t09sol - STA 302 H1F / 1001 HF Fall 2009 Test October 22,...

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Unformatted text preview: STA 302 H1F / 1001 HF Fall 2009 Test October 22, 2009 LAST NAME: SOLUTIONS FIRST NAME: STUDENT NUMBER: ENROLLED IN: (circle one) STA 302 STA 1001 INSTRUCTIONS: Time: 90 minutes Aids allowed: calculator. A table of values from the t distribution is on the last page (page 10). Total points: 50 Some formulae: b 1 = ( X i- X )( Y i- Y ) ( X i- X ) 2 = X i Y i- n X Y X 2 i- n X 2 b = Y- b 1 X Var( b 1 ) = 2 ( X i- X ) 2 Var( b ) = 2 1 n + X 2 ( X i- X ) 2 Cov( b ,b 1 ) =- 2 X ( X i- X ) 2 SSTO = ( Y i- Y ) 2 SSE = ( Y i- Y i ) 2 SSR = b 2 1 ( X i- X ) 2 = ( Y i- Y ) 2 2 { Y h } = Var( Y h ) = 2 1 n + ( X h- X ) 2 ( X i- X ) 2 2 { pred } = Var( Y h- Y h ) = 2 1 + 1 n + ( X h- X ) 2 ( X i- X ) 2 r = ( X i- X )( Y i- Y ) p ( X i- X ) 2 ( Y i- Y ) 2 S XX = ( X i- X ) 2 = X 2 i- n X 2 1abc 1de 2a 2bcd 2efgh 3ab 3c 3de 1 1. The method of least squares is used to fit a simple linear regression model Y = + 1 X + to n observations ( X i ,Y i ). The values of the X i s are not realizations of random variables but are fixed in advance by the researcher. Assume the following: the form of the model is appropriate, the Gauss-Markov conditions hold, and the distribution of the error terms is Normal. In your answers, you may use any of the formulae on the front page. (a) (4 marks) Which of the assumptions are needed to fit the model using least squares? How would you assess the necessary assumptions? For least squares, the only assumption needed is the form of the model. To check this look at plots of Y versus X and the residuals versus either X or the predicted values. Look for outliers / influential points and curvature. (b) (3 marks) What is E( Y )? Which of the assumptions did you use to determine your answer? E ( Y ) = + 1 X Assumptions used: form of model and E ( ) = 0 (c) (2 marks) Suppose the researcher is interested in the relationship between X and Y on a certain range of X s. She uses the smallest value in the range of X for half of the observations and the largest value in the range of X for the other half of the observations and fits the simple linear regression model to the resulting data. What is the advantage of fixing the X s to be these values? What is the disadvantage? Advantage: This choice of X s will make S XX as large as possible, giving more pre- cise estimates (smaller variance) of the model parameters. Disadvantage: Wont be able to determine if the form of the relationship really is linear without observations on more values of X ....
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t09sol - STA 302 H1F / 1001 HF Fall 2009 Test October 22,...

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