Z-Test for Means

Overview

The z-test is used to test hypotheses about population means when the population standard deviation ($\sigma$) is known or the sample size is large.

Conditions

1. Population $\sigma$ is known (or $n \geq 30$ and use $s$)
2. Random sample
3. Population is normal OR $n \geq 30$ (by CLT)

Test Statistic

$$Z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}$$

Where:

• $\bar{x}$ = sample mean
• $\mu_0$ = hypothesized population mean
• $\sigma$ = population standard deviation
• $n$ = sample size

One-Sample Z-Test

Hypotheses

Test Type	$H_0$	$H_1$
Two-tailed	$\mu = \mu_0$	$\mu \neq \mu_0$
Left-tailed	$\mu \geq \mu_0$	$\mu < \mu_0$
Right-tailed	$\mu \leq \mu_0$	$\mu > \mu_0$

Critical Values

$\alpha$	Two-tailed	One-tailed
0.10	$\pm 1.645$	1.28
0.05	$\pm 1.96$	1.645
0.01	$\pm 2.58$	2.33

Two-Sample Z-Test

For Independent Samples

$$Z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)_0}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}$$

Usually testing $H_0: \mu_1 - \mu_2 = 0$

Examples

Example 1: One-Sample Two-Tailed

A factory claims mean output is 500 units. Sample: $n = 64$, $\bar{x} = 510$, $\sigma = 40$. Test at $\alpha = 0.05$.

$$H_0: \mu = 500, \quad H_1: \mu \neq 500$$ $$Z = \frac{510 - 500}{40/\sqrt{64}} = \frac{10}{5} = 2.0$$

Critical values: $\pm 1.96$

$$\lvert 2.0 \rvert > 1.96 \Rightarrow \text{Reject } H_0$$ $$\text{p-value} = 2 \times P(Z > 2.0) = 2 \times 0.0228 = 0.0456$$

Evidence supports $\mu \neq 500$.

Example 2: One-Sample Left-Tailed

Claim: $\mu < 100$. Sample: $n = 50$, $\bar{x} = 97$, $\sigma = 15$. Test at $\alpha = 0.05$.

$$H_0: \mu \geq 100, \quad H_1: \mu < 100$$ $$Z = \frac{97 - 100}{15/\sqrt{50}} = \frac{-3}{2.12} = -1.42$$

Critical value: $-1.645$

$$-1.42 > -1.645 \Rightarrow \text{Fail to reject } H_0$$

Not enough evidence that $\mu < 100$.

Example 3: Two-Sample Test

Compare two populations:

• Group 1: $n_1 = 100$, $\bar{x}_1 = 75$, $\sigma_1 = 10$
• Group 2: $n_2 = 100$, $\bar{x}_2 = 72$, $\sigma_2 = 12$

Test $H_0: \mu_1 = \mu_2$ at $\alpha = 0.05$.

$$Z = \frac{(75 - 72) - 0}{\sqrt{\frac{100}{100} + \frac{144}{100}}} = \frac{3}{\sqrt{2.44}} = \frac{3}{1.56} = 1.92$$

Critical values: $\pm 1.96$

$$\lvert 1.92 \rvert < 1.96 \Rightarrow \text{Fail to reject } H_0$$

No significant difference between means.

Decision Summary

Using p-value

$$p \leq \alpha \Rightarrow \text{Reject } H_0$$ $$p > \alpha \Rightarrow \text{Fail to reject } H_0$$

Using critical value

$$\lvert Z \rvert > z_{\text{critical}} \Rightarrow \text{Reject } H_0 \quad \text{(two-tailed)}$$ $$Z > z_{\text{critical}} \Rightarrow \text{Reject } H_0 \quad \text{(right-tailed)}$$ $$Z < -z_{\text{critical}} \Rightarrow \text{Reject } H_0 \quad \text{(left-tailed)}$$

When to Use Z vs T

Situation	Use
$\sigma$ known	Z-test
$\sigma$ unknown, $n \geq 30$	Z or T (similar results)
$\sigma$ unknown, $n < 30$	T-test