The cost of a casting is believed to be proportional to its weight. Fifteen part
ID: 3224550 • Letter: T
Question
The cost of a casting is believed to be proportional to its weight. Fifteen parts have been collected and the weight and cost of each part is recorded in the table below. a) Using this data, determine the least squares regression equation to fit the data. You must show your calculations (by hand or in Excel). If you're using Excel, you can use the SUM, SUMSQ, and SUMPRODUCT functions to do the sums, but you can't just use LINEST to get the answers In your submission, you must indicate the values for the following parameters. Label them and put a box around them: sigma y_i x_i sigma x^2_i S_xx beta_0 x or sigma x_i y or sigma y_i S_xy beta_1 b) Conduct a hypothesis test (with alpha = 0.05) on your value of beta_1 to evaluate whether the regression is significant. Label all 7 steps of the hypothesis test. c) Determine a 90% confidence interval on beta_0. d) Determine the cost that is predicted by the regression equation for a casting that weighs 45 kg. e) Determine a 95% confidence interval on the mean cost of a casting that weighs 45 kg. f) Determine a 95% prediction interval on the cost of a casting that weighs 45 kg. g) Calculate R^2 for your regression equation.Explanation / Answer
A.a)
Weight(x)
Cost(y)
xy
x-square
Y-square
22.8
940
21432
519.84
883600
12.5
695
8687.5
156.25
483025
87.4
2368
206963.2
7638.76
5607424
27.3
1064
29047.2
745.29
1132096
77.4
2154
166719.6
5990.76
4639716
44.3
1433
63481.9
1962.49
2053489
27.9
1038
28960.2
778.41
1077444
11.1
668
7414.8
123.21
446224
37.5
1253
46987.5
1406.25
1570009
40.4
1246
50338.4
1632.16
1552516
40.9
1345
55010.5
1672.81
1809025
13.8
713
9839.4
190.44
508369
34.5
1188
40986
1190.25
1411344
70.9
2012
142650.8
5026.81
4048144
45
1419
63855
2025
2013561
Sum total
593.7
19536
942374
31058.73
29235986
from the table, we have
n=15
sum(x)=593.7
sum(y)=19536
sum(xy)=942374
sum(x-square)=31058.73
Sxx=sum(x-square)-{sum(x)}^2/n
solving we get
Sxx=7560
Sxy=sum(xy)-{sum(x)*sum(y)/n}
solving we get
Sxy=169139
b1=Sxy/Sxx
b1=169139/7560=22.37
b0=sum(y)-b1*sum(x)/n
solving we get
b0=416.89
A.d) weight(x)=45
b0=416.89
b1=22.37
to calculate, cost(y)
cost(y)=b0+b1*x
cost(y)=416.89+22.37*45
cost(y)=1423
A.e) given,
weight(x)=45
alpha=0.05
T-value(0.025,13)=2.160
SSE=29235986 - 416.89*19536-22.37*942374
SSE=8209
Se=sqrt(SSE/n-2)
Se=25.13
point estimate(y)=b0+b1*x
point estimate(y)=416.89+22.37*45
point estimate(y)=1423.54
Confidence interval is given by
y± t-value*Se*sqrt{1/n+(x0-x mean)^2/Sxx}
1423.54±2.16*25.13*sqrt(0.07)
therefore confidence interval is given as
CI=(1409.12, 1437.96)
A.g) to calculate R^2
Syy=sum(y-square)-sum(y)^2/n
Syy=3792300
R^2=b1^2*Sxx/Syy
=22.37*22.37*169139/3792300
R^2=22.32
Weight(x)
Cost(y)
xy
x-square
Y-square
22.8
940
21432
519.84
883600
12.5
695
8687.5
156.25
483025
87.4
2368
206963.2
7638.76
5607424
27.3
1064
29047.2
745.29
1132096
77.4
2154
166719.6
5990.76
4639716
44.3
1433
63481.9
1962.49
2053489
27.9
1038
28960.2
778.41
1077444
11.1
668
7414.8
123.21
446224
37.5
1253
46987.5
1406.25
1570009
40.4
1246
50338.4
1632.16
1552516
40.9
1345
55010.5
1672.81
1809025
13.8
713
9839.4
190.44
508369
34.5
1188
40986
1190.25
1411344
70.9
2012
142650.8
5026.81
4048144
45
1419
63855
2025
2013561
Sum total
593.7
19536
942374
31058.73
29235986
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