SESSION:11 The Chi-Square DistributionFormula Review

Session:11 The Chi-Square Distribution

Formula Review

Introductory Business Statistics | Leadership Development – Micro-Learning Session

Rice University 2020 | Michael Laverty, Colorado State University Global Chris Littel, North Carolina State University| https://openstax.org/details/books/introductory-business-statistics

11.1 Facts About the Chi-Square Distribution

χ² = (Z₁)² + (Z₂)² + … (Z_df)² chi-square distribution random variable

μ_χ² = df chi-square distribution population mean

$σ_{χ^{2}} = \sqrt{2 (d f)}$

$�_{�^{2}} = \sqrt{2 (� �)}$ Chi-Square distribution population standard deviation

11.2 Test of a Single Variance

$χ^{2} =$

$�^{2} =$ $\frac{(n - 1) s^{2}}{σ_{0}^{2}}$

$\frac{(� - 1) �^{2}}{�_{0}^{2}}$ Test of a single variance statistic where:
n: sample size
s: sample standard deviation
$σ_{0}$

$�_{0}$ : hypothesized value of the population standard deviation

df = n – 1 Degrees of freedom

Test of a Single Variance

Use the test to determine variation.
The degrees of freedom is the number of samples – 1.
The test statistic is $\frac{(n - 1) s^{2}}{σ_{0}^{2}}$

, where n = sample size, s² = sample variance, and σ² = population variance.
The test may be left-, right-, or two-tailed.

11.3 Goodness-of-Fit Test

$\sum_{k} \frac{{(O - E)}^{2}}{E}$

$\sum_{�} \frac{{(� - �)}^{2}}{�}$ goodness-of-fit test statistic where:

O: observed values
E: expected values

k: number of different data cells or categories

df = k − 1 degrees of freedom

11.4 Test of Independence

Test of Independence

The number of degrees of freedom is equal to (number of columns – 1)(number of rows – 1).
The test statistic is $\sum_{i \cdot j} \frac{{(O - E)}^{2}}{E}$

where O = observed values, E = expected values, i = the number of rows in the table, and j = the number of columns in the table.
If the null hypothesis is true, the expected number $E = \frac{(row total)(column total)}{total surveyed}$

.

11.5 Test for Homogeneity

$\sum_{i \cdot j} \frac{{(O - E)}^{2}}{E}$

$\sum_{� \cdot �} \frac{{(� - �)}^{2}}{�}$ Homogeneity test statistic where: O = observed values
E = expected values
i = number of rows in data contingency table
j = number of columns in data contingency table

df = (i −1)(j −1) Degrees of freedom