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Class 7 -chapter15

# Class 7 -chapter15 - Chapter 15 Goodness of Fit...

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Chapter 15 The procedures for this chapter require categorical or qualitative data. If the data are quantitative then they can be converted to categorical data. The Goodness of Fit Test basically tests: Consider the distribution of test scores for an example Range 0-59 60-69 70-79 80-89 90-100 Category F D C B A Hypothesized Probability 0.1 0.15 0.3 0.25 0.2 H 0 : The data come from a specified distribution H A : The data do not come from the specified distribution H 0 : p F = .1, p D = .15, p C = .3, p B = .25, p A = .2 H A : The data do not come from the specified distribution

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Consider the distribution of test scores for an example Range 0-59 60-69 70-79 80-89 90-100 Category F D C B A Hypothesized Probability 0.1 0.15 0.3 0.25 0.2 Expected Number 10 15 30 25 20 Actual Observed Number 15 10 27 28 20 If the observed is close to the Expected then the Null is probably true. If the observed is NOT close to the Expected then the Null is probably NOT true. k = Number of Categories The probability distribution for testing is the Chi-square with degrees of freedom = # Categories - # Parameters estimated from the data - 1 If no paramenters are estimated from the data then degrees of freedom = k-1 This is a 1-tail upper-test, since the test statistic becomes larger as the Expected and Observ Observe n=100 observations The Test Statistic measures the amount that the observed data depart from what is expected based on the null hypothesis Expected Nu (Total number) * Category) = n * (hypoth The probabili determined b ( 29 = - = k 1 j 2 Expected Expected Observed TS

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rved values get farther apart umber in a category = * (Probability of being in the thesized proportion) lity of being in a category is by the null hypothesis.
Consider the distribution of test scores for an example Range 0-59 60-69 70-79 80-89 90-100 Category F D C B A 0.1 0.15 0.3 0.25 0.2 10 15 30 25 20 Actual Observed Number 15 10 27 28 20 Cell values for the TS 2.5 1.67 0.3 0.36 0 Test Statistic = 4.83 9.49 p-value = 0.31 < Reject or Accept Null based on this p-value Use the Chi-square with degrees of freedom = # Categories - # Parameters estimated from the data - 1 Degrees of Freedom = 4 k = 5 = Number of Categories ### = 2010 p-value Reject the Null Fail to Reject the Null 4.83 ### Observe n=100 observations Hypothesized Probability,

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Class 7 -chapter15 - Chapter 15 Goodness of Fit...

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