Summary of Chapter 8 (Sections 8.3, )
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1 Summary of Chapter 8 (Sections 8.3, ) 8.3. Tests of the Equality of Two Normal Distributions (1) Hypothesis Test for Equality of Two Means Let X N(µ X, σx 2 ) and Y N(µ Y, σy 2 ). Assume that X and Y are independent. We have two samples: {X 1, X 2,..., X n } and {Y 1, Y 2,..., Y m }. Null hypothesis H 0 : µ X µ Y = 0. Test statistics: where S P = T = X Ȳ S P 1/n + 1/m, (n 1)SX 2 + (m 1)S2 Y. n + m 2 T has a t distribution with r = n + m 2 degrees of freedom. Table : Tests of Hypotheses for Equality of Two Means where t = H 0 H 1 Critical Region µ X = µ Y µ X > µ Y t t α (n + m 2) µ X = µ Y µ X < µ Y t t α (n + m 2) µ X = µ Y µ X µ Y t t α/2 (n + m 2) x ȳ s P 1/n+1/m is the observed value of the test statistics T. p-value can be computed by using formulas mentioned before. (2) Hypothesis Test for the Equality of Two Variances Null hypothesis H 0 : σ 2 X = σ2 Y. Test statistics: F = S2 X S 2 Y F (n 1, m 1). 1
2 Table : Tests of Hypotheses for Equality of Variances H 0 H 1 Critical Region σx 2 = σ2 Y σx 2 > σ2 Y s 2 x/s 2 y F α (n 1, m 1) σx 2 = σ2 Y σx 2 < σ2 Y s 2 y/s 2 x F α (m 1, n 1) σx 2 = σ2 Y σx 2 σ2 Y s 2 x/s 2 y F α/2 (n 1, m 1) or s 2 y/s 2 x F α/2 (m 1, n 1) 8.5. Chi-Square Goodness of Fit Tests Let an experiment have k mutually exclusive and exhaustive outcomes A 1,, A k. Denote p i = P (A i ), i = 1,, k. We would like to test the hypothesis H 0 : p i = p i0, i = 1,, k against all other alternative hypotheses H 1. Case 1: discrete distributions. Let the experiment be repeated n independent times. Let Y i be the observed number of times (frequency) that A i occurred. Then the expected frequency is np i0 (which should be at least 5). When H 0 is true, the test statistic is Q k 1 = k (Y i np i0 ) 2 χ 2 (k 1). np i0 The critical region is q k 1 χ 2 α(k 1), where α is the significance level. If there are d unknown parameters in the hypothesized distribution that need to be estimated from the given sample data, then we must calculate p i0 by using the estimates of the parameters. The test statistic is Q k 1 χ 2 (k 1 d) and the critical region would be q k 1 χ 2 α(k 1 d). Case 2: Continuous distributions. Let W be a continuous random variable with distribution function F (w). We would like to test H 0 : F (w) = F 0 (w) against all other alternatives H 1, where F 0 (w) is a known continuous distribution function. We partition the space of W into k class intervals: A 1 = (, a 1 ], A 2 = (a 1, a 2 ],, A k = (a k 1, ). Let p i = P (W A i ). Let Y i be the number of times that the observed values of W belong to A i, i = 1,, k in n independent repetitions of the experiment. Then, Y 1,, Y k have a multinomial distribution with parameters n, p 1,, p k 1. Let p i0 = P (W A i ) when the distribution function of W is F 0 (w). Then, H 0 is modified to be H 0 : p i = p i0, i = 1,, k. H 0 is rejected if k (y i np i0 ) 2 q k 1 = χ 2 α(k 1 d), np i0 where d is the number of unknown parameters in F 0 (w). 2
3 8.6. Contingency Tables (1) Test of Homogeneity (a) Testing the equality of h independent multinomial distributions. Let A 1,, A k be k mutually exclusive and exhaustive events of the h independent experiments. Let p ij = P (A i ), i = 1,, k, j = 1,, h. Test H 0 : p i1 = p i2 = = p ih = p i, i = 1,, k. We repeat the jth experiment n j independent times and let Y 1j, Y 2j,, Y kj be the frequencies of the respective events A 1,, A k. The test statistic is h k (Y ij n j p ij ) 2 χ 2 (h(k 1)). n j p ij Under H 0, We must estimate h ˆp i = Y ij h n, j = 1,, k 1, j and ˆp k = 1 ˆp 1 ˆp k 1. So, under H 0, we have h k (Y ij n j ˆp i ) 2 χ 2 ((h 1)(k 1)). n j ˆp i The critical region is q χ 2 α((h 1)(k 1)). (b) Testing the equality of two or more independent distributions which are not necessarily multinomial. We consider two random variables U and V with continuous distribution functions F (u) and G(v), respectively. We are interested in testing H 0 : F (x) = G(x) for each x. We partition the real line (, ) into k mutually exclusive and exhaustive sets A 1,, A k. Let p i1 = P (U A i ) and p i2 = P (V A i ), i = 1,, k. We replace H 0 : F (x) = G(x) with H 0 : p i1 = p i2, i = 1,, k. This is now equivalent to testing the equality of two multinomial distributions. Let n 1 and n 2 be the number of observations of U and V, respectively. For i = 1,, k, Let Y ij be the number of these observations of U and V, respectively, that fall into a set A i. Then, we can proceed to make the test of H 0 as discussed before. If H 0 is rejected at the significance level α, then H 0 will be rejected at the same significance level α. If we don not reject H 0, then we also do not reject H 0. 3
4 (2) Test of Independence Suppose that a random experiment results in an outcome that can be classified by two different attributes. We are interested in testing the independence of the two attributes. The first attribute falls in one and only one of k mutually exclusive and exhaustive events A 1,, A k, and the second attribute falls in one and only one of k mutually exclusive and exhaustive events B 1,, B h. Let p ij = P (A i B j ), i = 1,, k, j = 1,, h. We repeat the experiment n independent times, and denote Y ij the frequency of A j B j. Test H 0 : P (A i B j ) = P (A i )P (B j ), i = 1,, k, j = 1,, h. Notation: p i = P (A i ) = h p ij and p j = P (B j ) = k p ij. We can reformulate the null hypothesis H 0 : p ij = p i p j, i = 1,, k, j = 1,, h. We need to estimate p i and p j by using observed frequencies: and h ˆp i = y i /n, where y i = y ij, k ˆp j = y j /n, where y j = y ij. The test statistic is h k H 0 is rejected if q χ 2 α((h 1)(k 1)). [Y ij n(y i /n)(y j /n)] 2 n(y i /n)(y j /n) χ 2 ((h 1)(k 1)) One-Factor Analysis of Variance Testing the equality of the means of various distributions. Let X 1,, X m be iid normal random variables with distributions N(µ 1, σ 2 ),, N(µ m, σ 2 ), respectively. Test H 0 : µ 1 = µ 2 = = µ m against all possible alternative hypotheses H 1. Let X i1, X i2,, X ini be a random sample of size n i from the distribution N(µ i, σ 2 ) of X i, i = 1, 2,, m. Let and n i X i = X ij /n i, i = 1,, m (the ith sample mean) where n = n 1 + n n m. X = 1 n m n i X ij (grand mean), 4
5 Total sum of square decomposition: SS(T O) = SS(E) + SS(T ), where SS(T O) = m ni (X ij X ) 2 is called the total sum of squares, SS(E) = m ni (X ij X i ) 2 is called the error sum of squares, and SS(T ) = m n i( X i X ) 2 is called the between treatment sum of squares. Then, we have SS(T O)/σ 2 χ 2 (n 1), SS(E)/σ 2 χ 2 (n m), SS(T )/σ 2 χ 2 (m 1). Test statistic: F = SS(T )/(m 1) F (m 1, n m) SS(E)/(n m) when H 0 is true. H 0 is rejected if F F α (m 1, n m). The ANOVA table is given as follows: Source SS DF MS F Treatment SS(T ) m 1 MS(T ) = SS(T ) m 1 Error SS(E) n m MS(E) = SS(E) n m Total SS(T O) n 1 MS(T ) MS(E) 5
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