Section 8.3 - Section 8.3 Test for a Categorical Population...

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Section 8.3 Test for a Categorical Population Assume each unit in the population can be placed in to one of k non-overlapping categories. For example, we may want to study a characteristic of the people belonging to different age groups say 15-30, 40-50 and 50-70. Sometimes a unit may be classified with respect to 2 variables 1 x and 2 x are independent. To test a hypothesis here, we use a new probability distribution known as 2 χ - distribution . Definition. The random variables X i are k independent, normally distributed random variables with mean 0 and variance 1, then the random variable is distributed according to the chi-square distribution. This is usually written The chi-square distribution has one parameter: k - a positive integer that specifies the number of degrees of freedom (i.e. the number of X i ) 1
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The following graphs are some of the of the 2 χ - densities. Probability density function A probability density of the chi-square distribution is where Γ denotes the Gamma function which has closed-form values at the half-integers. 2
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Test for Univariate data : Assume each unit in the population can be classified into one of k non-overlapping categories. That is the characteristic X is a discrete variable and } ,..., 2 , 1 { k X with ) ( i X P i = = π . 1 = proportion of units belonging to 1 st category; . . . k = proportion of units belonging to k th category. Also, i = probability that a randomly chosen individual falls into i-th category. Then, 1 ... 1 = + + k . Consider 0 10 1 0 ,..., : k k H = = (specified) against 0 : H H a is not true. Take a sample of size n from the population and let 1 n = no. of individuals belonging to 1 st category . . . k n = no. of individuals belonging to k th category. so that = k i n n 1 . 3
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The value k n n ,..., 1 are called observed counts. The basic idea is to compare observed counts with expected (under 0 H ) counts 0 10 ,..., k n n π . How do we do that? A simple way is - k i i n n 1 2 0 ) ( But, for example, if 1 n = 95, 100 10 = n 2 n = 15, 20 20 = n Then both have same difference = 5. But the first frequency is less than 5%, while 0 H second one is 25% less than their expected frequencies. The 2 χ - test takes into account the percentage derivatives. 2 - test for Univariate Data Hypothesis 0 10 0 ,..., : k k H = = ( 0 i are specified) against 1 H : 0 H is not true. Test Statistic: = - = k i i i i n n n X 1 0 2 0 2 ) ( 4
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, ) ( 2 - = k i i i i E E O where i O = observed frequency, i E = expected frequency of the i th category. When 0 H is true, 2 X - follow chi-square distribution with (k-1) df, when all 5 0 i n π for all i. Also, P – value = ) ( 2 2 ) 1 ( x X P k - (Note 2 x is the observed value) Note : If for some 5 0 < i n , categorizes should be combined so that the conditions are satisfied. Example
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Section 8.3 - Section 8.3 Test for a Categorical Population...

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