Essay About Chi-Square Goodness And Frequency Counts Of Categories

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][pic 2]Where:Fo        = Frequency of observed valuesFe        = frequency of expected valuesK        = number of categoriesC        = number of parameters being estimated from the sample dataExample:Uniform TestThe table below shows the results when a die is rolled 120 timesScore123456Frequency152914182024Conduct 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[pic 3]Step 3:        Let α= 0.05Step 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= 6So, df= k-1-c = 6-1-0= 5The critical value is  = 11.071[pic 4]Step 5: Find out the expected frequency (fe)Score Expected Frequency (fe)1120/6 = 202120/6 = 203120/6 = 204120/6 = 205120/6 = 206120/6 = 20 n= 120Step 6:ScoreFofe[pic 5]115201.25229204.05314201.8418200.2520200624200.8Total1201208.1Step 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.