Question

A US consumer lobby wishes to develop a model to predict gasoline usage, as measured by miles per gallon, based on the horsepower of the car's engine. The Excel data file AUTO.xls (contained in a folder under the CML Quizzes tab) contains data on this for fifty recent models. Use Excel Data Analysis to estimate a linear model for the relationship, a 92% confidence interval for the slope coefficient and the residual plot. State all numerical answers below correct to four decimal places using the Excel output results. 1. the intercept , 2. the slope coefficient , and 3. the standard error of the estimate, 4. Using this model, predict the gasoline mileage for the car with 120 horsepower. Use all decimal places in your calculation by selecting and using the values of b0 and b1 in the output generated by Excel. 5. Does the prediction involve extrapolating the relationship? Type yes or no. State the 6. lower bound 7. upper bound for the 92% confidence interval for the slope coefficient.

Answer 1

ANSWER::

NOTE: Answers to the point are given below. For better understanding,

Back-up Theory and Details of Excel Calculations are given at the end.

Part (1)

the intercept = 50.0052 ANSWER 1

Part (2)

the slope coefficient – 0.2363 ANSWER 2

Part (3)

the standard error of the estimate

For intercept: 2.5235 ANSWER 3

For slope coefficient: 0.0226 ANSWER 4

Part (4)

Predicted value of the gasoline mileage for the car with 120 horsepower:

21.6522 mpg ANSWER 5

Part (5)

Does the prediction involve extrapolating the relationship? No. ANSWER 6

[because given x-values range from 48 to 165 and 120 is well within that range]

Part (6)

92% confidence interval for the slope coefficient

lower bound: - 0.2839 ANSWER 7

upper bound .- 0.1886 ANSWER 8

Back-up Theory

Let

y represent the gasoline usage, as measured by miles per gallon (i.e., MPG), and

x represent the horsepower of the car's engine.

The linear regression model Y = ?0 + ?1X + ?, ………………………………………..(1)

where ? is the error term, which is assumed to be Normally distributed with mean 0 and variance ?2.

Estimated Regression of Y on X is given by: Y = ?0cap + ?1capX, ………………………….(2)

where

?1cap = Sxy/Sxx and ?0cap = Ybar – ?1cap.Xbar..……………………………………………..(3)

Mean X = Xbar = (1/n)sum of xi ………………………………………….……………….(4)

Mean Y = Ybar = (1/n)sum of yi ………………………………………….……………….(5)

Sxx = sum of (xi – Xbar)2 …………………………………………………..………………………………..(6)

Syy = sum of (yi – Ybar)2 …………………………………………………..………………………………..(7)

Sxy = sum of {(xi – Xbar)(yi – Ybar)} …………………………………………………………………….………(8)

All above sums are over i = 1, 2, …., n

n = sample size ………………………………………………………………………………(9)

Estimate of ?2 is given by s2 = (Syy – ?1cap2Sxx)/(n - 2)……………………………………..(10)

Standard Error of ?1cap is sb, where sb2 = s2/Sxx

Standard Error of ?0cap is sa, where sa2 = sb2{(sum of xi2 over i = 1, 2, …., n)/n}

Standard Error of yicap = s?[(1/n) + {(xi – Xbar)2/Sxx}]

Now to work out the solution,

Details Excel Calculations:

n	50
xbar	90.84
ybar	28.542
Sxx	36408.72
Syy	3271.8418
Sxy	-8602.464
Slope coeff	-0.2363
intercept	50.0052
s^2	25.818669
s	5.0812
SE(slope coeff)	0.0266
SE(intercpt)	2.5235
tn-2,?/2	1.7885
CIbLB	-0.2839
CIbUB	-0.1886

DONE

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