Regression methods were used to analyze the data from a study investigating the

Question

Regression methods were used to analyze the data from a study investigating the relationship between roadway surface temperature and pavement deflection. Using the data provided in table 1, The goal is to develop a simple linear regression model to estimate pavement deflection at any given temperature. 1. Part a) Calculate the least squares estimates of the slope and intercept. Part b) Use the equation ofthe fitted line to predict what pavement deflection would be observed when the surface temperature is 85°F. Part c) Use the equation of the fitted line to find what change in mean pavement deflection would be expected for a 5 change in surface tem perature? Part d) Suppose that the surface temperature is 55°F, calculate the fitted value of pavement deflection and the corresponding residual Part e) What percentage of variation in the data can be explained by this model? Table 1 Pavement Deflection Datoa surface temperature 0.65 0.8 0.69 0.85 0.77 0.82 No.pavement deflection 45 47 62 52 60

Explanation / Answer

Here to produce the linear regressed fitted line we require following values. x,y,xy, x2 & y2

HEre the table to calculate the all these values is given below.

So here n = 6

so, if fitted line is y^ = a + bx

then,

a =  [(y) (x2 ) - (x) (xy)]/ [ n (x2 ) - (x)2 ]

a = [ 4.58 * 17407 - 321 *247.62 ]/ [6 * 17407 - 3212]

a = 238.04/ 1401

a = 0.1699

b = [ n(xy) - (x)((y)]/ [ n (x2 ) - (x)2 ]

b = [6 * 247.62 - 321 * 4.58]/  [6 * 17407 - 3212]

b = 15.54/1401 = 0.011

SO regression equation is

y^ = 0.0111x + 0.1699

(b) Here x = 850 F

y^ = 0.0111 * 85 +0.1699 = 1.1127

(c) When there is a change of 5F then the change in pavement deflection is 0.0111 * 5 = 0.05546

(d) Here surface temperature = 55F

y^ = 0.0111 * 55 +0.1699 = 0.78

Residual = Actual - residual = 0.8 - 0.78 = 0.02

(e) Here the coefficient of determination R2

R2 = [n(xy) - (x)((y)]2 / [ (nx2 - (x)2] [ ny2 - (y)2]

R2 = [6 * 247.62 - 321* 4.58]2 / [6 * 17407 - 3212] * [6 * 3.5264 - 4.582]

R2 = 241.4916 / (1401 * 0.182)

R2 = 0.9471

No. Surface temperature(X) Pavement deflection (y) x^2 xy y^2 1 45 0.65 2025 29.25 0.4225 2 55 0.8 3025 44 0.64 3 47 0.69 2209 32.43 0.4761 4 62 0.85 3844 52.7 0.7225 5 52 0.77 2704 40.04 0.5929 6 60 0.82 3600 49.2 0.6724 Sum 321 4.58 17407 247.62 3.5264

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