Chi Square Method

Chi Square

        In this module , we explore techniques for analyzing categorical data. Categorical data are nonnumerical data that are frequency counts of categories from one or more variables . For example, it is determined that of the 450 peoples attending high school reunion , 150 are entrepreneur, 200 are employee, 100 are housewife. The chi-square goodness-of fit test is used to analyze probabilities of multinomial distribution trials along a single dimensions. The chi-square goodness-of-fit test compares the expected, or theoretical, frequencies of categories from a population distribution to the observed , or actual, frequencies from a distribution  to determine whether there is difference between what was expected and what was observed.

[pic 1]

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Where:

Fo        = Frequency of observed values

Fe        = frequency of expected values

K        = number of categories

C        = number of parameters being estimated from the sample data

Example:

Uniform Test

The table below shows the results when a die is rolled 120 times

Conduct a chi square test to see whether the die is uniformly distributed or not. Use α= 0.05.

Step 1:

Find out the hypotheses for this example follows.

        Ho        = The observed distribution is the uniformly distributed.

        Ha        = The observed distribution is not uniformly distributed.

Step 2:

The statistical test being used is

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Step 3:

        Let α= 0.05

Step 4:

        Chi-square goodness-of-fit- tests are one tailed because a chi-square of zero indicates perfect agreement between distributions. Any deviation from zero difference occurs in the positive direction only because chi-square is determined by a sum of squared values and can never be negative. With six categories in this example (1,2,3,4,5,6) , k= 6

So, df= k-1-c = 6-1-0= 5

The critical value is  = 11.071[pic 4]

Step 5:

Find out the expected frequency (fe)

 n= 120

Step 6:

Step 7:

        Because the observed value of chi-square of 8.1 is lower than the critical table value 11.071, we will accept the null hypothesis.

Step 8:

        Business Implications: The die rolled uniformly.