251solngr2
4/12/06
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Graded Assignment 2
Name: Solution
Class days and time:
Student number:
There will be a penalty for papers that are unstapled or do not have the three information items
requested above.
For part 1) use the first joint probability table in Problem K4. Modify the table as follows: subtract the last
digit of your student number (divided by 100) from all three numbers on the diagonal, add the same number
to any 3 numbers off the diagonal, if the last digit of your student number is zero, use 10; Add the second to
last digit to the 15 in problem 2, using 10 if the last digit is zero. For example, if the last two digits of your
number are 30: in 1) the .40 on the diagonal becomes .40 - .10 = .30 and a zero will become .10. The sum
of the numbers in the table will not change; in 2) the 15 will became 15 + 3 = 18.
1) For the joint probability table (i) check for independence, (ii) Compute E x  and Varx  , (iii) Compute
Covx, y  or  xy and Corr x, y  or  xy , (iv) Compute Ex  y  and Var x  y  from the results in (ii)
 
 
and (iii), (v) Compute Cov3x  3, y  and Corr 3x  3, y  using the formulas in section K4 of 251v2out
or section C1 of 251var2. Note that y  1y  0 .
x
Solution to Problem K4a: Find  xy for
1
4
.10
5
0
6
0
y 2
0
.40
0
3
0
0
.50
(i) Check for independence: First you need to find Px  and P y  . Look at the upper left hand
probability below. Its value is .10 and it represents Px  4   y  1 . If x and y are independent, we
would have Px  4   y  1  Px  4 P y  1  .10.10   ..01 . Since this is not true, x and y
cannot be independent. Even one place where the joint probability is not the product of the marginal
probabilities is enough. If this one is not enough to convince you, how about Px  4   y  3  0
 Px  4 P y  3  .10.50   .05 . Actually the best way to prove non-independence is to look for
zeroes. If Px  4   y  3  0 and x and y are independent, then it must be true that Px  4  0 or
P y  3  0 . Notice that the second row is not proportional to the first row or any other row.
A zero covariance or correlation would be the consequence of independence, but it is not true that a zero
correlation or covariance would prove independence. We have already seen one example where there is a
zero correlation, but no independence.
x
4
 .10

 0
 0

0.1
1
y
2
3
Px 
xPx 
0.4 
5
0
0.4
6
0

0
.50 

0.5
2.0 
3.0 
.40
0
x 2 Px  1.6  10 .0  18 .0 
P y 
.10
yP y  y 2 P y 
0.1
0.1
.40
0.8
1.6
.50
1.5
4.5
1.00
2.4
6.2
5.4
29 .6
 Px   1 (a check),   Ex   xPx  5.4 , E x    x
 P y   1 ,   E  y    yP y   2.4 and E y    y P y   6.2 .
2
To summarize
x
2
2
Px   29 .6 ,
2
y
1
251solngr2
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(ii) Compute E x  and Varx  .
From the above  x  E x  
 xPx  5.4 ,
 
 x Px  
Var y   E y     y P y   
Varx  E x 2   x2 
2
2
2
y
2
x
2
2
y
 29.6  5.42  0.44 ,  y  E  y  
 6.2  2.42  0.44 .
 
 yP y   2.4 and
 
(iii) Compute Covx, y  or  xy and Corr x, y  or  xy
051
061  0.4 0
0 
 .10 41



E xy  
xyPxy    042 .40 52 062   0 4.0 0   13 .4
  043 .053 .50 63  0 0 9.0
 xy  Covxy  Exy   x  y  13.4  5.42.4  0.44 .

So that  xy 
 xy
 x y

0.44
 1.00 .
0.44 0.44
2
 1 is as
The correlation and covariance are positive, indicating a tendency of y to rise when x rises.  xy
high as you can get on a zero to one scale, indicating a relationship of proportionality. Note that
1   xy  1 always!
(iv) Compute Ex  y  and Var x  y  from the results in (ii) and (iii).
Ex  y   Ex  E y    x   y  5.4  2.4  7.8 and
Var x  y    x2   y2  2 xy  Var x   Var  y   2Covx, y   0.44  0.44  20.44   1.76
(v) Compute Cov3x  3, y  and Corr 3x  3, y 
251v2out says Cov(ax  b, cy  d )  acCov( x, y) and Corr (ax  b, cy  d )  (sign(ac))Corr ( x, y) , where
signac has the value 1 or 1 depending on whether the product of a and c is negative or
positive. a  3 and c  1 .  xy  Covx, y   0.44 and  xy  Corrx, y   1.00
Cov(3x  3, 1y  0)  31Cov( x, y)  30.44   1.32
Corr (3x  3,1y  0)  (sign(31))Corr ( x, y)  11.00   1.00.
Solution to Problem K4a with .05 added and subtracted: Find  xy for
x
4
5
0
6
.05
1
.05
y 2
0
.35
0
3
.05
.05
.45
(i) Check for independence: First you need to find Px  and P y  .
Look at the upper left hand
probability below. Its value is .05 and it represents Px  4   y  1 . If
x and y are independent, we
would have Px  4   y  1  Px  4 P y  1  .10.10   .01 . Since this is not true, x and y cannot
be independent. Even one place where the joint probability is not the product of the marginal probabilities
is enough. If this one is not enough to convince you, how about Px  4   y  3  0
 Px  4 P y  3  .10.35   .035 . Actually the best way to prove non-independence is to look for
zeroes. If Px  4   y  3  0 and x and y are independent, then it must be true that Px  4  0 or
P y  3  0 . Notice that the second row is not proportional to the first row or any other row.
2
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A zero covariance or correlation would be the consequence of independence, but it is not true that a zero
correlation or covariance would prove independence. We have already seen one example where there is a
zero correlation, but no independence.
x
4
1
y
2
3
Px 
xPx 
5
0
0.4
6
.05 

0
.45 

0.5
2.0 
3.0 
 .05

 0
 .05

0.1
.35
.05
0.4 
P y 
.10
yP y  y 2 P y 
0.10
0.10
.35
0.70
1.40
.55
1.65
4.95
1.00
2.45
6.45
5.4
x Px  1.6  10 .0  18 .0 
2
29 .6
 Px   1 (a check),   Ex   xPx  5.4 , E x    x
 P y   1 ,   E y    yP y   2.45 and E y    y P y   6.45
2
To summarize
2
x
2
Px   29 .6 ,
2
y
(ii) Compute E x  and Varx  .
From the above  x  E x  
 xPx  5.4 ,
 
 x Px  
Var y   Ey     y P y   
Varx  E x 2   x2 
2
2
2
y
2
x
2
2
y
 29.6  5.42  0.44 ,  y  E  y  
 6.45  2.452  0.4475 .
 
 yP y   2.45 and
 
(iii) Compute Covx, y  or  xy and Corr x, y  or  xy
051 .0561  0.20
0
0.30 
 .0541
xyPxy     042 .3552 062   0
3.50
0   13 .45
 .0543 .0553 .4563 0.60 0.75 8.10 
 Covxy  Exy   x  y  13.45  5.42.45  0.22 .
E xy  
 xy

So that  xy 
 xy
 x y

0.22
 0.4958
0.44 0.4475
2
 .2458
The correlation and covariance are positive, indicating a tendency of y to rise when x rises.  xy
is relatively low on a zero to one scale, indicating a relatively weak relationship. Note that 1   xy  1
always!
(iv) Compute Ex  y  and Var x  y  from the results in (ii) and (iii).
Ex  y   Ex  E y    x   y  5.4  2.45  7.85 and
Var x  y    x2   y2  2 xy  Var x   Var  y   2Covx, y   0.44  0.4475  20.22   1.3275
(v) Compute Cov3x  3, y  and Corr 3x  3, y 
251v2out says Cov(ax  b, cy  d )  acCov( x, y) and Corr (ax  b, cy  d )  (sign(ac))Corr ( x, y) , where
signac has the value 1 or 1 depending on whether the product of a and c is negative or
positive. a  3 and c  1 .  xy  Covx, y   0.22 and  xy  Corrx, y   .4958
Cov(3x  3, 1y  0)  31Cov( x, y)  30.22   0.66
Corr (3x  3,1y  0)  (sign(31))Corr ( x, y)  10.4958   0.4958 .
3
251solngr2
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Solution to Problem K4a with .10 added and subtracted: Find  xy for
x
4
5
0
6
.10
1
.00
y 2
0
.30
0
3
.10
.10
.40
(i) Check for independence: First you need to find Px  and P y  . Look at the upper left hand
probability below. Its value is .00 and it represents Px  4   y  1 . If x and y are independent , we
would have Px  4   y  1  Px  4 P y  1  .10.10   .01 . Since this is not true, x and y
cannot be independent. Even one place where the joint probability is not the product of the marginal
probabilities is enough. Actually the best way to prove non-independence is to look for zeroes. If
Px  4   y  1  0 and x and y are independent , then it must be true that Px  4  0 or
P y  1  0 . If this one is not enough to convince you, how about Px  4   y  3  .10
 Px  4 P y  3  .10.60   .06 . Notice that the second row is not proportional to the first row or any
other row.
A zero covariance or correlation would be the consequence of independence, but it is not true that a zero
correlation or covariance would prove independence. We have already seen one example where there is a
zero correlation, but no independence.
x
1
y
2
3
Px 
xPx 
4
 .00

 0
 .10

0.1
0.4 
5
0
0.4
6
.10 

0
.40 

0.5
2.0 
3.0 
.30
.10
x Px  1.6  10 .0  18 .0 
2
P y 
.10
yP y  y 2 P y 
0.1
0.1
.30
0.6
1.2
.60
1.8
5.4
1.00
2.5
6.7
5.4
29 .6
 Px   1 (a check),   Ex   xPx  5.4 , E x    x
 P y   1 ,   E  y    yP y   2.5 and E y    y P y   6.7 .
2
To summarize
x
2
2
Px   29 .6 ,
2
y
(ii) Compute E x  and Varx  .
 xPx  5.4 ,
From the above  x  E x  
 
 x Px  
Var y   E y     y P y   
Varx  E x 2   x2 
2
2
2
y
2
2
x
2
y
 29.6  5.42  0.44 ,  y  E  y  
 6.7  2.52  0.45 .
 
 yP y   2.5 and
 
(iii) Compute Covx, y  or  xy and Corr x, y  or  xy
051 .10 61  0
0 0.6
 .00 41



E xy  
xyPxy     042 .30 52 062   0 3.0 0   13 .5
 .10 43 .10 53 .40 63 1.2 1.5 7.2
 xy  Covxy  Exy   x  y  13.5  5.42.5  0.00 .

So that  xy 
 xy
 x y

0.00
 0.00 .
0.44 0.45
4
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2
 0 is as low
The correlation and covariance are zero, indicating no linear relation between x and y .  xy
as you can get on a zero to one scale, indicating no relationship. Note that 1   xy  1 always!
(iv) Compute Ex  y  and Var x  y  from the results in (ii) and (iii).
Ex  y   Ex  E y    x   y  5.4  2.5  7.9 and
Var x  y    x2   y2  2 xy  Var x   Var  y   2Covx, y   0.44  0.45  20.00   0.89 .
(v) Compute Cov3x  3, y  and Corr 3x  3, y 
251v2out says Cov(ax  b, cy  d )  acCov( x, y) and Corr (ax  b, cy  d )  (sign(ac))Corr ( x, y) , where
signac has the value 1 or 1 depending on whether the product of a and c is negative or
positive. a  3 and c  1 .  xy  Covx, y   0.22 and  xy  Corrx, y   .4958
Cov(3x  3, 1y  0)  31Cov( x, y)  30  0
Corr (3x  3,1y  0)  (sign(31))Corr ( x, y)  10  0.
Minitab Computations: For the third version of the problem.
————— 4/3/2006 5:16:02 PM ————————————————————
Welcome to Minitab, press F1 for help.
Results for: 1gr2-061.MTW
MTB > WSave "C:\Documents and Settings\rbove\My Documents\Minitab\2gr2061.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\1gr2-061.MTW'
MTB > Execute "C:\Documents and Settings\rbove\My
Documents\Minitab\251popcorr.txt" 1.
Executing from file: C:\Documents and Settings\rbove\My
Documents\Minitab\251popcorr.txt
Executing from file: 251name973.txt
Executing from file: marg973.txt
Data Display
Row
1
2
3
4
5
P1
0.0
0.0
0.1
0.0
0.0
P2
0.0
0.3
0.1
0.0
0.0
#This is the original data with zeroes in the unused columns.
P3
0.1
0.0
0.4
0.0
0.0
P4
0
0
0
0
0
P5
0
0
0
0
0
x
4
5
6
0
0
C16
0.1
0.4
0.5
0.0
0.0
y
1
2
3
0
0
C18
0.1
0.3
0.6
0.0
0.0
Executing from file: meany973.txt
Data Display
sumpx
sumpy
#Just a check to see if probabilities add to 1.
1.00000
1.00000
Executing from file: meanz973.txt
Executing from file: exysq973.txt
Data Display
Row
1
2
C20
0.0
0.0
C21
0.0
3.0
#In C20-24 products for E(xy), x products in C25-26, y products in C27-28.
C22
0.6
0.0
C23
0
0
C24
0
0
C25
0.4
2.0
C26
1.6
10.0
C27
0.1
0.6
C28
0.1
1.2
5
251solngr2
3
4
5
4/12/06
1.2
0.0
0.0
1.5
0.0
0.0
(Open this document in 'Page Layout' view!)
7.2
0.0
0.0
0
0
0
0
0
0
3.0
0.0
0.0
18.0
0.0
0.0
1.8
0.0
0.0
5.4
0.0
0.0
Data Display
Ex
Ex2
Ey
Ey2
5.40000
29.6000
2.50000
6.70000
Executing from file: xyvar973.txt
Data Display
varx
vary
0.440000
0.450000
Data Display
K20
K21
K22
K23
K24
Exy
#Column sums for E(xy) and E(xy)
1.20000
4.50000
7.80000
0
0
13.5000
Executing from file: cov973.txt
Data Display
covxy
corr
sdx
sumpx
sdy
sumpy
#Covariances, correlation, standard deviations.
0
0
0.663325
1.00000
0.670820
1.00000
Executing from file: tb2973.txt
Executing from file: tb2s973.MTB
Data Display
Row
1
2
3
4
5
6
7
8
C30
0.0
0.0
0.1
0.0
0.0
0.1
0.4
1.6
C31
0.0
0.3
0.1
0.0
0.0
0.4
2.0
10.0
#Our table
C32
0.1
0.0
0.4
0.0
0.0
0.5
3.0
18.0
C33
0
0
0
0
0
0
0
0
C34
0
0
0
0
0
0
0
0
C35
0.1
0.3
0.6
0.0
0.0
1.0
5.4
29.6
C36
0.1
0.6
1.8
0.0
0.0
2.5
C37
0.1
1.2
5.4
0.0
0.0
6.7
Data Display (WRITE)
0.0
0.0
0.1
0.0
0.0
0.1
0.4
1.6
0.0
0.3
0.1
0.0
0.0
0.4
2.0
10.0
0.1
0.0
0.4
0.0
0.0
0.5
3.0
18.0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0.1
0.3
0.6
0.0
0.0
1.0
5.4
29.6
0.1
0.6
1.8
0.0
0.0
2.5
*
*
0.1
1.2
5.4
0.0
0.0
6.7
*
*
* NOTE * Column lengths not equal.
6
251solngr2
4/12/06
(Open this document in 'Page Layout' view!)
Executing from file: tb3973.txt
Data Display (WRITE)
0.0
0.0
1.2
0.0
0.0
0.0
3.0
1.5
0.0
0.0
0.6
0.0
7.2
0.0
0.0
0
0
0
0
0
#Products for E(xy)
0
0
0
0
0
2) For the following sample (i) Compute the sample mean and variance of x , (ii) Compute Covx, y  or
s xy and Corr x, y  or rxy , (iii) Compute the sample mean and variance of x  y  from the results in (i)
and (ii). (iv) Compute Cov7 x  3,2 y  and Corr 7 x  3,2 y  using the formulas in section K4 of
251v2out or section C1 of 251var2. Note that 2 y  2 y  0 .
Original Version:
y
x
15
-2
4
4
6
2
2
5
1
7
1
6
3
5
10
-4
(i) Compute the sample mean and variance of x .
The entire table is below with required computations.
Row
x
x2
xy
y
y2
1
2
3
4
5
6
7
8
15
4
6
2
1
1
3
10
42
225
16
36
4
1
1
9
100
392
-2
4
2
5
7
6
5
-4
23
4
16
4
25
49
36
25
16
175
-30
16
12
10
7
6
15
-40
-4
 x  42,  x  392 ,  y  23,  y  175 ,
 x  42  5.250 and y   y  23  2.875 .
Then x 
So n  8,
s x2 
2
x
n
2
 nx
2
8
2
n 1

n
8
392  85.250 
 24 .50000 , s 2y 
7
2
y
2
and
 xy
 ny 2
n 1

  4. .
175  82.875 2
 15 .55357 .
7
( s x  24 .50000  4.94975 and s x  15 .55357  3.94380 ).
(ii) Compute Covx, y  or s xy and Corr x, y  or rxy .
s xy  Covx, y  
rxy 
s xy
sx s y
 xy  nxy   4  85.250 2.875   17.82143
n 1
 Corr x, y  
7
 17 .82143
and
  0.833468  .0.9129 . The correlation and
24 .50000 15 .55357
7
251solngr2
4/12/06
(Open this document in 'Page Layout' view!)
covariance are negative, indicating a tendency of y to fall when x rises. rxy2  .8335 is fairly large on a
zero to one scale, indicating that the relationship is moderately strong. Note that 1  rxy  1 always!
(iii) Compute the sample mean and variance of
x y  x y 
x  y  from the results in (i) and (ii).
42 23 65


 5.250  2.875  8.125 .
8
8
8
s x2 y  s x2  s 2y  2s xy  24 .50000  15 .55357  217 .82143   4.4107 .
(iv) Compute Cov7 x  3,2 y  and Corr 7 x  3,2 y  using the formulas in section K4 of 251v2out or
section C1 of 251var2. Note that 2 y  2 y  0 .
251v2out says Cov(ax  b, cy  d )  acCov( x, y)
and Corr (ax  b, cy  d )  (sign(ac))Corr ( x, y) , where signac has the value 1 or 1 depending
on whether the product of a and c is negative or positive. a  7 and c  2 .
s xy  Covx, y   17.82143 and rxy  Corrx, y   .0.9129.
Cov(7 x  3,  2 y  0)  72Cov( x, y)  14 17.82143   249 .5000
Corr (7 x  3,  2 y  0)  (sign(72))Corr ( x, y)  10.9129   0.9129 .
Version with 5 added:
y
x
15
-2
4
4
6
2
2
5
1
7
1
6
3
5
10
-4
(i) Compute the sample mean and variance of x .
The entire table is below with required computations.
Row
x
x2
xy
y
y2
1
2
3
4
5
6
7
8
20
4
6
2
1
1
3
10
47
400
16
36
4
1
1
9
100
567
-2
4
2
5
7
6
5
-4
23
4
16
4
25
49
36
25
16
175
-40
16
12
10
7
6
15
-40
-14
 x  47,  x  567 ,  y  23,  y  175 ,
 x  47  5.875 and y   y  23  2.875 .
Then x 
So n  8,
2
n
8
2
n
and
 xy
 14 . .
8
8
251solngr2
s x2 
x
2
4/12/06
 nx 2
n 1

(Open this document in 'Page Layout' view!)
567  85.875 2
 41 .55357 , s 2y 
7
y
2
 ny 2
n 1

175  82.875 2
 15 .55357 .
7
( s x  41 .55357  6.44621 and s x  15 .55357  3.94380 ).
(ii) Compute Covx, y  or s xy and Corr x, y  or rxy .
s xy  Covx, y  
rxy 
s xy
sx s y
 xy  nxy   14  85.875 2.875   21.30357
n 1
 Corr x, y  
7
 21 .30357
and
  0.702209  .0.8380 . The correlation and
41 .55357 15 .55357
covariance are negative, indicating a tendency of y to fall when x rises. rxy2  .7022 is fairly large on a
zero to one scale, indicating that the relationship is moderately strong. Note that 1  rxy  1 always!
(iii) Compute the sample mean and variance of
x y  x y 
x  y  from the results in (i) and (ii).
47 23 70


 5.875  2.875  8.750 .
8
8
8
s x2 y  s x2  s 2y  2s xy  41 .55357  15 .55357  221 .30357   14 .5000 .
(iv) Compute Cov7 x  3,2 y  and Corr 7 x  3,2 y  using the formulas in section K4 of 251v2out or
section C1 of 251var2. Note that 2 y  2 y  0 .
251v2out says Cov(ax  b, cy  d )  acCov( x, y)
and Corr (ax  b, cy  d )  (sign(ac))Corr ( x, y) , where signac has the value 1 or 1 depending
on whether the product of a and c is negative or positive. a  7 and c  2 .
s xy  Covx, y   21.30357 and rxy  Corrx, y   .0.8380.
Cov(7 x  3,  2 y  0)  72Cov( x, y)  14 21.30357   298 .2450
Corr (7 x  3,  2 y  0)  (sign(72))Corr ( x, y)  10.9129   0.8380 .
Version with 10 added:
y
x
25
-2
4
4
6
2
2
5
1
7
1
6
3
5
10
-4
(i) Compute the sample mean and variance of x .
The entire table is below with required computations.
x
x2
Row
xy
y
y2
1
2
25
4
625
16
-2
4
4
16
-50
16
9
251solngr2
3
4
5
6
7
8
4/12/06
6
2
1
1
3
10
52
(Open this document in 'Page Layout' view!)
36
4
1
1
9
100
792
2
5
7
6
5
-4
23
4
25
49
36
25
16
175
12
10
7
6
15
-40
-24
 x  52,  x  792 ,  y  23,  y  175 ,
 x  52  6.500 and y   y  23  2.875 .
Then x 
So n  8,
s x2 
2
x
n
2
2
8
 nx
2
n 1

n
8
792  86.500 2
 64 .85714 , s 2y 
7
y
2
and
 xy
 ny 2
n 1

 24 . .
175  82.875 2
 15 .55357 .
7
( s x  65 .85714  8.05339 and s x  15 .55357  3.94380 ).
(ii) Compute Covx, y  or s xy and Corr x, y  or rxy .
s xy  Covx, y  
rxy 
s xy
sx s y
 xy  nxy   24  86.500 2.875   24.78571 and
n 1
 Corr x, y  
7
 24 .78571
  0.6089966  0.7804 . The correlation and
64 .85714 15 .55357
covariance are negative, indicating a tendency of y to fall when x rises. rxy2  .6090 is near the middle
on a zero to one scale, indicating that the relationship is moderately strong. Note that 1  rxy  1 always!
(iii) Compute the sample mean and variance of
x y  x y 
x  y  from the results in (i) and (ii).
52 23 75


 6.500  2.875  9.375 .
8
8
8
s x2 y  s x2  s 2y  2s xy  64 .85714  15 .55357  224 .78571   30 .8392 .
(iv) Compute Cov7 x  3,2 y  and Corr 7 x  3,2 y  using the formulas in section K4 of 251v2out or
section C1 of 251var2. Note that 2 y  2 y  0 .
251v2out says Cov(ax  b, cy  d )  acCov( x, y)
and Corr (ax  b, cy  d )  (sign(ac))Corr ( x, y) , where signac has the value 1 or 1 depending
on whether the product of a and c is negative or positive. a  7 and c  2. .
s xy  Covx, y   24.78571 and rxy  Corrx, y   .0.7804.
Cov(7 x  3,  2 y  0)  72Cov( x, y)  14 24.78571   346 .9999
Corr (7 x  3,  2 y  0)  (sign(72))Corr ( x, y)  10.7804   0.7804 .
Minitab run of all three problems.
————— 4/3/2006 8:08:56 PM ————————————————————
Welcome to Minitab, press F1 for help.
Results for: 1gr2-063.MTW
#Results for first version.
10
251solngr2
4/12/06
(Open this document in 'Page Layout' view!)
MTB > exec '251samcov'
Executing from file: 251samcov.MTB
Executing from file: var973.MTB
Data Display
K5
171.500
Descriptive Statistics: C1
Variable
C1
N
8
N*
0
Mean
5.25
SE Mean
1.75
Data Display
Row
1
2
3
4
5
6
7
8
C1
15
4
6
2
1
1
3
10
StDev
4.95
Minimum
1.00
Q1
1.25
Median
3.50
Q3
9.00
Maximum
15.00
#Input to compute Var(x)
C2
225
16
36
4
1
1
9
100
Data Display
sum
sumsq
count
smean
svar
sdev
DF
sterr
42.0000
392.000
8.00000
5.25000
24.5000
4.94975
7.00000
1.75000
Executing from file: var973.MTB
Data Display
svar
108.875
Descriptive Statistics: C1
Variable
C1
N
8
N*
0
Data Display
Row
1
2
3
4
5
6
7
8
C1
-2
4
2
5
7
6
5
-4
Mean
2.88
SE Mean
1.39
StDev
3.94
Minimum
-4.00
Q1
-1.00
Median
4.50
Q3
5.75
Maximum
7.00
#Input to compute Var(y)
C2
4
16
4
25
49
36
25
16
Data Display
sum
sumsq
count
smean
svar
sdev
DF
sterr
23.0000
175.000
8.00000
2.87500
15.5536
3.94380
7.00000
1.39434
11
251solngr2
4/12/06
(Open this document in 'Page Layout' view!)
Data Display
Row
1
2
3
4
5
6
7
8
x
15
4
6
2
1
1
3
10
xsq
225
16
36
4
1
1
9
100
y
-2
4
2
5
7
6
5
-4
ysq
4
16
4
25
49
36
25
16
xy
-30
16
12
10
7
6
15
-40
#Table for computation.
Data Display
sumx
sumx2
sumy
sumy2
sumxy
n
xbar
ybar
svarx
svary
scovxy
sx
sy
rxy
rxy2
42.0000
392.000
23.0000
175.000
-4.00000
8.00000
5.25000
2.87500
24.5000
15.5536
-17.8214
4.94975
3.94380
-0.912945
0.833468
#Final Output.
MTB > WOpen "C:\Documents and Settings\rbove\My Documents\Minitab\1gr2064.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\rbove\My
Documents\Minitab\1gr2-064.MTW'
Worksheet was saved on Mon Apr 03 2006
Results for: 1gr2-064.MTW
#Results for second version.
MTB > exec '251samcov'
Executing from file: 251samcov.MTB
Executing from file: var973.MTB
Data Display
K5
290.875
Descriptive Statistics: C1
Variable
C1
N
8
N*
0
Mean
5.88
SE Mean
2.28
StDev
6.45
Minimum
1.00
Q1
1.25
Median
3.50
Q3
9.00
Maximum
20.00
Data Display
Row
1
2
3
C1
20
4
6
C2
400
16
36
12
251solngr2
4
5
6
7
8
2
1
1
3
10
4/12/06
(Open this document in 'Page Layout' view!)
4
1
1
9
100
Data Display
sum
sumsq
count
smean
svar
sdev
DF
sterr
47.0000
567.000
8.00000
5.87500
41.5536
6.44621
7.00000
2.27908
Executing from file: var973.MTB
Data Display
svar
108.875
Descriptive Statistics: C1
Variable
C1
N
8
N*
0
Mean
2.88
SE Mean
1.39
StDev
3.94
Minimum
-4.00
Q1
-1.00
Median
4.50
Q3
5.75
Maximum
7.00
Data Display
Row
1
2
3
4
5
6
7
8
C1
-2
4
2
5
7
6
5
-4
C2
4
16
4
25
49
36
25
16
Data Display
sum
sumsq
count
smean
svar
sdev
DF
sterr
23.0000
175.000
8.00000
2.87500
15.5536
3.94380
7.00000
1.39434
Data Display
Row
1
2
3
4
5
6
7
8
x
20
4
6
2
1
1
3
10
xsq
400
16
36
4
1
1
9
100
y
-2
4
2
5
7
6
5
-4
ysq
4
16
4
25
49
36
25
16
xy
-40
16
12
10
7
6
15
-40
13
251solngr2
4/12/06
(Open this document in 'Page Layout' view!)
Data Display
sumx
sumx2
sumy
sumy2
sumxy
n
xbar
ybar
svarx
svary
scovxy
sx
sy
rxy
rxy2
47.0000
567.000
23.0000
175.000
-14.0000
8.00000
5.87500
2.87500
41.5536
15.5536
-21.3036
6.44621
3.94380
-0.837979
0.702209
Results for: 1gr2-065.MTW
#Results for third version.
MTB > exec '251samcov'
Executing from file: 251samcov.MTB
Executing from file: var973.MTB
Data Display
K5
454.000
Descriptive Statistics: C1
Variable
C1
N
8
N*
0
Mean
6.50
SE Mean
2.85
StDev
8.05
Minimum
1.00
Q1
1.25
StDev
3.94
Minimum
-4.00
Q1
-1.00
Median
3.50
Q3
9.00
Maximum
25.00
Data Display
Row
1
2
3
4
5
6
7
8
C1
25
4
6
2
1
1
3
10
C2
625
16
36
4
1
1
9
100
Data Display
sum
sumsq
count
smean
svar
sdev
DF
sterr
52.0000
792.000
8.00000
6.50000
64.8571
8.05339
7.00000
2.84730
Executing from file: var973.MTB
Data Display
svar
108.875
Descriptive Statistics: C1
Variable
C1
N
8
N*
0
Mean
2.88
SE Mean
1.39
Median
4.50
Q3
5.75
Maximum
7.00
Data Display
Row
1
C1
-2
C2
4
14
251solngr2
2
3
4
5
6
7
8
4
2
5
7
6
5
-4
4/12/06
(Open this document in 'Page Layout' view!)
16
4
25
49
36
25
16
Data Display
sum
sumsq
count
smean
svar
sdev
DF
sterr
23.0000
175.000
8.00000
2.87500
15.5536
3.94380
7.00000
1.39434
Data Display
Row
1
2
3
4
5
6
7
8
x
25
4
6
2
1
1
3
10
xsq
625
16
36
4
1
1
9
100
y
-2
4
2
5
7
6
5
-4
ysq
4
16
4
25
49
36
25
16
xy
-50
16
12
10
7
6
15
-40
Data Display
sumx
sumx2
sumy
sumy2
sumxy
n
xbar
ybar
svarx
svary
scovxy
sx
sy
rxy
rxy2
52.0000
792.000
23.0000
175.000
-24.0000
8.00000
6.50000
2.87500
64.8571
15.5536
-24.7857
8.05339
3.94380
-0.780382
0.608997
15