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Multiple-Choice Test
Background
Regression
COMPLETE SOLUTION SET
1.
The average and standard deviation of the following numbers are
2 4 10
(A) 6.0, 4.0857
(B) 6.0, 4.2783
(C) 7.2, 4.0857
(D) 7.2, 4.4757
Solution
The correct answer is (B)
n
Average = x 
x
i 1
i
n
2  4  10  12  1.6  6.4

6
 6.0
n

 x
i 1
i
2
n 1
6

 x
 x
i 1
i
 x
6 1
2
12
1.6 6.4

(2  6) 2  (4  6) 2  (10  6) 2  (12  6) 2  (1.6  6) 2  (6.4  6) 2
5

16  4  16  36  19.36  0.16
5
91.52
5
 4.2783

2.
The average of 7 numbers is given 12.6. If 6 of the numbers are 5, 7, 9, 12,17 and 10,
the remaining number is
(A) -47.9
(B) -47.4
(C) 15.6
(D) 28.2
Solution
The correct answer is (D)
If x is the remaining number, then
5  7  9  12  17  10  x
 12.6
7
x  12.6  7   5  7  9  12  17  10 
 88.2  (60)
 28.2
3.
The average and standard deviation of 7 numbers is given a 8.142 and 5.005,
respectively. If 5 numbers are 5, 7, 9, 12 and 17, the other two numbers are
(A) -0.1738, 7.175
(B) 3.396, 12.890
(C) 3.500, 3.500
(D) 4.488, 2.512
Solution
The correct answer is (D)
From the average of the numbers being 8.142, we have
5  7  9  12  17  x  y
 8.142
7
x  y  8.142  7   5  7  9  12  17 
x y 7
From the standard deviation being 5.005, we have
5  8.1422  7  8.1422  9  8.1422  12  8.1422
2
2
2
 17  8.142    x  8.142    y  8.142 
7 1
(1)
 5.005
x  8.1422   y  8.1422 
5.005 2  6  5  8.1422  7  8.1422  9  8.1422  12  8.1422  17  8.1422 
x 2  16.284 x  66.292  y 2  16.284 y  66.292  45.039
x 2  y 2  16.284( x  y )  87.544
From equation (1)
x 7 y
then
(7  y ) 2  y 2  16.284(7  y )  16.2840 y  87.544
y 2  14 y  49  y 2  16.284 y  16.284 y  113.988  87.544
2 y 2  14 y  64.988  87.544
2 y 2  14 y  22.556  0
Using the quadratic equation solution
(2)
y

gives
 (14) 
 142  4  2  22.556
2 2
14  15.552
4
y  4.488
or
y  2.512
Thus,
x  2.512
or
x  4.488
Hence x, y   (4.488,2.512) or (2.512, 4.488) , which are the same pair of numbers.
4.
The sum of the square of the difference between data point and its average for the
data is
2 5 10 12 2.5 6.7
(A) 4.023
(B) 13.49
(C) 16.19
(D) 80.93
Solution
The correct answer is (D)
The sum of the squares of the differences, also called summed squared error (SSE) St is given by
n
S t    yi  y 
2
i 1
where
2  5  10  12  2.5  6.7
6
 6.366
y
Thus,
n
 2    y i  y 2
i 1
 (2  6.366) 2  (5  6.366) 2  (10  6.366) 2  (12  6.366) 2  (2.5  6.366) 2  (6.7  6.366) 2
 19.06  1.865  13.20  31.74  14.94  0.1115
 80.91
5.
Two medications are tried to heal esophageal ulcers in patients. The time to heal is
reported as the time the patient reports 1 or less heartburn episode per week.
Pacalo Reggon
26
25
23
31
21
32
25
23
32
19
37
26
The medication with less recovery time with standard deviation and mean is
(A) Pacalo, x  27.33,   6.022 2
(B) Reggon, x  26.00,   4.900
(C) Pacalo, x  27.33,   5.497
(D) Pacalo, x  27.33,   6.022
Solution
The correct answer is (B)
The mean of Pacalo is
26  23  21  25  32  37
x
6
164

6
 27.33
The standard deviation of Pacalo is


26  27.332  23  27.332  21  27.332  25  27.332  32  27.332  37  27.332
6 1
1.768  18.74  40.06  5.428  21.80  93.50
5
181.3
5
 6.022

The mean of Reggon is
25  31  32  23  19  26
x
6
156

6
 26
The standard deviation of Reggon is


25  262  31  262  32  262  23  262  19  262  26  262
6 1
1  25  36  9  49  0
5
120
5
 4.900

6.
A very large number of data points are chosen on a function y  3e 2 x from x  0.2 to
2.1. The average value of these values most nearly is
(A) 51.5
(B) 78.2
(C) 97.8
(D) 102
Solution
The correct answer is (A)
Since a very large number of points are chosen, the average will be close to the mean of
the function.
b
f 
 f x dx
a
ba
2.1

 3e
2x
0.2
2.1  0.2
97.79

1.9
 51.5