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##### Page 8
ENGR 13200 - Exam 3
April 16, 2013
8
if d_end <= d_start
set(handles.error_mess_st,'String','High value must be
greater than low value')
else
wire_diam = linspace(d_start,d_end,10)
Mass = pi^2/4*rho*Diam*wire_diam.^2*NumCoils
plot(wire_diam,Mass,'bs-')
ylabel('Mass (kg)')
xlabel('Wire Diameter (m)')
title('Mass of a 10 coil high carbon steel wire')
end

##### Page 9
ENGR 13200 - Exam 3
April 16, 2013
9
Problem #6 (35 points)
A Purdue research team is developing a new implant device using biocompatible and swellable
materials. Recently they performed animal testing of the device. For this test, the disc-shaped
device was implanted under the skin of a pig and the diameter of the device was monitored as a
function of time using a Magnetic Resonance Imaging (MRI) method.
The results are shown in
the table below.
Time
[Day]
Diameter of the Device
[mm]
1
1
2
5
3
3
5
4
7
7
Pat created a scatter plot of the diameter of the implanted device as a function of time and drew a
best fit line by eye as shown.
0
1
2
3
4
5
6
7
8
0
1
2
3
4
5
6
7
8
Diameter of Device (mm)
Time (days)
Implant diameter as a function of time
after implant in pig

##### Page 10
ENGR 13200 - Exam 3
April 16, 2013
10
A. Compute the coefficient of determination (r
2
) for the regression line drawn by Pat.
Show your work in the table and r
2
box provided on the answer sheet.
Solution
i
Time
(days)
Diameter
(mm)
f
(x
i
)
[y
i
-
f
(x
i
)]
2
[y
i
-y
bar
]
2
1
1
1
1
0
9
2
2
5
2
9
1
3
3
3
3
0
1
4
5
4
5
1
0
5
7
7
7
0
9
Sums
20
10
20
Mean
4
r
2
=
1
10/20 = 0.5
B. Use the least squares method to set up the simultaneous equations that could be solved for the
slope (a) and intercept (b).
Work has been started in the table on the answer sheet; show any
additional work that may be needed in the table. Fill in the boxes in the two equations below
the table.
You do NOT need to solve these equations.
Solution
i
Time
(days)
Diameter
(mm)
Time
2
Time*Diameter
Diameter
2
1
1
1
1
1
1
2
2
5
4
10
10
3
3
3
9
9
9
4
5
4
25
20
20
5
7
7
49
49
49
Sums
18
20
88
89
Not needed
Mean
88
a
+
18
b =
89
n
i
i
n
i
i
y
bn
x
a
1
1
18
a
+
5
b =
20
n
i
i
i
n
i
i
n
i
i
y
x
x
b
x
a
1
1
1
2

##### Page 11
ENGR 13200 - Exam 3
April 16, 2013
11
C. The coefficient of determination (r
2
) for the regression line found using least squares method
is 0.62.
Assuming this value is higher than the coefficient found using the 2-point method,
which of the following is/are true?
[Circle
ALL
that apply]
a.
The Sum of Squares of Deviations (SST), when computed using least squares is
higher
than that found by regression by eye.
b.
The Sum of Squares of Deviations (SST), when computed using least squares is
lower
than that found by regression by eye.
c.
The Sum of Squares of Errors (SSE), when computed using least squares is
higher
than
that found by regression by eye.
d.
The Sum of Squares of Errors (SSE), when computed using least squares is
lower
than
that found by regression by eye.
Solution:
d
D. The best-fit line resulting from the least squares method is
d
= 0.7
t
+ 1.4, where
d
is diameter
(mm) and
t
is time (days). Predict the diameter of the implanted device at 8 days.
The diameter after day 8 cannot be predicted as “8 days” is not within the range of time
examined for the test.
E.
The research team found after reporting the results that the device size measured at 7 days
was not recorded correctly. The actual diameter at 7 days was 4 mm instead of 7 mm.
With regards to the least squares method results and the examination of the goodness of fit,
what will happen?
[Circle
ONLY ONE
response for each question]
a.
slope will
Increase, Decrease, or Remain the same
b.
SST will
Increase, Decrease, or Remain the same
c.
r
2
will
Increase, Decrease, or Remain the same
Solution
a.
slope will
Increase,
Decrease
, or Remain the same
b.
SST will
Increase,
Decrease
, or Remain the same
c.
r
2
will
Increase,
Decrease
, or Remain the same