesis test as a fire alarm:

• Type I Error (\(\alpha\)): The alarm goes off, but there is no fire (false alarm)
• Type II Error (\(\beta\)): There is a fire, but the alarm does not go off (missed detection)

Summary: Errors at a Glance

Example 1: Test for the Mean (\(\sigma\) known)

A bottling company fills bottles with a nominal volume of \(\mu_0 = 500\) mL. The filling process is known to have a standard deviation of \(\sigma = 8\) mL. A quality inspector takes a random sample of \(n = 36\) bottles and obtains a sample mean of \(\bar{x} = 497\) mL.

At the 5% significance level, is there evidence that the machine is underfilling?

Example 1: Solution — Step 1

Step 1: State the hypotheses.

The inspector suspects underfilling, i.e., the mean is less than 500.

\[H_0: \mu \geq 500 \quad \text{vs.} \quad H_1: \mu < 500\]

Example 1: Solution — Step 2

Step 2: Significance level.

\[\alpha = 0.05\]

Example 1: Solution — Step 3

Step 3: Test statistic.

The population standard deviation \(\sigma = 8\) is known, so:

\[Z_0 = \frac{\bar{X} - \mu_0}{\sigma / \sqrt{n}} \underset{\text{under } H_0}{\sim} N(0, 1)\]

Example 1: Solution — Step 4

Step 4: Critical region.

For a left-tailed test at \(\alpha = 0.05\):

\[\text{Rejection region: } \left]-\infty\, ;\, -z_{\alpha}\right] = \left]-\infty\, ;\, -1.645\right]\]

Example 1: Solution — Step 5

Step 5: Compute the observed value.

\[z_{obs} = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} = \frac{497 - 500}{8 / \sqrt{36}} = \frac{-3}{8/6} = \frac{-3}{1.333} \approx -2.25\]

Example 1: Solution — Step 6

Step 6: Decision.

\[z_{obs} = -2.25 \leq -1.645\]

Example 1: Solution — Step 7

Step 7: Conclusion.

At the 5% significance level, there is sufficient statistical evidence to conclude that the machine is underfilling the bottles (i.e., the mean volume is less than 500 mL).

Example 2: Formulating Hypotheses — Practice

For each scenario, write \(H_0\) and \(H_1\), and state the type of test:

1. A bank claims the average time to process a loan is 5 days. A consultant believes it takes longer.
2. A manufacturer states the defect rate is at most 3%. An auditor suspects it is higher.
3. A researcher wants to test if a training program changes employee productivity from the current average of 40 units/day.

Example 2: Solutions

1. \(H_0: \mu \leq 5 \quad \text{vs.} \quad H_1: \mu > 5\) → Right-tailed

Quick Recap Before the Break

Motivation

Before the break, we used the critical region approach:

1. Fix \(\alpha\)
2. Find the critical value(s)
3. Compare \(z_{obs}\) (or \(t_{obs}\), \(q_{obs}\)) with the critical value

Definition of the \(p\)-value

\(p\)-value Formulas

The formula depends on the type of test:

Here \(T\) denotes the test statistic (could be \(Z\), \(T\), \(Q\), etc.) and \(t_{obs}\) is its observed value.

Decision Rule Using the \(p\)-value

\(p\)-value: Graphical Interpretation

The shaded blue areas represent the \(p\)-value: the probability of obtaining a result as extreme as \(t_{obs}\), under \(H_0\).

Example 4: \(p\)-value with the \(t\)-distribution

A random sample of \(n = 20\) observations is drawn from a Normal population. The sample mean is \(\bar{x} = 101\) and the corrected sample standard deviation is \(s' = 3\).

Test \(H_0: \mu \leq 100\) vs. \(H_1: \mu > 100\).

Example 4: Solution

Test statistic (\(\sigma\) unknown, Normal population):

\[T_0 = \frac{\bar{X} - \mu_0}{S'/\sqrt{n}} \underset{\text{under } H_0}{\sim} t_{(n-1)} = t_{(19)}\]

Example 4: Solution (cont.)

\(p\)-value (right-tailed test):

\[p = P(T > 1.49 \mid H_0)\]

Reading the \(p\)-value from Tables

When using the \(t\)-Student or \(\chi^2\) tables, we typically cannot compute the exact \(p\)-value. Instead, we bracket it:

Example 5: Two-tailed \(p\)-value

A manufacturer specifies that the mean diameter of a component is \(\mu_0 = 25\) mm. The population variance is known: \(\sigma^2 = 4\). A random sample of \(n = 49\) components has \(\bar{x} = 25.6\) mm.

Test \(H_0: \mu = 25\) vs. \(H_1: \mu \neq 25\). Compute the \(p\)-value.

Equivalence: Critical Region vs. \(p\)-value

Both approaches always give the same conclusion for a given \(\alpha\):

Common Misinterpretations of the \(p\)-value

The Connection

There is a direct relationship between a two-tailed hypothesis test and a confidence interval:

Example 6: CI and Test Equivalence

Recall Example 5: \(\mu_0 = 25\), \(\sigma = 2\), \(n = 49\), \(\bar{x} = 25.6\).

The 95% confidence interval for \(\mu\) is:

\[\bar{x} \pm z_{0.025} \cdot \frac{\sigma}{\sqrt{n}} = 25.6 \pm 1.96 \times \frac{2}{7} = 25.6 \pm 0.56\]

Exercise 1

A manufacturer claims that light bulbs last, on average, at least 1000 hours. A random sample of 25 bulbs from a Normal population gives \(\bar{x} = 980\) hours and \(s' = 40\) hours.

At \(\alpha = 0.05\), is there evidence against the manufacturer’s claim?

Exercise 1: Solution

Step 1: \(H_0: \mu \geq 1000\) vs. \(H_1: \mu < 1000\) (left-tailed)

Exercise 1: Solution (cont.)

Step 4: Left-tailed, \(\alpha = 0.05\), \(\nu = 24\)

\[\text{Rejection region: } \left]-\infty\,;\, -t_{0.05,(24)}\right] = \left]-\infty\,;\, -1.711\right]\]

(From Table 7: \(t_{0.05,(24)} = 1.711\))

Exercise 1: \(p\)-value

\[p = P(T \leq -2.50) = P(T \geq 2.50)\]

From Table 7, \(\nu = 24\): \(t_{0.01,(24)} = 2.492\) and \(t_{0.005,(24)} = 2.797\).

Exercise 2

A food processing company fills cereal boxes labeled as 500 g. The variance of the filling process is known to be \(\sigma^2 = 25 \text{ g}^2\). A random sample of \(n = 64\) boxes yields \(\bar{x} = 498\) g.

1. Test whether the mean weight differs from 500 g at \(\alpha = 0.05\).

2. Compute the \(p\)-value.

Exercise 2: Solution

a) \(H_0: \mu = 500\) vs. \(H_1: \mu \neq 500\) (two-tailed)

\(\sigma = 5\) is known, \(n = 64\):

\[Z_0 = \frac{\bar{X} - 500}{\sigma/\sqrt{n}} \sim N(0,1) \qquad z_{obs} = \frac{498 - 500}{5/\sqrt{64}} = \frac{-2}{5/8} = \frac{-2}{0.625} = -3.20\]

Exercise 2: Solution (cont.)

b) \(p\)-value (two-tailed):

\[p = 2 \times P(Z \geq |{-3.20}|) = 2 \times P(Z \geq 3.20)\]

Exercise 3

From a random sample of \(n = 20\) observations drawn from a Normal population, one obtains \(\bar{x} = 101\) and \(s' = 3\).

Consider the test: \(H_0: \mu \leq 100\) vs. \(H_1: \mu > 100\).

1. For \(\alpha = 0.05\), determine the rejection region.

2. If, for another sample of the same size, \(t_{obs} = 2.861\), what is the \(p\)-value?

Exercise 3: Solution

a) Right-tailed test, \(T_0 \sim t_{(19)}\):

\[\text{Rejection region: } \left[t_{0.05,(19)}\,;\, +\infty\right[ = \left[1.729\,;\, +\infty\right[\]

(Table 7: \(t_{0.05,(19)} = 1.729\))

Summary of Today’s Lecture

Next Lecture (April 21)

We will apply the testing framework to specific cases:

• Section 3.3: Tests for means and variances of Normal populations
• Section 3.4: Tests for the difference of means of two Normal populations
• Section 3.5: Large-sample tests (asymptotic normality)

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.