UNIT 5 • TESTING OF HYPOTHESIS

Z-Test
Large Sample Significance Test

The Z-Test determines whether an observed sample mean $\bar{x}$ differs significantly from a claimed population baseline $\mu_0$ when sample size is large ( $n \ge 30$ ) or population standard deviation $\sigma$ is known!

Z-STATISTIC COMPUTATION FORMULANORMAL DISTRIBUTION

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

1. Mean DeviationDifference between sample mean and target:

\bar{x} - \mu_0

2. Standard ErrorSampling variability denominator:

SE = \frac{\sigma}{\sqrt{n}}

3. Decision RuleReject

H_0

|Z_{calc}| > Z_{\alpha/2}

(e.g. 1.96 for 5% level).

Step-by-Step Worked Example

GUIDED STORY

OUR PROBLEM TO SOLVE

Test the manufacturer's lifetime claim:

\mu_0 = 1000 \text{ hours}, \quad \sigma = 50 \text{ hours}

A sample of $n = 100$ lightbulbs yields $\bar{x} = 1012$ hours. Test at $\alpha = 0.05$ level!

STEP 01 • HYPOTHESES & STANDARD ERROR

State null ( $H_0$ ) and alternative ( $H_1$ ) hypotheses:

H_0: \mu = 1000, \quad H_1: \mu \neq 1000 \quad (\text{Two-tailed test})

SE = \frac{\sigma}{\sqrt{n}} = \frac{50}{\sqrt{100}} = \frac{50}{10} = 5.0

STEP 02 • COMPUTE Z-STATISTIC & CRITICAL THRESHOLD

Calculate test statistic $Z_{calc}$ :

Z_{\text{calc}} = \frac{1012 - 1000}{5.0} = \frac{12}{5.0} = 2.40

|Z_{\text{calc}}| = 2.40 > Z_{0.025} = 1.96

FINAL RESULT

Decision: Reject H₀ (Statistically Significant Difference)

\boxed{\text{Reject } H_0: |Z_{\text{calc}}| = 2.40 > 1.96}

Because calculated $|Z_{calc}| = 2.40$ exceeds the critical threshold $1.96$ at 5% significance level, we reject the null hypothesis and conclude the mean lifetime differs significantly from 1000 hours!

LABORATORY

Interactive Live Calculator & Visualizer

LARGE SAMPLE HYPOTHESIS TESTING

Z-Test Engine (Test of Significance for Means)

PreviousLeast Squares Method

Unit 5•25 of 28

NextStudent's t-Test

∑

f(x)∫dxπ

NETZ Engine

Evaluating numerical algorithm

UNIT 5 • TESTING OF HYPOTHESIS

Z-Test
Large Sample Significance Test

Z-STATISTIC COMPUTATION FORMULANORMAL DISTRIBUTION

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

1. Mean DeviationDifference between sample mean and target:

\bar{x} - \mu_0

2. Standard ErrorSampling variability denominator:

SE = \frac{\sigma}{\sqrt{n}}

3. Decision RuleReject

H_0

|Z_{calc}| > Z_{\alpha/2}

(e.g. 1.96 for 5% level).

Step-by-Step Worked Example

GUIDED STORY

OUR PROBLEM TO SOLVE

Test the manufacturer's lifetime claim:

\mu_0 = 1000 \text{ hours}, \quad \sigma = 50 \text{ hours}

A sample of $n = 100$ lightbulbs yields $\bar{x} = 1012$ hours. Test at $\alpha = 0.05$ level!

STEP 01 • HYPOTHESES & STANDARD ERROR

State null ( $H_0$ ) and alternative ( $H_1$ ) hypotheses:

H_0: \mu = 1000, \quad H_1: \mu \neq 1000 \quad (\text{Two-tailed test})

SE = \frac{\sigma}{\sqrt{n}} = \frac{50}{\sqrt{100}} = \frac{50}{10} = 5.0

STEP 02 • COMPUTE Z-STATISTIC & CRITICAL THRESHOLD

Calculate test statistic $Z_{calc}$ :

Z_{\text{calc}} = \frac{1012 - 1000}{5.0} = \frac{12}{5.0} = 2.40

|Z_{\text{calc}}| = 2.40 > Z_{0.025} = 1.96

FINAL RESULT

Decision: Reject H₀ (Statistically Significant Difference)

\boxed{\text{Reject } H_0: |Z_{\text{calc}}| = 2.40 > 1.96}

LABORATORY

Interactive Live Calculator & Visualizer

LARGE SAMPLE HYPOTHESIS TESTING

Z-Test Engine (Test of Significance for Means)

PreviousLeast Squares Method

Unit 5•25 of 28

NextStudent's t-Test

Z-Test Large Sample Significance Test

Step-by-Step Worked Example

Decision: Reject H₀ (Statistically Significant Difference)

Interactive Live Calculator & Visualizer

Z-Test Engine (Test of Significance for Means)

NETZ Engine

Z-Test Large Sample Significance Test

Step-by-Step Worked Example

Decision: Reject H₀ (Statistically Significant Difference)

Interactive Live Calculator & Visualizer

Z-Test Engine (Test of Significance for Means)

Z-Test
Large Sample Significance Test

Z-Test
Large Sample Significance Test