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11.9: Formula Review

Last updated

Mar 17, 2025
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- 11.8: Chapter Review
- 11.10: Practice

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11.2 Facts About the Chi-Square Distribution

$X^2=\left(Z_1\right)^2+\left(Z_2\right)^2+\ldots\left(Z_{d f}\right)^2$ chi-square distribution random variable
$\mu_{X^2}=d f$ chi-square distribution population mean
$\sigma_{\chi^2}=\sqrt{2(d f)}$ Chi-Square distribution population standard deviation

11.3 Test of a Single Variance

$\chi^2=\dfrac{(n-1) s^2}{\sigma_0^2}$ Test of a single variance statistic where:

$n:$ sample size
s: sample standard deviation
$\sigma_0$ : hypothesized value of the population standard deviation
$d f=n-1$ 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 $\dfrac{(n-1) s^2}{\sigma_0^2}$ , where $n=$ sample size, $s^2=$ sample variance, and $\sigma^2=$ population variance.
The test may be left-, right-, or two-tailed.

11.4 Goodness-of-Fit Test

$\sum_k \dfrac{(O-E)^2}{E}$ goodness-of-fit test statistic where:

O: observed values
$E$ : expected values

$k$ : number of different data cells or categories
$d f=k-1$ degrees of freedom

11.5 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 $\sum_{i \cdot j} \dfrac{(O-E)^2}{E}$ 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 $E=\dfrac{\text { (row total)(column total) }}{\text { total surveyed }}$ .

11.6 Test for Homogeneity.

$\sum_{i \cdot j} \dfrac{(O-E)^2}{E}$ 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
$d f=(i-1)(j-1)$ Degrees of freedom