Mass (kg) Fuel Efficiency 1283 27 1548 15 2508 25 1777 29 2075 16 2207 16 1233 3
ID: 3367993 • Letter: M
Question
Mass (kg)
Fuel Efficiency
1283
27
1548
15
2508
25
1777
29
2075
16
2207
16
1233
30
1392
24
1114
29
1928
17
1463
25
1248
27
1392
20
1574
23
1538
20
1276
25
1980
19
1353
23
1956
19
1913
18
1. Which variable should be the independent variable and which should be the dependent variable?
2. Calculate the following to 4 decimal places:
correlation coefficient =
regression equation y-hat =
Using LinRegTTest on your TI-83/83, the p-value =
3. Is the correlation significant? Why or why not?
4. In one or two complete sentences, what is the practical interpretation of the slope of the least squares line in terms of fuel efficiency and mass? Your answer should include the value of the slope of the line and correct units.
5. For a car that has mass 1700 kilograms, predict its fuel efficiency to 2 decimal places. Include units.
6. Can we predict the fuel efficiency of a car with mass 1000 kg using the least squares line? Explain why or why not.
7. What percent of variation in fuel efficiency can be explained by the variation in mass using the regression line? [Hint: Coefficient of determination]
8. Are there any outliers? ___________ (yes / no) If so, which point (or points) is an outlier?
9. Give the SSE (sum of square errors) and s (the standard deviation of the |y-y-hat| values) to 4 decimal places.
Mass (kg)
Fuel Efficiency
1283
27
1548
15
2508
25
1777
29
2075
16
2207
16
1233
30
1392
24
1114
29
1928
17
1463
25
1248
27
1392
20
1574
23
1538
20
1276
25
1980
19
1353
23
1956
19
1913
18
Explanation / Answer
Solution [Answers to the point are given below. Detailed theory and calculations follow at the end. ]
1. Independent variable: x = Mass (kg) Answer 1
Dependent variable: y = Fuel Efficiency Answer 2
2. Calculate the following to 4 decimal places:
Correlation Coefficient = - 0.5426 Answer 3
Regression Equation y-hat = 33.43 – 0.006763x Answer 4
T Test Statistic (for testing significance of slope coefficient) = - 2.7409 Answer 5
p-value = 0.0134 Answer 6
3. The correlation is significant at 5% level of significance. Answer 7
Because: p-value < 0.05 Answer 8
4. Practical interpretation of the slope of the least squares line: For increase of every kg in the mass, the fuel efficiency would come down by 0.006763 or in other words, for increase of every 1000 kg in the mass, the fuel efficiency would come down by 6.763 Answer 9
5. For a car that has mass 1700 kilograms, its predicted fuel efficiency is 21.93 Answer 10
6. Predicting the fuel efficiency of a car with mass 1000 kg cannot be done using the least squares line, because 1000 is beyond the range x-values (1114 - 2578) and extrapolation is not permitted in least square line. ANSWER 11
7. Percent of variation in fuel efficiency that can be explained by the variation in mass using the regression line is 29.45% [Coefficient of determination = r2]
Back-up Theory
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 ……………………………………..(11)
Test statistic for testing significance of ?1 is: t = ?1/SE(?1) ~ tn – 2 and
p-value P(tn – 2 |tcal|)……………………………………………….…………………..(12)
Details of Excel Calculations
n
20
Xbar
1637.90
ybar
22.35
Sxx
2771611.8
Syy
430.55
Sxy
-18745.3
?1cap
-0.0067633
?0cap
33.4276433
s^2
16.8760845
sb^2
6.0889E-06
r
-0.5426432
r^2
0.29446169
tb
-2.7408821
x0
1700
ycap at x0
21.9299978
p-value
0.01343099
DONE
n
20
Xbar
1637.90
ybar
22.35
Sxx
2771611.8
Syy
430.55
Sxy
-18745.3
?1cap
-0.0067633
?0cap
33.4276433
s^2
16.8760845
sb^2
6.0889E-06
r
-0.5426432
r^2
0.29446169
tb
-2.7408821
x0
1700
ycap at x0
21.9299978
p-value
0.01343099