) An engineer would like to use simple linear regression to investigate the rela

Question

) An engineer would like to use simple linear regression to investigate the relationship between roadway surface temperature in °F (x) and pavement deflection (y). A sample of n=20 pairs of observations was obtained and the following summary quantities were calculated: yi = 12.75, yi2 =8.86, xi=1478, xi2 =143,215.8, and xiyi =1083.67 (all summations are for i = 1…n).

(a) Find the least-squares estimates of the slope and the intercept.

(b) Find a 95% confidence interval for the slope and for the intercept.

(c) Assess the utility of the model by formulating and testing an appropriate hypothesis regarding the slope parameter. Use =0.05. Report the p-value corresponding to this test. What are your conclusions?

(d) What change in mean pavement deflection would be expected for a 2°F increase in surface temperature?

(e) What is the predicted value of pavement deflection when the surface temperature is 85°F? What is the respective 99% prediction interval?

Explanation / Answer

Solution:

We are given

n = 20

X = 1478

Y = 12.75

X^2 = 143215.8

Y^2 = 8.86

XY = 1083.67

Xbar = X/n = 1478/20 = 73.9

Ybar = Y/n = 12.75/20 = 0.6375

Part a

We have to find least squares estimates of the slope and intercept.

Formulas for least squares estimates are given as below:

b = (XY – n*Xbar*Ybar)/(X^2 – n*Xbar^2)

a = Ybar – b*Xbar

First we have to find slope b

b = (1083.67 - 20*73.9*0.6375) / (143215.8 - 20*73.9^2)

b = slope = 0.004161

Now, we have to find value intercept a

a = Ybar – b*Xbar

a = 0.6375 - 0.004161*73.9

a = intercept = 0.330002

Part b

Confidence interval for slope is given as below:

Confidence interval = 1 -/+ t*SE

Where, SE = Sb1 = sqrt(SSE/(n - 1))/sqrt(x^2 - (x)^2/n)

SSE = Y^2 – a*Y – b*XY

SSE = 8.86 - 0.330002*12.75 - 0.004161*1083.67

SSE = 0.143324

SE = Sb1 = sqrt(0.143324/(20 - 1))/sqrt(143215.8 - 1478^2/20)

SE = Sb1 = 0.086852567/ 184.36811

SE = Sb1 = 0.000471

We are given

Confidence level = 95%

n = 20

df = n – 1 = 20 – 1 = 19

Critical t value = 2.0930

(By using t-table or excel)

Confidence interval = 1 -/+ t*SE

Confidence interval = 0.004161 -/+ 2.0930*0.000471

Lower limit = 0.004161 - 2.0930*0.000471 = 0.003175

Upper limit = 0.004161 + 2.0930*0.000471 = 0.005147

Part c

Here, we have to use t test for regression slope 1

We are given = 0.05

H0: 1 = 0 versus Ha: 1 0

Two tailed test

Test statistic = t = 1 /SE(1) = 0.004161/0.000471 = 8.834395

df = n – 1 = 20 – 1 = 19

P-value = 0.00

P-value < = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that regression slope is statistically significant.

Part d

Change in mean pavement deflection = 2*b

Change in mean pavement deflection = 2*0.004161

Change in mean pavement deflection = 0.008322

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