# Formula Review

## 11.1Facts About the Chi-Square Distribution

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

μχ2 = df chi-square distribution population mean

$σ χ 2 = 2( df ) σ χ 2 = 2( df )$ Chi-Square distribution population standard deviation

## 11.2Test of a Single Variance

$χ 2 = χ 2 =$ $(n−1) s 2 σ 0 2 (n−1) s 2 σ 0 2$ Test of a single variance statistic where:
n: sample size
s: sample standard deviation
$σ0σ0$: 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 $(n–1) s 2 σ 0 2 (n–1) s 2 σ 0 2$, where n = sample size, s2 = sample variance, and σ2 = population variance.
• The test may be left-, right-, or two-tailed.

## 11.3Goodness-of-Fit Test

$∑ k (O−E) 2 E ∑ k (O−E) 2 E$ 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

## 11.4Test 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 $∑ i⋅j (O−E) 2 E ∑ i⋅j (O−E) 2 E$ 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 $E = (row total)(column total) total surveyed E= (row total)(column total) total surveyed$.

## 11.5Test for Homogeneity

$∑ i⋅j (O−E) 2 E ∑ i⋅j (O−E) 2 E$ 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