Final-Practice-Solutions

STAT 3470-AU 23 Solutions to Practice Questions - Final December 6, 2023 Nasser Sadeghkhani Q.1 Regression methods were used to analyze the data from a study investigating the relationship between roadway surface temperature in F ( x ) and pavement defection ( y ). Summary quantities were n = 20, X y i = 12 . 75 , X y 2 i = 8 . 86 , X x i = 1478 , X x 2 i = 143 , 215 . 8 X x i y i = 1083 . 67 . (a) Calculate the least squares estimates of the slope and intercept. Estimate σ 2 . (b) Use the equation of the fitted line to predict what pavement deflection would be observed when the surface temperature is 90F. (c) Give a point estimate of the mean pavement deflection when the surface is 85F. (d) What change in mean pavement deflection would be expected for a 1F change in surface temperature? (e) Test for significance of regression using α = 0 . 05. What conclusion can you draw? (f) Estimate the standard errors of the slope and intercept. (g) Find a 95% CI for β 0 , and β 1 . (h) Find a 95% CI for expected value of pavement deflection (or true value of Y ) when the surface tem- perature is 85F. (i) Find 95% PI when the for the future pavement deflection when the surface temperature is 85F. (j) Complete the ANOVA table. What are your (null and alternative) hypotheses here. (k) What is your conclusion in (j). 1

Sol. (a) We have ˆ β 1 = S xy S xx , ˆ β 0 = y − ˆ β 1 x, ˆ σ 2 = MSE = S yy − ˆ β 1 S xy n − 2 , where S xy = ∑ x i y i − 1 n ( ∑ x i )( ∑ y i ) = 141 . 445 S xx = ∑ x 2 i − 1 n ( ∑ x i ) 2 = 33991 . 6 S yy = ∑ y 2 i − 1 n ( ∑ y i ) 2 = 0 . 731875 , so that ˆ β 1 = 0 . 00416, ˆ β 0 = 0 . 32999, and ˆ σ 2 = 0 . 00797 (b) ˆ y (90) = ˆ β 0 + ˆ β 1 · 90 = 0 . 70 (c) The question can be rephrased as “use the equation of the fitted line to predict what pavement deflection would be observed when the surface temperature is 85F”, i.e.ˆ y (85) = ˆ β 0 + ˆ β 1 · 85 = 0 . 68. (d) That is the definition of the slope: ˆ β 1 = 0 . 00416. (e) We test for H 0 : β 1 = 0 , against H 1 : β 1 ̸ = 0. The test statistic is T 0 = ˆ β 1 − 0 p ˆ σ 2 /S xx = 0 . 00416 p 0 . 00797 / 33991 . 6 = 8 . 6 , and we reject H 0 in favour of a linear relationship between x and y , since | 8 . 6 | > t (18 , . 025) = 2 . 1. (f) The standard errors are se( ˆ β 1 ) = s ˆ σ 2 S xx , se( ˆ β 0 ) = s ˆ σ 2 1 n + ¯ x 2 S xx . So, se( ˆ β 1 ) = 0 . 00048, se( ˆ β 0 ) = 0 . 04098. (g) CI for β 1 : 0 . 00416 ± 0 . 00048 × 2 . 1 where 2 . 1 is the critical value of t (18 , 0 . 025). CI for β 0 : 0 . 32999 ± 0 . 04098 × 2 . 1. (h) 0 . 68 ± 2 . 1 q 0 . 00797[ 1 20 + (85 − 73 . 9) 2 33991 . 6 ], where 73 . 9 is the ¯ x . (i) 0 . 68 ± 2 . 1 q 0 . 00797[1 + 1 20 + (85 − 73 . 9) 2 33991 . 6 ]. (j) We use ANOVA for testing H 0 = β 1 = 0 vs. H 0 = β 1 ̸ = 0. In here, • SST = 0 . 731875, with df = 19 • SSR = 0 . 00416 × 141 . 445 = 0 . 5884 , with df = 1 • SSE = 0 . 731875 − 0 . 5884 = 0 . 14347 , with df = 18, Therefore • MSR = 0 . 5884 • MSE = 0 . 14347 18 = 0 . 008 therefore • F 0 = 0 . 5884 0 . 008 = 73 . 55 (k) F 0 is (much) greater than F (1 , 18 , 0 . 05) = 4 . 41, we reject H 0 in favor of a linear relationship between x, and y. 2

Q.2 The number of emails arriving at a server per minute is claimed to follow a Poisson distribution . To test this claim, the number of emails arriving in 70 randomly chosen 1-minute intervals is recorded. The table below summarises the results. Test the hypothesis that the number of emails per minute follows a Poisson distribution? Use α = 0 . 05. # emails freq. 0 13 1 22 2 23 3 12 ≥ 4 0 3