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Z Test

TL;DR

Section titled “TL;DR”
  • Tests whether a sample mean differs significantly from a known population mean.
  • Requires the sample’s standard deviation and a Z-score computed from the difference in means.
  • Use a Z-table to convert the Z-score into a probability to assess significance.

Definition

Section titled “Definition”

A Z-Test is a statistical test used to determine if there is a significant difference between the mean of a sample and a known population mean. It tests the hypothesis that the sample mean is different from the population mean.

Explanation

Section titled “Explanation”

To conduct a Z-Test, follow the steps described in the source material:

  • Calculate the standard deviation of the sample, which measures how spread out the data are.
  • Compute the Z-score by subtracting the population mean from the sample mean and dividing the result by the standard deviation of the sample. In general form:
Z-score=sample mean−population meanstandard deviation of the sample\text{Z-score} = \frac{\text{sample mean} - \text{population mean}}{\text{standard deviation of the sample}}
  • Use a Z-table to determine the probability of observing a result as extreme as the computed Z-score assuming the sample mean and the population mean are the same. A low probability indicates the difference is statistically significant.

Examples

Section titled “Examples”

Example 1

Section titled “Example 1”

A company samples 50 employees and finds a sample mean salary of 50,000.Thenationalaveragesalaryfortheindustryis50,000. The national average salary for the industry is 48,000. To determine if the difference is significant, the company calculates the sample standard deviation and then the Z-score:

Z-score=50,000−48,000standard deviation of the sample\text{Z-score} = \frac{50,000 - 48,000}{\text{standard deviation of the sample}}

The company then uses a Z-table to find the probability of obtaining such an extreme result if the sample mean and the population mean are the same. A low probability indicates statistical significance.

Example 2

Section titled “Example 2”

A high school teacher samples 20 students and finds a sample mean test score of 75. The district average test score is 80. To assess significance, the teacher calculates the sample standard deviation and the Z-score:

Z-score=75−80standard deviation of the sample\text{Z-score} = \frac{75 - 80}{\text{standard deviation of the sample}}

The teacher consults a Z-table to determine the probability of observing such a result if the sample mean and the population mean are the same. A low probability indicates the difference is statistically significant.

Section titled “Related terms”
  • Z-score
  • Z-table
  • Standard deviation
  • Sample mean
  • Population mean