Z-Test for Large Sample Hypothesis Testing

The Z-Test is a parametric statistical test used to determine whether there is a statistically significant difference between a sample mean $\bar{x}$ and a population mean $\mu_0$ when the sample size is large ( $n \ge 30$ ) or when the population standard deviation $\sigma$ is known.

Z-Statistic Formula:

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

Where:

$\bar{x}$ is the calculated sample mean.
$\mu_0$ is the hypothesized population mean under $H_0$ .
$SE = \frac{\sigma}{\sqrt{n}}$ is the standard error.

Example of Z-Test

Testing Claimed Mean

\mu_0 = 1000, \quad \sigma = 50

Sample of $n = 100$ gives $\bar{x} = 1012$ at $\alpha = 0.05$

Step 1 & 2: Formulate Hypotheses & Standard Error

H_0: \mu = 1000, \quad H_1: \mu \neq 1000

SE = \frac{50}{\sqrt{100}} = 5.0

Step 3 & 4: Compute Z & Compare with Critical Value

Z = \frac{1012 - 1000}{5.0} = 2.40

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

Conclusion

\boxed{\text{Reject } H_0 \text{ at } \alpha = 0.05}

Therefore, because $|Z_{\text{calc}}| = 2.40 > 1.96$ , we reject the null hypothesis and conclude that the mean lifetime differs significantly from 1000 hours.

Interactive Z-Test Calculator

Z-Test (Large Sample Significance Test)

PreviousLeast Squares Method

Unit 5•25 of 28

NextStudent's t-Test

Z-Test for Large Sample Hypothesis Testing

Z-Statistic Formula:

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

Where:

$\bar{x}$ is the calculated sample mean.
$\mu_0$ is the hypothesized population mean under $H_0$ .
$SE = \frac{\sigma}{\sqrt{n}}$ is the standard error.

Example of Z-Test

Testing Claimed Mean

\mu_0 = 1000, \quad \sigma = 50

Sample of $n = 100$ gives $\bar{x} = 1012$ at $\alpha = 0.05$

Step 1 & 2: Formulate Hypotheses & Standard Error

H_0: \mu = 1000, \quad H_1: \mu \neq 1000

SE = \frac{50}{\sqrt{100}} = 5.0

Step 3 & 4: Compute Z & Compare with Critical Value

Z = \frac{1012 - 1000}{5.0} = 2.40

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

Conclusion

\boxed{\text{Reject } H_0 \text{ at } \alpha = 0.05}

Therefore, because $|Z_{\text{calc}}| = 2.40 > 1.96$ , we reject the null hypothesis and conclude that the mean lifetime differs significantly from 1000 hours.

Interactive Z-Test Calculator

Z-Test (Large Sample Significance Test)

PreviousLeast Squares Method

Unit 5•25 of 28

NextStudent's t-Test

Z-Test for Large Sample Hypothesis Testing

Example of Z-Test

Step 1 & 2: Formulate Hypotheses & Standard Error

Step 3 & 4: Compute Z & Compare with Critical Value

Conclusion

Interactive Z-Test Calculator

Z-Test (Large Sample Significance Test)

NETZ Engine

Z-Test for Large Sample Hypothesis Testing

Example of Z-Test

Step 1 & 2: Formulate Hypotheses & Standard Error

Step 3 & 4: Compute Z & Compare with Critical Value

Conclusion

Interactive Z-Test Calculator

Z-Test (Large Sample Significance Test)