Direction: This test is worth 250 points and each problem worth points. DO ANY SIX

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1 Term Test 3 December 5, 2003 Name Math 52 Student Number Direction: This test is worth 250 points and each problem worth 4 points DO ANY SIX PROBLEMS You are required to complete this test within 50 minutes In order to receive full credit, answer each problem completely and must show all work Please make sure that you have all the 9 pages GOOD LUCK! Let X, X 2,, X 2 be a random sample from a normal population with mean zero and variance What is the form of the LRT critical region for testing the null hypothesis H o : 0 versus H a : 5? Answer: Here o 0 and a 5 By the above definition, the form of the critical region is given by with σ o 2 0 and σ a 2 5 x, x 2,, x 2 R 2 L σ o 2, x, x 2,, x 2 C L σ a2, x, x 2,, x 2 k x, x 2,, x 2 R 2 2 e 2 x i σo 2 2πσ 2 o k e 2 x i σa 2 2π a } x, x 2,, x 2 R 2 e 2 20 x2 i k 2 } 2 x, x 2,, x 2 R 2 x 2 i a, where a is some constant Hence the likelihood ratio test is of the form: Reject H o if } 2 X 2 i a 2 Let X, X 2, X 3 denote a random sample of size 3 from a population X with probability density function θ for 0 x θ fx; θ 0 otherwise, where 0 < θ < is a parameter What is the likelihood ratio critical region of size 7 25 for testing H o : θ 5 versus H a : θ 5?

2 Answer: The ML estimator of θ is θ X 3 The likelihood function of the sample is The statistics W x, x 2, x 3 is given by Lθ θ 3 W x, x 2, x 3 Lθ o, x, x 2, x 3 max Lθ, x, x 2, x 3 3 X3 25 Therefore the critical region is C x, x 2, x 3 R 3 X 3 a }, where a is some constant Since α 7 25, we get 7 25 α P X 3 a θ 5 a a3 25 Therefore a 3 7 and the critical region is C 0 3 x2 25 dx x, x 2, x 3 R 3 X } 3 Suppose that X is a random variable about which the hypothesis H o : X UNIF 0, against H a : X N0, is to be tested What is the most powerful test with a significance level α 005 based on one observation of X? Answer: By Neyman-Pearson Theorem, the form of the critical region is given by C x R L } o x L a x k x R } 2π e 2 x2 k } k x R x 2 2 ln 2π x R x a, } where a is some constant Hence the most powerful or best test is of the form: Reject H o if X a Since, the significance level of the test is given to be α 005, the constant a can be determined Now we proceed to find a Since 005 α P Reject H o / H o is true} P X a / X UNIF 0, a 0 dx a, 2

3 hence a 005 Thus, the most powerful critical region is given by C x R 0 < x 005} based on the support of the uniform distribution on the open interval 0, Since the support of this uniform distribution is the interval 0,, the acceptance region or the complement of C in 0, is C c x R 005 < x < } However, since the support of the standard normal distribution is R, the actual critical region should be the complement of C c in R Therefore, the critical region of this hypothesis test is the set x R x 005 or x } The most powerful test for α 005 is: Reject H o if X 005 or X 4 Let X, X 2,, X be a random sample from a distribution with density function θ x θ for 0 < x < fx 0 otherwise, where θ > 0 The null hypothesis H o : θ 2 is to be rejected in favor of the alternative H a : θ > 2 if and only if at least 5 of the sample observations are larger than 08 What is the significance level of the test? Answer: The probability of a sample observation, X, greater than 08 is given by P X > θ x θ dx 08 θ If the null hypothesis, H o : θ 2, is true, then the probability of a sample observation, X, greater than 08 is P X > The significance level of the hypothesis test is given by α P reject H o, / H o is true

4 5 Observations y, y 2,, y n are assumed to come from a model with EY/x i α + βx i, where α and β are unknown parameters, and x, x 2,, x n are given constants What are the least squares estimate of the parameters α and β in terms of S x x and S xy? Answer: The sum of the squares of the errors is given by Eα, β ɛ 2 i y i α βx i 2 Differentiating Eα, β with respect to α and β respectively, we get and n α Eα, β 2 y i α βx i n β Eα, β 2 y i α βx i x i 2 Setting these partial derivatives αeα, β and β Eα, β to 0, we get y i α βx i 0 3 and From 3, we obtain which is y i α βx i x i 0 4 y i nα + β x i y α + β x 5 Similarly, from 4, we have which can be rewritten as follows x i y i α x i + β x 2 i x i xy i y + nx y n α x + β x i xx i x + nβ x 2 Defining we see that reduces to S xy : x i xy i y S xy + nx y α n x + β [ + nx 2] 7 4

5 Substituting 5 into 7, we have S xy + nx y [y β x] n x + β [ + nx 2] Simplifying the last equation, we get S xy β which is In view of 8 and 5, we get β S xy α y S xy x Thus the least squares estimates of α and β are respectively α y S xy x and β S xy, Define or precisely explain the followings: i Statistical hypothesis A hypothesis is a conjecture about the parameter of the population ii Power function of a hypothesis test The power or power function of a test is defined as P Type I Error if H o is true πθ P Type II Error if H a is true iii Simple hypothesis A hypothesis is said to be a simple hypothesis if it completely specifies the density fx; θ of the population iv Type I error of a hypothesis test The type I error is an error of rejecting a true null hypothesis 5

6 v Type II error of a hypothesis test The type II error is an error of accepting a false null hypothesis vi Significance level of a hypothesis test The significance level of a hypothesis test is the probability of rejecting a true null hypothesis vii Normal regression analysis A regression analysis is called a normal regression analysis if the conditional density of Y i given X i x i is of the form fy i /x i 2πσ 2 e 2 yi α βx i 2 σ, where denotes the variance, and α and β are the regression coefficients viii Composite hypothesis A hypothesis is said to be a composite hypothesis if it does not completely specifies the density fx; θ of the population ix Neyman-Pearson Theorem Let X, X 2,, X n be a random sample from a population with probability density function fx; θ Let Lθ, x,, x n n fx i ; θ be the likelihood function of the sample Then any critical region C of the form C x, x 2,, x n } L θ o, x,, x n L θ a, x,, x n k for some constant 0 k < is best or uniformly most powerful of its size for testing H o : θ θ o against H a : θ θ a 7 Show that in normal regression analysis, the distributions of the estimators β and α are given by β N β, and α N α, n + x2,

7 where x i x 2 Answer: Since β S xy x i x Y i Y xi x Y i, the β is a linear combination of Y i s As Y i N α + βx i,, we see that β is also a normal random variable By Theorem 92, β is an unbiased estimator of β The variance of β is given by V ar β 2 xi x V ar Y i /x i 2 xi x S 2 xx σ2 x i x 2 Hence β is a normal random variable with mean or expected value β and variance σ2 That is σ β N β, 2 Now determine the distribution of α Since each Y i Nα + βx i,, the distribution of Y is given by Since the distribution of x β is given by Y N β N α + βx, σ2 n β, x β N x β, x 2 σ2 Since α Y x β and Y and x β being two normal random variables, α is also a normal random variable with mean equal to α + β x β x α and variance variance equal to σ2 α N α, n + x2 n + x2 That is 7

Hypothesis Testing Chap 10p460

Hypothesis Testing Chap 10p460 Hypothesis Testing Chap 1p46 Elements of a statistical test p462 - Null hypothesis - Alternative hypothesis - Test Statistic - Rejection region Rejection Region p462 The rejection region (RR) specifies