Hypothesis Testing (Extended)

Hypothesis Testing (Extended Treatment)

This document provides a rigorous treatment of hypothesis testing methodology, including null and Alternative hypotheses, significance levels, Type I and II errors, one-tailed and two-tailed tests, And critical regions.

1. The Hypothesis Testing Framework

1.1 Null and alternative hypotheses

The null hypothesis $H_0$ is the default assumption — a statement of “no effect” or “no difference.” It is assumed to be true unless the evidence is sufficiently compelling to reject It.

The alternative hypothesis $H_1$ specifies what we believe might be true instead. It is only Accepted if the evidence against $H_0$ is strong enough.

1.2 Test statistic

A test statistic is a function of the sample data whose distribution is known under $H_0$ . Common test statistics include sample proportions, sample means, and sample correlation Coefficients.

1.3 Significance level

The significance level $\alpha$ is the maximum probability of rejecting $H_0$ when it is Actually true. Common values are $\alpha = 0.05$ (5%), $\alpha = 0.01$ (1%), and $\alpha = 0.10$ (10%).

1.4 The decision rule

Assume $H_0$ is true.
Calculate the probability of obtaining a test statistic at least as extreme as the observed value, assuming $H_0$ .
If this probability (the $p$ -value) is less than $\alpha$ Reject $H_0$ . Otherwise, do not reject $H_0$ .

1.5 Critical value approach

Alternatively, find the critical value $c$ such that $P(\mathrm{test\ statistic} \geq c \mid H_0) = \alpha$ (for an upper-tailed test). If the observed test statistic exceeds $c$ Reject $H_0$ .

1.6 Interpreting the conclusion

“Reject $H_0$ ”: there is sufficient evidence at the $\alpha$ significance level to support $H_1$ .
“Do not reject $H_0$ ”: there is insufficient evidence to reject $H_0$ . This does not mean $H_0$ is true.

2. Type I and Type II Errors

2.1 Definitions

Decision	$H_0$ true	$H_0$ false
Reject $H_0$	Type I error	Correct decision
Do not reject	Correct decision	Type II error

Type I error: Rejecting $H_0$ when it is true (false positive).

P(\mathrm{Type\ I}) = \alpha

Type II error: Failing to reject $H_0$ when it is false (false negative).

P(\mathrm{Type\ II}) = \beta

2.2 The power of a test

The power of a test is the probability of correctly rejecting $H_0$ when it is false:

\mathrm{Power} = 1 - \beta

The power depends on:

The significance level $\alpha$ (increasing $\alpha$ increases power).
The sample size $n$ (increasing $n$ increases power).
The true value of the parameter (the further from $H_0$ The greater the power).

2.3 Worked example

Problem. A machine produces bolts with mean length $50\;\mathrm{mm}$ . The standard deviation is $0.5\;\mathrm{mm}$ . A sample of 16 bolts has mean $50.18\;\mathrm{mm}$ . Test at the 5% significance Level whether the mean length has changed.

$H_0: \mu = 50$ , $H_1: \mu \neq 50$ (two-tailed).

Under $H_0$ : $\bar{X} \sim N(50, 0.5^2/16) = N(50, 0.015625)$ .

Critical values: $\bar{x}$ such that $P(|\bar{X} - 50| \geq c) = 0.05$ .

$z = \pm 1.96$ for a two-tailed 5% test.

$c = 50 \pm 1.96 \times 0.125 = 50 \pm 0.245$

Critical region: $\bar{X} \lt 49.755$ or $\bar{X} \gt 50.245$ .

Test statistic: $z = \dfrac{50.18 - 50}{0.125} = 1.44$ .

Since $1.44 \lt 1.96$ We do not reject $H_0$ . There is insufficient evidence at the 5% level To conclude that the mean length has changed.

2.4 Finding the probability of Type II error

Continuing the example above, suppose the true mean is $\mu = 50.2$ .

$\beta = P(\mathrm{do\ not\ reject}\ H_0 \mid \mu = 50.2)$

$= P(49.755 \lt \bar{X} \lt 50.245 \mid \bar{X} \sim N(50.2, 0.015625))$

$= P\!\left(\dfrac{49.755 - 50.2}{0.125} \lt Z \lt \dfrac{50.245 - 50.2}{0.125}\right) = P(-3.56 \lt Z \lt 0.36)$

$= \Phi(0.36) - \Phi(-3.56) \approx 0.6406 - 0.0002 = 0.6404$

Power $= 1 - 0.6404 = 0.3596$ .

3. One-Tailed and Two-Tailed Tests

3.1 One-tailed tests

A one-tailed test is used when $H_1$ specifies a direction:

$H_1: \mu \gt \mu_0$ (upper-tailed): critical region in the upper tail.
$H_1: \mu \lt \mu_0$ (lower-tailed): critical region in the lower tail.

The entire significance level $\alpha$ is in one tail, making it easier to detect an effect in the Specified direction.

3.2 Two-tailed tests

A two-tailed test is used when $H_1$ does not specify a direction:

$H_1: \mu \neq \mu_0$ : critical region split between both tails, with $\alpha/2$ in each.

3.3 Choosing one-tailed vs two-tailed

Use a two-tailed test unless there is a strong prior reason to expect a specific direction.
A one-tailed test has greater power in the specified direction but cannot detect effects in the opposite direction.
The choice must be made before examining the data.

3.4 Critical values table

For a standard normal test at significance level $\alpha$ :

Test type	$\alpha = 0.10$	$\alpha = 0.05$	$\alpha = 0.01$
Two-tailed	$\pm 1.645$	$\pm 1.960$	$\pm 2.576$
Upper-tailed	$1.282$	$1.645$	$2.326$
Lower-tailed	$-1.282$	$-1.645$	$-2.326$

4. Hypothesis Tests for the Binomial Proportion

4.1 Setting up the test

To test whether a population proportion $p$ equals a specified value $p_0$ :

H_0: p = p_0, \qquad H_1: p \neq p_0\ (\mathrm{or}\ p \gt p_0\ \mathrm{or}\ p \lt p_0)

Under $H_0$ If $X$ is the number of successes in $n$ trials, then $X \sim B(n, p_0)$ .

4.2 Worked example: binomial test

Problem. A coin is tossed 20 times and lands heads 15 times. Test at the 5% significance level Whether the coin is biased.

$H_0: p = 0.5$ , $H_1: p \neq 0.5$ (two-tailed).

Under $H_0$ : $X \sim B(20, 0.5)$ .

For a two-tailed test at 5%, we need the critical region in each tail to have probability $\leq 0.025$ .

Lower tail: $P(X \leq k) \leq 0.025$ .

$P(X \leq 5) = 0.0207 \leq 0.025$ . So $k = 5$ (critical region: $X \leq 5$ ).

Upper tail: $P(X \geq k) \leq 0.025$ .

$P(X \geq 15) = P(X \leq 5) = 0.0207 \leq 0.025$ . So $k = 15$ (critical region: $X \geq 15$ ).

Since $X = 15$ falls in the critical region, we reject $H_0$ . There is sufficient evidence at the 5% level to conclude the coin is biased.

4.3 Finding the actual significance level

The actual significance level is the probability of being in the critical region under $H_0$ :

\alpha_{\mathrm{actual}} = P(X \leq 5) + P(X \geq 15) = 2(0.0207) = 0.0414

This is approximately 4.14%, which is the closest we can get to 5% with a discrete distribution.

nominal level. The Critical region is chosen so that $P(\mathrm{critical\ region} \mid H_0)$ does not exceed $\alpha$ And is as close as possible to $\alpha$ .

5. Critical Regions

5.1 Definition

The critical region (or rejection region) is the set of values of the test statistic that lead to rejection of $H_0$ . The acceptance region is its complement.

5.2 Finding the critical region

Procedure:

Identify the distribution of the test statistic under $H_0$ .
Determine whether the test is one-tailed or two-tailed.
Find the smallest region containing the most extreme values whose total probability under $H_0$ does not exceed $\alpha$ .

5.3 Worked example: Poisson critical region

Problem. A receptionist receives on average 2 calls per 5 minutes. Over a 5-minute period, she Receives 7 calls. Test at the 5% level whether the rate has increased.

$H_0: \lambda = 2$ , $H_1: \lambda \gt 2$ (upper-tailed).

Under $H_0$ : $X \sim \mathrm{Po}(2)$ .

Critical region: smallest $k$ such that $P(X \geq k) \leq 0.05$ .

$P(X \geq 5) = 1 - P(X \leq 4) = 1 - e^{-2}\!\left(1 + 2 + 2 + \frac{4}{3} + \frac{2}{3}\right) = 1 - e^{-2} \times 7.667 = 1 - 0.9473 = 0.0527$

$P(X \geq 6) = 1 - P(X \leq 5) = 1 - 0.9835 = 0.0165 \leq 0.05$

Critical region: $X \geq 6$ . Actual significance level: $1.65\%$ .

Since $X = 7 \geq 6$ We reject $H_0$ . There is sufficient evidence that the call rate has Increased.

5.4 Worked example: normal critical region

Problem. The masses of packets of biscuits are normally distributed with standard deviation $3\;\mathrm{g}$ . A sample of 10 packets has mean mass $248\;\mathrm{g}$ . Find the critical region for testing whether the mean mass is less than $250\;\mathrm{g}$ at the 1% significance level.

$H_0: \mu = 250$ , $H_1: \mu \lt 250$ .

Under $H_0$ : $\bar{X} \sim N(250, 3^2/10) = N(250, 0.9)$ .

$P(\bar{X} \lt c) = 0.01 \implies \dfrac{c - 250}{\sqrt{0.9}} = -2.326$

$c = 250 - 2.326\sqrt{0.9} = 250 - 2.208 = 247.79$

Critical region: $\bar{X} \lt 247.79$ .

Since $\bar{x} = 248 \gt 247.79$ We do not reject $H_0$ at the 1% level.

6. Practice Problems

Problem 1

A die is rolled 30 times and a six appears 9 times. Test at the 5% significance level whether the Die is biased towards showing a six.

Solution

$H_0: p = 1/6$ , $H_1: p \gt 1/6$ (upper-tailed).

$X \sim B(30, 1/6)$ .

Find smallest $k$ : $P(X \geq k) \leq 0.05$ .

$P(X \geq 9) = 1 - P(X \leq 8)$ .

$P(X \leq 8) = \displaystyle\sum_{r=0}^{8}\binom{30}{r}(1/6)^r(5/6)^{30-r} \approx 0.9502$ .

$P(X \geq 9) \approx 0.0498 \leq 0.05$ .

Since $X = 9$ is in the critical region, reject $H_0$ . Sufficient evidence the die is biased Towards six.

Problem 2

A manufacturer claims that the mean lifetime of a component is 500 hours. A sample of 25 components has mean lifetime 490 hours with standard deviation 15 hours. Test the claim at the 5% significance Level.

Solution

$H_0: \mu = 500$ , $H_1: \mu \neq 500$ (two-tailed).

$\bar{X} \sim N(500, 15^2/25) = N(500, 9)$ approximately.

$z = \dfrac{490 - 500}{3} = -3.33$ .

Critical values: $\pm 1.96$ .

Since $|-3.33| = 3.33 \gt 1.96$ Reject $H_0$ . Sufficient evidence the mean lifetime differs from 500 hours.

Problem 3

The number of accidents per week at a factory is thought to follow a Poisson distribution with mean 3. In a particular week, 8 accidents occur. Test at the 5% level whether the accident rate has Increased.

Solution

$H_0: \lambda = 3$ , $H_1: \lambda \gt 3$ .

$X \sim \mathrm{Po}(3)$ .

$P(X \geq 7) = 1 - P(X \leq 6) \approx 1 - 0.9665 = 0.0335 \leq 0.05$ .

$P(X \geq 6) = 1 - P(X \leq 5) \approx 1 - 0.9161 = 0.0839 \gt 0.05$ .

Critical region: $X \geq 7$ . Since $X = 8 \geq 7$ Reject $H_0$ . Sufficient evidence the rate Has increased.

Problem 4

In a hypothesis test with $H_0: p = 0.4$ and $H_1: p \gt 0.4$ based on a sample of size 20, Find: (a) the critical region at the 5% level; (b) the actual significance level; (c) the Probability of a Type II error if the true $p = 0.6$ .

Solution

Under $H_0$ : $X \sim B(20, 0.4)$ .

(a) $P(X \geq 11) = 1 - P(X \leq 10) \approx 1 - 0.9435 = 0.0565 \gt 0.05$ .

$P(X \geq 12) = 1 - P(X \leq 11) \approx 1 - 0.9790 = 0.0210 \leq 0.05$ .

Critical region: $X \geq 12$ .

(b) Actual significance level $\approx 2.10\%$ .

$\beta = P(X \leq 11 \mid p = 0.6) \approx 0.4044$ .

Common Pitfalls

Confusing the domain and range of functions, or not considering restrictions (e.g., denominator cannot be zero).
Losing marks by not showing sufficient working. Always write out each step, especially in proof questions.
Misreading the question, particularly with “hence’ vs ‘hence or otherwise’. The former requires using previous work.
Rounding too early in multi-step calculations. Carry full precision through and round only the final answer.

Worked Examples

Example 1: One-Tailed Binomial Test

Problem. A coin is tossed 20 times and 15 heads are observed. Test at the 5% significance level whether the coin is biased towards heads.

Solution. $H_0: p = 0.5$ , $H_1: p > 0.5$ . Under $H_0$ : $X \sim B(20, 0.5)$ .

$P(X \geq 15) = P(X = 15) + P(X = 16) + \ldots + P(X = 20)$ .

$P(X \geq 15) = \binom{20}{15}(0.5)^{20} + \binom{20}{16}(0.5)^{20} + \ldots + \binom{20}{20}(0.5)^{20}$ .

P(X \geq 15) = (15504 + 4845 + 1140 + 190 + 20 + 1) \times (0.5)^{20} = 21700 \times 9.537 \times 10^{-7} \approx 0.0207

Since $0.0207 < 0.05$ , we reject $H_0$ . There is significant evidence that the coin is biased towards heads.

$\blacksquare$

Example 2: Normal Hypothesis Test for the Mean

Problem. A machine fills packets with a mean weight of $500\ \mathrm{g}$ and standard deviation $12\ \mathrm{g}$ . A sample of 36 packets has mean $503.5\ \mathrm{g}$ . Test at the 1% level whether the mean has increased.

Solution. $H_0: \mu = 500$ , $H_1: \mu > 500$ .

Under $H_0$ : $\bar{X} \sim N\!\left(500, \frac{144}{36}\right) = N(500, 4)$ .

Test statistic: $z = \frac{503.5 - 500}{\sqrt{4}} = \frac{3.5}{2} = 1.75$ .

One-tailed critical value at 1%: $z_{0.01} = 2.326$ .

Since $1.75 < 2.326$ , we do not reject $H_0$ . There is insufficient evidence that the mean has increased.

$\blacksquare$

flowchart TD
    A[Hypothesis Testing Depth] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Summary

Hypothesis testing: state $H_0$ and $H_1$ , choose significance level, calculate test statistic, compare with critical value.
Type I error: rejecting $H_0$ when it is true (probability $= \alpha$ ).
Type II error: failing to reject $H_0$ when it is false (probability $= \beta$ ).
For binomial tests, use cumulative probabilities; for normal tests, use the $z$ -score.
The critical region is the set of values that lead to rejection of $H_0$ .

Intuition

Mathematics is the study of structure, quantity, and change. Algebra provides symbols for unknown quantities, geometry describes spatial relationships, and calculus captures motion and growth. Together, these branches form a powerful toolkit for solving problems that range from calculating areas to predicting population dynamics. Mathematical literacy is essential for science, technology, and informed citizenship.

Cross-References

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