Sample Midterm with solutions -3775-CORRIGE

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3775

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Mathematics

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Jan 9, 2024

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MAT 3375 Midterm test solutions Octobre 22 , 2019 Professor M. Alvo Time: 80 minutes Student number : Given name : Family name : This is an open book exam. Calculators are permitted. Answer all questions. The test is out of 20. 1. (10 points) The Tri-City O ffi ce Equipment Corporation sells an im- ported copier on a franchise basis and performs preventive maintenance and repair service on this copier. The data below have been collected from 45 recent calls on users to perform routine preventive maintenance service; for each call, X is the number of copiers serviced and Y is the total number of minutes spent by the service person. Assume a first-order regression model is appropriate. Y i = β 0 + β 1 X i + " i a. Complete the analysis of variance table below. Solution: The SS add to the total as do the df. MS=SS/df. b. Compute the test statistic to test the null hypothesis that β 1 = 0 against the alternative that β 1 6 = 0. What is its distribution? 1 <un jeu de données avait été fourni> cette solution continue dans la table à la page suivante How would you carry out the test with significance level alpha? En bleu, des commentaires de ma part. En rouge, des ajouts aux solutions.
F = MS ( R ) /MSE = 325 . 97 / 0 . 3365 = 968 . 7 > F (1 , 43) . Here F (1 , 43) is the upper 100(1 - )% point of an F distribution with 1 and 43 degrees of freedom. c. Interpret β 0 in your estimated regression function. β 0 is the intercept in the regression model. It indicated the value of Y when X=0 and is a start up value. d. Obtain a point estimate of the mean service time when X = 5 copiers are serviced. ˆ Y = b 0 + 5 b 1 e. What are the usual assumptions made in linear model? We assume Y i = β 0 + β 1 X i + " i where the { " i } are i.i.d. normal (0 , σ 2 ) error terms. Analysis of Variance Table Source SS df MS Time 325.97 1 325.97 Error 14.47 43 0.3365 Total 340.44 44 2. (5 points) The plot below refers to the rate of crime against Number of the population having at least a high school education for several counties in the US. Carry out a test on whether or not Education is important as a predictor by using the extra sum of squares principle. Make the usual as- sumptions for the linear regression model. Use the output data from Rstudio below. Model SSE df Rate˜0 522098 84 Rate˜Education 2665 82 a. State the full and reduced models. Solution: Full model: Y i = β 0 + β 1 X i + " i Reduced model: Y i = β 0 + " i b. Obtain the test statistic F* for the general linear test and state its distribution. F = SSE ( R ) - SSE ( F ) df R - df F SSE ( F ) df F 2 solution (a) continue ici……. F 968.7 <une visualization avait été fournie> Reduced. Full SS(“Reduced”) .. c’est la norme carrée du vecteur (yi) - \bar y (1n) qui montre la carence du modèle “réduit”, c’est à dire le modèle sans \beta_1 “Full”.. c’est ce qu’on a appelé résidu, c’est à dire la norme carrée du vecteur (yi) - (\hat yi), i.e. l’erreur du grand modèle SS(Reduced) - SS(Full) = ce qu’on a appelé SS(Regr). Faire le test avec niveau de confiance alpha veux dire rejeter l’hypothèse si la condition soulignée est vraie. Text y_i = epsilon_i (on reconnait ça d’après le titre “Rate~0” et aussi le fait qu’il y a 2 df, c’est à dire 2 paramètres à estimer, pour passer au Full model.)
= 522098 - 2665 84 - 82 2665 82 = 7990 . 8 F has an F distribution with 2 and 82 degrees of freedom 3. (5 points)The summary data as well as the ANOVA table that follow show assessed value for property tax purposes (Y in thousand dollars) and sales price (X, in thousand dollars) for a sample of 15 parcels of land for industrial development sold recently in arm’s length transactions in a tax district. The following R commands was used > fit= lm(formula = Sales ˜ Assessed) > summary(fit) a. Obtain the regression equation. What is the estimate of the variance? Solution: ˆ Y = - 5 . 812 + 2 . 542 X b. Construct 95% confidence intervals for the coe ffi cients. Express your answers in terms of the student coe ffi cients and specify the appropriate de- grees of freedom. Coe ffi cients Estimate S.E. t-value Intercept -5.812 3.065 -1.896 Assessed 2.542 0.2245 11.3222 ANOVA Source SS df MS F Assessed 401.81 1 401.81 128.18 Residuals 40.75 13 3.13 CI for β 0 : - 5 . 812 ± t / 2 (3 . 065) CI for β 1 : 2 . 542 ± t / 2 (0 . 2245) where is the t / 2 is the upper 100(1 - / 2)% point of a Student-t distri- bution with 13 degrees of freedom. 3
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