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These are the important key points of solved assignment of Applied Regression Analysis are: Regression Line, Mean Response, Scatter Plot, Computer Output, Complicated Model, Confidence Interval, Linear Relationship, Admission Might be Interested, Population Standard Deviation, Random Variable

Typology: Exercises

2012/2013

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Download Regression Line - Applied Regression Analysis - Solved Assignment and more Exercises Mathematical Statistics in PDF only on Docsity! Question #1: (a) The least-square estimate of the regression line when Y regressed on X is: Ŷ = −1.69956 + 0.83991X The change in the mean response when X increases by one unit is just β1. Therefore, the estimate of the change in the mean response when X increases by one point is β̂1 = 0.83991. (b) The scatter plot is given on the computer output on pages 2. The estimated regression line seems ok, but other complicated model might fit better. (c) Note that 1− α = 0.99→ α 2 = 0.005 , df = n− 2 = 20− 2 = 18→ tα 2 = 2.8784 A 99% confidence interval for β1 is: β̂1 ± t0.005 se(β̂1) = 0.83991± 2.8784 (0.14405) or (0.42528, 1.25454). This confidence interval is also given in the computer output on page 1. Therefore, we are 99% sure that the unknown β1 is between 0.42528 to 1.25454. The confidence interval does not include zero, therefore, we reject H0 : β1 = 0 versus H1 : β1 6= 0 at α = 1 − 0.99 = 0.01. The director of admission might be interested in β = 0, since it implies that there is no linear relationship between X and Y . By rejecting H0, we conclude that there is a linear relationship between X and Y . Therefore, X is useful to predict Y . (Note that a more complicated model might predict Y better. We should check this part in our research.) (d) If x = 5.0, we have Ŷ = −1.69956 + 0.83991(5) = 2.49999 (e) S is an unbiased estimate of σ. Then S = σ̂ = √ MSRes = √ 0.18924 = .4350 Note that σ is population standard deviation of random variable Y and it is expressed in the same unit that Y is expressed. Since S is an estimate of Y , it has the same unit as σ and Y . 1 Docsity.com (f) If x0 = 4.7, we have Ŷ = −1.69956 + 0.83991(4.7) = 2.248017 In addition, x̄ = 5 and 1− α = 0.05→ α 2 = 0.025 , df = n− 2 = 20− 2 = 18→ tα 2 = 2.101 Therefore, a 95% confidence interval for µY when x0 = 4.7 is 2.248017± 2.101 √ .18924 [ 1 20 + (4.7− 5)2 9.12 ] = 2.248017± 0.223631 or (2.024386, 2.471648) (g) A 95% confidence interval for Y when x0 = 4.7 is 2.248017± 2.101 √ .18924 [ 1 + 1 20 + (4.7− 5)2 9.12 ] = 2.248017± 0.94093 or (1.3071, 3.1889) (h) Yes. The prediction interval in part (f) is wider than the confidence interval in part (g). Note that the prediction interval in part (h) is an interval for a single future observation while the confidence interval in part (f) is an interval for average of Y . Since the last one has less variation, it has shorter confidence interval too. (i) The test statistic for testing H0 : β1 = 0 is TS = β̂1 − 0 se(β̂1) = 0.83991− 0 0.14405 = 5.83 The p-value, for H1 : β1 6= 0, is P − value = 2P (TS > 5.83) < 0.0001 Since P − value < α = 0.05, we reject H0 : β1 = 0. (j) The test statistic for testing H0 : β0 = 0 is TS = β̂0 − 0 se(β̂0) = −1.69956− 0 0.72678 = −2.34 The p-value, for H1 : β0 6= 0, is P − value = 2P (TS > −2.34) = 0.0311 2 Docsity.com (c) The test statistic for testing H0 : µM = µF or H0 : µM − µF = 0 is TS = (ȲM − ȲF )− 0 SP √ 1 nM + 1nF where SP = √ (nM − 1)S2M + (nF − 1)S2F nM + nF − 2 = √ (97− 1)(6.46323) + (47− 1)(1.843256) 47 + 97− 2 = 2.2285 Therefore, the test statistics is: TS = (11.322680− 9.202128)− 0 2.2285 √ 1 97 + 1 47 = 2.120552 0.39607 = 5.35 In addition, the test statistic for testing H0 : β1 = 0 is TS = β̂1 − 0 se(β̂1) = 2.12055− 0 0.39607 = 5.35 The value of two test are exactly the same. Recall that our model implies that µF = β0 and µM = β0 + β1 Therefore, β1 = 0 implies that µF = µM . So, H0 : β1 = 0 is equivalent to H0 : µM = µF . The two test statistics are exactly the same since both null hypothesis are the same. This assignment show that the usual t-test for comparing two means that we learned in an elementary statistical course (such as STAT 151) is a special case of regression model. In general, experimental design is special case of regression analysis when all independent variables are indicator variables. We will talk about indicator variable such as Z in future. Question #4: Recall that β̂1 = n∑ i=1 ciYi where n∑ i=1 ci = 0. Then Cov ( Ȳ , β̂1 ) = Cov ( Ȳ , n∑ i=1 ciYi ) = n∑ i=1 ci Cov ( Ȳ , Yi ) 5 Docsity.com = n∑ i=1 ci Cov 1 n n∑ j=1 Yj , Yi = 1 n n∑ i=1 n∑ j=1 ci Cov (Yj , Yi) Since E uncorrelated, we have Cov (Yj , Yi) = 0, for all i 6= j. Therefore Cov ( Ȳ , β̂1 ) = 1 n n∑ i=1 ciCov (Yi, Yi) = 1 n n∑ i=1 ciVar (Yi) = 1 n n∑ i=1 ci σ 2 = 1 n σ2 n∑ i=1 ci = 0, since ∑ ci = 0. In addition, Cov ( β̂0, β̂1 ) = Cov ( Ȳ − β̂1X̄ , β̂1 ) = Cov ( Ȳ β̂1 ) − Cov ( β̂1X̄ , β̂1 ) = 0− X̄ Cov ( β̂1 , β̂1 ) = −X̄ Var ( β̂1 ) . We proved, in the class, that Var ( β̂1 ) = σ2 Sxx . Therefore Cov ( β̂0, β̂1 ) = −X̄ σ 2 Sxx . 6 Docsity.com