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Units, Constants and Useful Formulas
χ^{2} Goodness of Fit
------------------
Goodness of fit involves the comparison between an observed
frequency distribution and the corresponding values of a
theoretical frequency distribution. The test is appropriate when:
- The sampling method is simple random sampling.
- The population is at least 10 times as large as the sample.
- The variable under study is categorical.
- The expected value of the number of sample observations in
each level of the variable is at least 5.
χ_{v}^{2} = Σ[O_{i} - E_{i})^{2}/E_{i}]
where:
k = number of levels of the categorical variable.
v = degrees of freedom = k - 1
O_{i} = Observed frequency of observations in level i
E_{i} = np_{i} = Expected frequency of observations in level i
p_{i} = proportion of observations in level i
Example
-------
A company claims that 30% of baseball cards are rookies, 60%
veterans, and 10% are All-Stars. The cards are sold in packages
of 100. Suppose a randomly-selected package of cards has 50
rookies, 45 veterans, and 5 All-Stars. Is this consistent with the
company's claim? Use a 0.05 level of significance?
- H_{0}: The proportion of rookies, veterans, and All-Stars is 30%,
60% and 10%, respectively.
- H_{1}: At least one of the proportions in the null hypothesis is
false.
v = k - 1 = 3 - 1 = 2
E_{i} = np_{i}
E_{1} = 100 * 0.30 = 30
E_{2} = 100 * 0.60 = 60
E_{2} = 100 * 0.10 = 10
χ^{2} = Σ[(O_{i} - E_{i})^{2}/E_{i}]
= [(50 - 30)^{2}/30 ] + [(45 - 60)^{2}/60] + [(5 - 10)^{2}/10]
= (400/30) + (225/60) + (25/10) = 13.33 + 3.75 + 2.50 = 19.58
From tabless, the p-value corresponding to 19.58 = 0.0001.
Therefore we can reject H_{0}