Use the z-test when you know the population standard deviation

or or enter summarized data (x̄, n, σ, S) below

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When entering raw data, the tool will run the Shapiro-Wilk normality test and calculate outliers, as part of the test calculation.

validation message

Target: To check if the assumed μ_{0} is statistically correct, based on a sample average.

You know the standard deviation from previous researches.

Example1: A farmer calculated last year the average of the apples' weight in his apple orchard μ_{0} equals 17 kg, based on the entire population.

The current year he checked a small sample of apples and the sample average x equals 18 kg

Has the average of the apples' weight changed this year?

The farmer know the standard deviation of the apple's weight from previous researches.

Hypotheses

H_{0}: μ ≥ = ≤ μ_{0}

H_{1}: μ < ≠ > μ_{0}

Test statistic

$Z=\frac{{\mathrm{x\u0304}}_{}-\mathrm{\mu \u2080}}{{\sigma}_{}/\sqrt{n}}$

$Z=\frac{{\mathrm{x\u0304}}_{}-\mathrm{\mu \u2080}}{{\sigma}_{}/\sqrt{n}}$

Normal distribution

Normal distribution | |

The standard deviations of the population is known | |

Population expected mean is known |

Sample average | |

Sample size |

The following R code should produce the same results:

Currently, there is no direct R function for the one-sample z test.

__Examples__

1. Two-tailed test

A farmer calculated last year the average of the apples' weight in his apple orchard μ0 equals 17kg, based on a big sample.

The current year the sample average x̄ equals 16kg.

Was the average of the apple's weight in the entire orchard changed this year? or is it just a random difference?

2. Left-tailed test.

In the same example as above, the farmer only cares to know if the entire average is lesser this year.