←
Formula Review

Formula Review

11.1 Facts 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

11.2 Test 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.

11.3 Goodness-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

11.4 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 ∑ 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 .

11.5 Test 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