Hypothesis Testing (Extended)

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.

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.

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.

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

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 $H_0$.

• “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.

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).

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

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).

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.

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 $= 1 - 0.6404 = 0.3596$.

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.

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.

• 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.

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

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 $X \sim B(n, p_0)$.

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.

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.

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