# Goodness Of Fit. For example, suppose there are four entrances to a building. You want to know if the four entrances are equally used. You observe 400.

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Goodness Of Fit

For example, suppose there are four entrances to a building. You want to know if the four entrances are equally used. You observe 400 people entering the building on a random basis: Goodness Of Fit If the entrances are equally utilized, we would expect each entrance to be used approximately 25% of the time. Is the difference shown above statistically significant? The purpose of a chi-square goodness-of-fit test is to compare an observed distribution to an expected distribution. H 0 : p M = p B = p S1 = p S2 H 1 : The proportions are not all equal.

If the observed frequencies are obtained from a random sample and each expected frequency is at least 5, the sampling distribution for the goodness- of-fit test is a chi-square distribution with k-1 degrees of freedom. (where k = the number of categories) O = observed frequency in each category E = expected frequency in each category Chi Square Test Test Statistic

Goodness-of-Fit Test: Equal Expected Frequencies Let f 0 and f e be the observed and expected frequencies, respectively. H 0 : There is no difference between the observed and expected frequencies. H 1 : There is a difference between the observed and the expected frequencies. H 0 : p 1 = p 2 = p 3 = p 4 H 1 : The proportions are not all equal.

df = 3 df = 5 df = 10  k-1 degrees of freedom. (where k = the number of categories) See Table P.495

EXAMPLE The following information shows the number of employees absent by day of the week at a large manufacturing plant. At the.05 level of significance, is there a difference in the absence rate by day of the week? Day Frequency Monday 120 Tuesday 45 Wednesday 60 Thursday 90 Friday 130 Total 445

EXAMPLE continued The expected frequency is: (120+45+60+90+130)/5=89. The degrees of freedom is (5-1)=4. The critical value is 9.488. (Appendix B, P.495)

Example continued Day Freq. Expec. (f o – f e ) 2 /f e Monday120 89 10.80 Tuesday 45 89 21.75 Wednesday 60 89 9.45 Thursday 90 89 0.01 Friday 130 89 18.89 Total 445445 60.90 Because the computed value of chi-square is greater than the critical value, H 0 is rejected. We conclude that there is a difference in the number of workers absent by day of the week.

A seller of baseball cards wants to know if the demand for the following 6 cards is the same. Example Goodness of Fit MegaStat

Goodness Of Fit (unequal frequencies)

The Bank of America (BoA) credit card department knows from national US government records that 5% of all US VISA card holders have no high school diploma, 15% have a high school diploma, 25% have some college, and 55% have a college degree. Given the information below, at the 1% level of significance can we conclude that (BoA) card holders are significantly different from the rest of the nation? Example - Goodness Of Fit (unequal frequencies) = (500)(.05) = (500)(.15) = (500)(.25) = (500)(.55)

Reject H 0 df = (4 - 1) = 3

Limitations of Chi-Square

1.) If there are only 2 cells, the expected frequency in each cell should be at least 5. Limitations of Chi-Square 2.) For more than 2 cells, chi-square should not be used if more than 20% of f e cells have expected frequencies less than 5.

MegaStat Two-thirds of the computed chi-square value is accounted for by just two categories (outcomes). Although the expected frequency is not less than 5, too much weight may be given to these categories. More experimental trials should be conducted to increase the number of observations.

Independence & Contingency Tables

Contingency Table Analysis A contingency table is used to investigate whether two traits or characteristics are related. Each observation is classified according to two criteria. The degrees of freedom is equal to: df = (# rows - 1)(# columns - 1). The expected frequency is computed as: Expected Frequency = (row total)(column total)/Grand Total

EXAMPLE Is there a relationship between the location of an accident and the gender of the person involved in the accident? A sample of 150 accidents reported to the police were classified by type and gender. At the.05 level of significance, can we conclude that gender and the location of the accident are related?

EXAMPLE continued The expected frequency for the work-male intersection is computed as (90)(80)/150=48. Similarly, you can compute the expected frequencies for the other cells. H 0 : Gender and location are not related. H 1 : Gender and location are related.

EXAMPLE continued H 0 is rejected if the computed value of χ 2 is greater than 5.991. There are (3- 1)(2-1) = 2 degrees of freedom. Find the value of χ 2. H 0 is rejected. We conclude that gender and location are related.

A crime agency wants to know if a male released from prison and returned to his hometown has an easier (or more difficult) time adjusting to civilian life. MegaStat Example Contingency Tables MegaStat

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