Testing on proportions


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1 Testing on proportions Textbook Section 5.4 April 7, 2011 Example 1. X 1,, X n Bernolli(p). Wish to test H 0 : p p 0 H 1 : p > p 0 (1) Consider a related problem The likelihood ratio test is where c is determined by H 0 : p = p 0 H 1 : p = p 1 (> p 0 ) n pxi 1 q1 xi 1 n pxi 0 q1 xi 0 P { c ( ) n p1 q xi 0 c q 0 p 0 n ( ) p1 q 0 x i log c q 1 p 0 n x i c, n X i c p = p 0 } = α. Note that the test is independent of p 1 and is a level α test for (1) (to be shown). Hence it is a UMP for (1). e.g. Let S = n X i, n = 900, p 0 = 0.5 and the observed statistic s obs = n x i = 500. Then, S Bin(n, p) N(np, npq). From { S np0 P c np } 0 p = p 0 = α, np0 q 0 np0 q 0 c np 0 np0q 0 z α or c = for α = Since s obs > 474.5, we reject H 0 and conclude that the candidate will win. This method does not provide the evidence against H 0. A more common way is to compute the pvalue, which is give by very strong evidence against H 0. max P {T (X) > T (x) p} = P (S > 500 p = 0.5) 0.04%, p 0.5 1
2 Example 2. (Approximate pivotal method) Bernoulli trails: X 1,, X n i.i.d.bernoulli(p). Approximate pivot: Z 1 = Z 2 = X p pq/n a N(0, 1), X p ˆpˆq/n a N(0, 1). Using Z 2, an approximate (1 α) CI is Using Z 1, P ( z α/2 X p ˆpˆq/n z α/2 ) 1 α [ˆp z α/2 ˆpˆq/n, ˆp + zα/2 ˆpˆq/n]. P ( ) ( ˆp p ) 2 zα/2 2 pq/n = 1 α, The function g(p) has two roots: g(p) = (ˆp p) 2 cpq 0, c = n 1 z 2 α/2. ˆp 1 = (ˆp + c/2 c 2 /4 + cˆpˆq)/(1 + c) ˆp 2 = (ˆp + c/2 + c 2 /4 + cˆpˆq)/(1 + c) where ˆq = 1 ˆp. Thus 1 α CI is [ˆp 1, ˆp 2 ]. As an illustration, assume that we wish to estimate the probability of a die facing 6. Roll a die 100 times and get Then, ˆp = Using the first approximation, the 95% confidence interval is 0.2 ± /100 = 0.2 ± Using the second method, the 95% confidence interval is ± Both intervals contain 1/6. Hence, the die is fair. 2
3 Goodness of Fit Textbook Section 5.7 April 6, 2010 One of well known real life examples occurred when the University of California at Berkeley was sued for bias against women who had applied for admission to graduate schools there. The admission figures for the fall of 1973 showed that men applying were more likely than women to be admitted, and the difference was so large that it was unlikely to be due to chance. Admitted Not admitted Applicants Men Women A goodness of fit test examine the case of a sequence if independent experiments each of which can have 1 of k possible outcomes. In terms of hypothesis testing, let π = (π 1,..., π k ) be postulated values of the probability P π {experiment takes on the ith outcome} = π i and let p = (p 1,..., p n ) denote the actual state of nature. Then, the parameter space is the n 1 simplex Θ = {p = (p 1,..., p n ); p i 0 for all i = 1,..., k, p i = 1}. The hypothesis test is H 0 : p i = π i, for all i = 1,..., k versus H 1 : p i π i, for some i = 1,..., k, The data x is the outcome of the n experiments. A sufficient statistic is n = (n 1,..., n k ) where n i is the number of time that outcome i occurs in n experiments. Thus, n = n i. The likelihood function L(p n) = p n1 1 pn k k. 1
4 Its logarithm ln L(p n) = n i ln p i. We maximize this using the method of Lagrange multipliers with constraint s(p) = p i = 1. Thus, at the maximum likelihood estimator (ˆp 1,..., ˆp k ), So, n i /ˆp i = λ, n i = λˆp i. Now sum on i to obtain Consequently, The likelihood ratio test p ln L(ˆp n) = λ ˆp s(p). ( n1,..., n ) k = λ(1,..., 1) ˆp 1 ˆp k n i = λ ˆp i and n = λ. n 1 ˆp i = n and ˆp i = n i n. Λ n (n) = L(n ˆp) ( ) n1 ( ) nk L(n π) = n1 nk. nπ 1 nπ k Recall that as the number of experiments n, 2 ln Λ n (N) = 2 ln N i = 2 ln nπ i. nπ i converges to a χ 2 k 1 random variable. Here N = (N 1,..., N k ) is the observed number of occurrences of outcome i. The traditional method was introduced between 1985 and 1900 by Karl Pearson and consequently has been in use for longer that the idea of likelihood ratio tests. To show the connection between the two tests, recall that ln a (a 1) 1 (a 1)2 2 is the quadratic Taylor polynomial approximation of ln a. Apply this to the logarithm of the likelihood ratio, we find that 2 ln Λ n (N) = 2 = 2 = 0 + ( (nπi ) 1 (nπ i ) + (nπ i ) ( ) ) 2 nπi 1 ( ) 2 nπi 1
5 The is generally rewritten by writing O i = to be the number of observed occurrences of i and E i = nπ i to be the number of expected occurrences of i as given by H 0. The data can be stored in a table Then, (nπ i ) 2 i 1 2 k observed O 1 O 2 O k expected E 1 E 2 E k (nπ i ) 2 nπ i (O i E i ) 2 = Qk 1. E i Example 1. (Textbook example 5.7.1) Roll a die, let A i = {x : x = i}, i = 1,, 6. The hypothesis H 0 : P (A i ) = π i = 1/6 H 1 : all the alternatives will be test at significance level 5%. n = 60, k = 6. Let X i denote the frequency with which the random experiment terminates with the outcome in A i. The data is as follows: Apply the above idea, Outcome A 1 A 2 A 3 A 4 A 5 A 6 Freq (O i 60 1/6) 2 Q 5 = 60 1/6 = (13 10) (4 10)2 10 = Hence the pvalue = P (Q 5 > 15.6 H 0 ) = < We conclude that we reject the null hypothesis. 1 Contingency tables For an r c contingency table, we consider two classifications for an experiment. Thus, we can partition the outcome of each experiment into two groups: A 1,... A c and B 1,... B r. Here, we write O ij to denote the number of occurences of the outcome A i B j are organize the results in a twoway table. A 1 A 2 A c total B 1 O 11 O 12 O 1c O 1 B 2 O 21 O 22 O 2c O B r O r1 O r2 O rc O r total O 1 O 2 O c n 3
6 The null hypothesis is that the classifications A and B are independent. To set the parameter space for this model, we have the rc 1 simplex Θ = {p = (p ij, 1 i r, 1 j c); p ij 0 for all i, j = 1, r c p ij=1 }. j=1 Write The hypothesis test is c r p i = p ij and p j = p ij. j=1 H 0 : p ij = p i p j, for all i, j versus H 1 : p ij p i p j, for some i, j. Follow the procedure as before for the goodness of fit test to end with the test statistic r c j=1 O ij ln E ij O ij r j=1 c (O ij E ij ) 2 is asymptotically distributed χ 2 ((r 1)(c 1)), where E ij = O i O j /n. For 2 2 contingency table, the above formula can be simplified as: Q 1 = n(o 11 O 22 O 21 O 12 ) 2 O 1 O 2 O 1 O 2. E ij = Q(r 1)(c 1) (1) Example 2. (Association test) In the cancer study, we would like to know whether smoke is associated with lung cancer. We collected the following data (hypothetical data). No cancer Cancer Total Nonsmoker Smoker Using the simplified formula above, we get Q 1 = 105 [ ] 2 /( ) = Or you may want to use equation (1), We first need to find E ij [ Then Q 1 = ( ) 2 / ( ) 2 / ( ) 2 / ( ) 2 /20.43 = Hence pvalue = P (χ 2 (1) > 3.42) = We conclude that we fail to reject the independence hypothesis at significance level 5%. Example 3. (Sex bias in Graduate admission at Berkeley) Back to the question at the beginning, is there really sex bias for graduate admission? You can now apply the test we just discuss and find that pvalue is quite small. So we reject the null hypothesis that the sex is independent of number of student admitted to graduate school. However when examining the individual departments, it was found that no department was significantly biased against women. In fact, most departments had a small but statistically significant bias in favor of women ]. 4
7 Department Men Women Applicants % admitted Applicants % admitted A % % B % 25 68% C % % D % % E % % F 272 6% 341 7% The research paper by Peter Bickel, et al (1975) concluded that women tended to apply to competitive departments with low rates of admission even among qualified applicants (such as in the English Department), whereas men tended to apply to lesscompetitive departments with high rates of admission among the qualified applicants (such as in engineering and chemistry). 5
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