Summary of Chapter 7 (Sections ) and Chapter 8 (Section 8.1)

Transcription

1 Summary of Chapter 7 (Sections ) and Chapter 8 (Section 8.1) Chapter 7. Tests of Statistical Hypotheses 7.2. Tests about One Mean (1) Test about One Mean Case 1: σ is known. Assume that X N(µ, σ 2 ), X 1,..., X n is a random sample from the distribution N(µ, σ 2 ). Null hypothesis H 0 : µ = µ 0. Z = X µ 0 σ/ n. Table : Tests of Hypotheses about One Mean, Variance Known where z = x µ0 σ/ n µ = µ 0 µ > µ 0 z z α µ = µ 0 µ < µ 0 z z α µ = µ 0 µ µ 0 z z α/2 is the observed value of the test statistics Z. Let z be the observed value of the test statistics Z. We can compute the p-value via the formula: 2P (Z z ), if H 1 : µ µ 0 ; p value = 1 P (Z < z), if H 1 : µ > µ 0 ; P (Z z), if H 1 : µ < µ 0. Case 2: σ is unknown. T = X µ 0 S/ n. T has a t distribution with r = n 1 degrees of freedom. 1

2 Table : Tests of Hypotheses about One Mean, Variance Unknown where t = x µ0 s/ n µ = µ 0 µ > µ 0 t t α (n 1) µ = µ 0 µ < µ 0 t t α (n 1) µ = µ 0 µ µ 0 t t α/2 (n 1) is the observed value of the test statistics T. Let t be the observed value of the test statistics T. We can compute the p-value via the formula: 2P (T t ), if H 1 : µ µ 0 ; p value = 1 P (T < t), if H 1 : µ > µ 0 ; P (T t), if H 1 : µ < µ 0. (2) Comparison of Two Means (Paired t-test) Assume that X and Y are dependent. Let W = X Y. From the original data {(x 1, y 1 ), (x 2, y 2 ),..., (x n, y n )}, one can get the difference data {w 1, w 2,..., w n }. Null hypothesis H 0 : µ X = µ Y is equivalent to H 0 : µ W = 0. T = W 0 S W / n. Test procedures are the same as previous case Tests of the 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. where S P = T = X Ȳ S P 1/n + 1/m, (n 1)SX 2 + (m 1)S2 Y. n + m 2 2

3 T has a t distribution with r = n + m 2 degrees of freedom. Table : Tests of Hypotheses for Equality of Two Means where t = 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 Tests for Variances (1) Test about the variance σ 2 Null hypothesis H 0 : σ 2 = σ 2 0. χ 2 = (n 1)S2, which has a chi-square distribution with r = n 1 degrees of σ0 2 freedom. Table : Tests of Hypotheses about the Variance σ 2 = σ0 2 σ 2 > σ0 2 χ 2 χ 2 α(n 1) σ 2 = σ0 2 σ 2 < σ0 2 χ 2 χ 2 1 α(n 1) σ 2 = σ0 2 σ 2 σ0 2 χ 2 χ 2 1 α/2 (n 1) or χ2 χ 2 α/2 (n 1) where χ 2 = (n 1)s2 σ 2 0 is the observed value of the test statistics χ 2. (2) Hypothesis Tests for the Equality of Two Variances Null hypothesis H 0 : σ 2 X = σ2 Y. F = S2 X S 2 Y F (n 1, m 1). Table : Tests of Hypotheses for Equality of Variances 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) 3

4 7.5. 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 n i X i = X ij /n i, i = 1,, m (the ith sample mean) j=1 and X = 1 n where n = n 1 + n n m. m n i i=1 j=1 X ij (grand mean), Total sum of square decomposition: SS(T O) = SS(E) + SS(T ), where SS(T O) = m ni i=1 j=1 (X ij X ) 2 is called the total sum of squares, SS(E) = m ni i=1 j=1 (X ij X i ) 2 is called the error sum of squares, and SS(T ) = m i=1 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) 4

5 Chapter 8. Non-parametric Methods 8.1. 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 i=1 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), i=1 np i0 where d is the number of unknown parameters in F 0 (w). 5