Which of the two regression function would you suggest Nolan used for prediction
ID: 3198352 • Letter: W
Question
Which of the two regression function would you suggest Nolan used for prediction purposes.?
Customers (in 1000s), X
Line Maintenance Expense (in $1000s), Y
X^2
Y^2
XY
1
25.3
484.6
640.09
234837.2
12260.38
2
36.4
672.3
1324.96
451987.3
24471.72
3
37.9
839.4
1436.41
704592.4
31813.26
4
45.9
694.9
2106.81
482886
31895.91
5
53.4
836.4
2851.56
699565
44663.76
6
66.8
681.9
4462.24
464987.6
45550.92
7
78.4
1037
6146.56
1075369
81300.8
8
82.6
1095.6
6822.76
1200339
90496.56
9
93.8
1563.1
8798.44
2443282
146618.8
10
97.5
1377.9
9506.25
1898608
134345.3
11
105.7
1711.7
11172.49
2929917
180926.7
12
124.3
2138.6
15450.49
4573610
265828
Total
848
13133.4
70719.06
17159981
1090172
Slope:
?=(n?(xy)??x?y)/(n?x2?(?x)2)
Offset:
?=(?y???x)/n
regression equation:
y=?x+?
?
15.02
?
33.32
regression equation: Line Maintenance Expense = 15.02*Customers+33.32
Customers
75
(in 000's)
Line Maintenance Expense (in $1000s)
$ 1,159.52
total level of line maintenance expense=
$ 1,159,519.26
Customers (in 1000s), X
Line Maintenance Expense (in $1000s), Y
X^2
Y^2
XY
25.3
484.6
640.09
234837.16
12260.38
36.4
672.3
1324.96
451987.29
24471.72
37.9
839.4
1436.41
704592.36
31813.26
45.9
694.9
2106.81
482886.01
31895.91
53.4
836.4
2851.56
699564.96
44663.76
66.8
681.9
4462.24
464987.61
45550.92
78.4
1037
6146.56
1075369
81300.8
82.6
1095.6
6822.76
1200339.36
90496.56
93.8
1563.1
8798.44
2443281.61
146618.78
97.5
1377.9
9506.25
1898608.41
134345.25
105.7
1711.7
11172.49
2929916.89
180926.69
124.3
2138.6
15450.49
4573609.96
265827.98
Total
848
13133.4
70719.06
17159980.62
1090172.01
Customers (in 1000s), X
Line Maintenance Expense (in $1000s), Y
X^2
Y^2
XY
1
25.3
484.6
640.09
234837.2
12260.38
2
36.4
672.3
1324.96
451987.3
24471.72
3
37.9
839.4
1436.41
704592.4
31813.26
4
45.9
694.9
2106.81
482886
31895.91
5
53.4
836.4
2851.56
699565
44663.76
6
66.8
681.9
4462.24
464987.6
45550.92
7
78.4
1037
6146.56
1075369
81300.8
8
82.6
1095.6
6822.76
1200339
90496.56
9
93.8
1563.1
8798.44
2443282
146618.8
10
97.5
1377.9
9506.25
1898608
134345.3
11
105.7
1711.7
11172.49
2929917
180926.7
12
124.3
2138.6
15450.49
4573610
265828
Total
848
13133.4
70719.06
17159981
1090172
Slope:
?=(n?(xy)??x?y)/(n?x2?(?x)2)
Offset:
?=(?y???x)/n
regression equation:
y=?x+?
?
15.02
?
33.32
regression equation: Line Maintenance Expense = 15.02*Customers+33.32
Customers
75
(in 000's)
Line Maintenance Expense (in $1000s)
$ 1,159.52
total level of line maintenance expense=
$ 1,159,519.26
Explanation / Answer
The regression equation is
The regression equation is , where m is the slope and c is the y-intercept.
And m is given by the method of least squares as
.
Then the y-intercept ‘c’ is given by .
There are two tables of calculations given in the question , in which the first table have some error, and the table given in the end of the question is correct. In the first table the headings of each coloumn are misplaced.
The second table of calculations and all other calculations are correct.
ie.
Here X=Customers (in 1000s), Y=Line Maintenance Expense (in $1000s),and hence we can the regression equation
Line Maintenance Expense = 15.02*Customers+33.32 .
Then
Customers (in 1000s), X Line Maintenance Expense (in $1000s), Y X^2 Y^2 XY 25.3 484.6 640.09 234837.2 12260.38 36.4 672.3 1324.96 451987.3 24471.72 37.9 839.4 1436.41 704592.4 31813.26 45.9 694.9 2106.81 482886 31895.91 53.4 836.4 2851.56 699565 44663.76 66.8 681.9 4462.24 464987.6 45550.92 78.4 1037 6146.56 1075369 81300.8 82.6 1095.6 6822.76 1200339 90496.56 93.8 1563.1 8798.44 2443282 146618.78 97.5 1377.9 9506.25 1898608 134345.25 105.7 1711.7 11172.49 2929917 180926.69 124.3 2138.6 15450.49 4573610 265827.98 Total 848 13133.4 70719.06 17159981 1090172.01Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.