251grass2-081
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Graded Assignment 2
Solution (13 pages)
Name: KEY
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. Note that neatness means paper neatly trimmed on the left side if it has been torn,
multiple pages stapled and paper written on only one side. The stapling is for your protection –
putting your name on every page helps too, I still have some unclaimed pages from old papers. Make
copies of your work before you hand it in.
1) Use the joint probability table below, which resembles the table in Problem K4. You will find a solved
problem somewhat like this in 251solnK0.
x
4
5
1
0
0
6
.40
y 2
0
.20
0
3 .40
0
0
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. For example, if the last two digits of your number are 30, the .40 on the
diagonal becomes .40 - .10 = .30 and a zero will become .10. The sum of the numbers in the table will still
be one.
For this 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 .
Solution: Only two solutions are given. The other solutions lie between them.
x
4
The table in question is
5
6 . a) If I use .01, I might get
.40
1
0
0
y 2
0
.20
0
3 .40
0
0
x
4
x
1
.01
5
0
6 . b) If I use .10, I might get
.39
4
1
.10
5
0
y 2
0
.19
0
y 2
0
.10
0
3
.39
.01
.01
3
.30
.10
.10
(i) Check for independence: First you need to find Px  and P y  .
6 .
.30
Look at the upper left hand
probability below. Its value is a).01 or b) .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. . We need to find out what these
probabilities are so we add the rows and columns to get marginal or total probabilities.
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x
1
a)
y
2
3
Px 





4
.01
x
P y 
.40
5
0
6
.39 

0
.19
0
.39
.01
.01

0.4 0.2 0.4
.19
.41
1.00
1
b)
y
2
3
Px 
`





4
.10
P y 
.40
5
0
6
.30 

0
.10
0
.30
.10
.10 

0.4 0.2 0.4
.10
.50
1.00
`
Thus we have a) Px  4 P y  1  .4.0.4  .16 and b) Px  4 P y  1  .4.0.4  .16 . Since these are
not equal to .01 in a) or .10 in b), 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  5   y  1  0  Px  5 P y  1  .2.40   .08 . Actually the fastest way to
prove non-independence is to look for zeroes. If Px  5   y  3  0 and x and y are independent,
then it must be true that Px  5  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.
Px   1 (a check for a valid distribution),
Lets finish the job we did in (i) by computing
x

 E x    xPx  , E x    x Px  ,  P y   1 , 
2
2
y
 E y  
 yP y  and E y    y
2
2
P y  .
The easiest way to do this is to multiply the items in the P y  column by the items in the y column to get
the yP y  column and then to multiply the items in the yP y  column by the items in the y column to get
the y 2 P y  column. Then multiply the items in the Px  row by the items in the x row to get the
xPx  row and then multiply the items in the xPx  row by the items in the x row to get the x 2 Px  row.
Then add up all the rows and columns outside the original table.
x
4
5
6
P y  yP y  y 2 P y 
 .01
1
0
.39 
.40 0.40
0.40


y
2
0
.
19
0
.
19
0
.
38
0.76


a)


3
.39
.01
.01
.41 1.23
3.69


Px 
0.4
0.2
0.4
1.00 2.01
4.85
xPx 
`1.6  1.0  2.4 
5.0
x 2 Px 
6.4  5.0  14 .4 
25 .8
x
4
5
6
P y  yP y  y 2 P y 
 .10
1
0
.30 
.40
0.4
0.4


y
2
0
.10
0
.10
0.2
0.8

b)


3
.30
.10
.10
.50
1.5
4.5


Px 
0.4
0.2
0.4 1.00
2.1
5.7
xPx 
`1.6  1.0  2.4 
5.0
x 2 Px  6.4  5.0  14 .4  25 .8
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 Px   1 (a check),   Ex   xPx  5.0 , E x    x Px  25.8 ,
 P y   1 ,   E y    yP y   2.01 and E y    y P y   4.85 .
b)  Px   1 (a check),   E x    xPx   5.0 , E x    x Px   25 .8 ,  P y   1 ,
  E  y    yP y   2.1 and E y    y P y   5.7 .
2
To summarize a)
2
x
2
2
y
2
2
x
2
2
y
(ii) Compute E x  and Varx  : Remember that variances and standard deviations are never negative.
We actually need means and variances for both x and y . From the above
 xPx  5.0 ,  x2  Varx  Ex 2   x2   x 2 Px   x2  25.8  5.02  0.8
0.8  0.8944 ),   E  y    yP y   2.01 and   Var  y   E y      y P y   
a)  x  E x  
( x 
2
y
y
2
2
y
2
2
y
 4.85  2.012  0.8099 (  y  0.8099  0.8999 );
 xPx  5.0 ,  x2  Varx  Ex 2   x2   x 2 Px   x2  25.8  5.02  0.8
0.8  0.8944 ),   E  y    yP y   2.1 and   Var  y   E y      y P y   
b)  x  E x  
( x 

y
2
y
y
2
2
y
2
2
y
P y    y2  5.7  2.12  1.29 (  y  1.29  1.1358 ).
2
 
 
(iii) Compute Covx, y  or  xy and Corr x, y  or  xy : a) We must now compute E xy  by multiplying
each pair of values of x and y by their joint probabilities. We had  x  5.0 ,  x2  0.8 (  x  0.8944 ),
x
1
 y  2.01 ,  y2  0.8099 (  y 0.8999 ) and
y
2
3
Px 





4
.01
5
0
6
.39 

0
.19
0
.39
.01
.01

0.4 0.2 0.4
P y 
.40
.19
.41
.
1.00
`
051 .39 61  0.04
0
2.34 
 .0141



E xy  
xyPxy     042 .19 52 062   0
1.90
0   9.29
 .39 43 .0153 .0163 4.68 0.15 0.18 
To complete what we have done, write  xy  Covxy  Exy   x  y  9.29  5.02.01  0.78 .

So that  xy 
 xy
 0.76
  0.9014  .9494 .
0.8 0.8099
b) We must now compute E xy  by multiplying each pair of values of x and y by their joint probabilities.
 x y

We had  x  5.0 ,  x2  0.8 (  x  0.8944 ),  y  2.1 and  y2  1.29 (  y  1.1358 ). and
x
1
y
2
3
Px 





4
.10
6
.30 

0
.10
0
.30
.10
.10 

0.4 0.2 0.4
`
5
0
P y 
.40
.10
.50
1.00
.
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051 .30 61  0.4 0 1.8
 .10 41

E xy  
xyPxy     042 .10 52 062   0 1.0 0   10 .1
 .30 43 .10 53 .10 63 3.6 1.5 1.8
To complete what we have done, write  xy  Covxy  Exy   x  y  10.1  5.02.1  0.40 .

So that  xy 
 xy

 0.40
  0.1550  .3937 . In general, joint probability tables with only the
0.8 1.29
diagonals filled produce correlations close to +1 or -1. A Southwest to Northeast diagonal produces a
negative correlation and you can see that a) is much closer to a diagonal than b).
Remember that the correlation must be between -1 and +1!
Note that the strength of a correlation is found by squaring the correlation and measuring
 x y
2
the strength on a zero to one scale. In a) we had  xy  .9494 , so  xy
  .94942  .9014 and we can
say that there is a relatively strong tendency for y to rise as x falls. In b) we had  xy  .3937 , so
2
 xy
  .39372  .1550 and we can say that there is a relatively weak tendency for y to rise as x falls.
(iv) Compute Ex  y  and Var x  y  from the results in (ii) and (iii). How many of you ignored the
instructions and wrote down each value of x  y with its probability. What a great way to waste
time!
The formulas that you were given were Ex  y   Ex  E y    x   y and Var x  y 
  x2   y2  2 xy  Var x   Var  y   2Covx, y 
a) We had  x  5.0 ,  x2  0.8 (  x  0.8944 ),  y  2.01 ,  y2  0.8099 (  y 0.8999 ) and
 xy  0.78 .
Ex  y    x   y  5.0  2.01  7.01 and Var x  y    x2   y2  2 xy  0.8  0.8099  20.78   .0499
(  x y  .0499  .2234 )
b) We had  x  5.0 ,  x2  0.8 (  x  0.8944 ),  y  2.1 and  y2  1.29 (  y  1.1358 ). and
 xy  0.40 .
Ex  y    x   y  5.0  2.1  7.1 and Var x  y    x2   y2  2 xy  0.8  1.29  20.40   1.29
(  x y  1.29  1.136 )
(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 .
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 . Cov(3x  3, 1y  0)  31Cov( x, y)  3Covx, y 
Corr (3x  3,1y  0)  (sign(31))Corr ( x, y)  sign 3Corr x, y    1 Corrx, y 
a) We had  xy  0.78 and  xy  .9494 . So Cov(3x  3, 1y  0)  3Covx, y   30.78   2.34 and
Corr (3x  3,1y  0)  1 Corr x, y   .9494 .
b) We had  xy  0.40 and  xy  .3937 . So Cov(3x  3, 1y  0)  3Covx, y   30.40   1.20 and
Corr (3x  3,1y  0)  1 Corr x, y   .3937 .
251grass2-081
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2) The PHLX Gold/Silver SectorSM (XAUSM) is a capitalization-weighted index composed of 16 companies
involved in the gold and silver mining industry. XAU was set to an initial value of 100 in January 1979;
options commenced trading on December 19, 1983. The Dow-Jones Utility average is an average based on
the prices of 16 (I think) utility stocks. Both gold and utilities can attract cautious investors under certain
stock market conditions, so it is interesting to look at how they move relative to one another. The values of
these two indices for 20 very recent trading days are given on the next page.
You are expected to work with 11 of the 20
x
y
observations shown above. Use the first 10 rows
Row pick Date PHLXGS
DJUT
of data and pick one more row by finding the
1
* 03/05 203.32 496.60
row marked with the second-to-last digit of your
2
* 03/04 195.62 489.97
student number. (i) Compute the sample mean
3
* 03/03 202.98 482.76
4
* 02/29 196.58 477.50
and standard deviation of x , (ii) Compute
5
* 02/28 202.84 492.40
Covx, y  or s xy and Corr x, y  or rxy , (iii)
6
* 02/27 197.84 496.04
7
8
9
10
11
12
13
14
15
16
17
18
19
20
*
*
*
*
0
1
2
3
4
5
6
7
8
9
02/26
02/25
02/22
02/21
02/20
02/19
02/15
02/14
02/13
02/12
02/11
02/08
02/07
02/06
193.13
188.12
189.94
196.56
189.56
186.10
177.32
176.87
179.43
176.65
182.06
181.25
174.88
129.98
504.63
500.78
497.54
491.82
499.95
499.85
500.41
498.79
504.05
502.08
497.90
494.39
496.96
498.66
Compute the sample mean and variance of
x  y  from the results in (i) and (ii). (iv) The
coefficient of variation is computed by dividing
the standard deviation by the mean. Compute a
coefficient of variation for x , y and x  y and
compare the relative safety of investing in
precious metal stocks, investing in utilities and
doing both. (v) Just for practice, compute
Cov6 x  3, y  and Corr 6 x  3, y  using the
formulas in section K4 of 251v2out or section
C1 of 251var2. Note that y  1y  0 .
Solution: I really don’t have time to do all of these by hand. Fortunately, (and not fortuitously!), when I
developed this problem I had the computer do all of the individual means, variances, sums, sums of
squares, xy columns, covariances and correlations. I will present them as soon as feasible. The results are
all relatively similar. So, using Minitab as a calculator, I will do the requested calculations for the original
data.
(i) Compute the sample mean and standard deviation of x .
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
x
03/05
03/04
03/03
02/29
02/28
02/27
02/26
02/25
02/22
02/21
02/20
02/19
02/15
02/14
02/13
02/12
02/11
02/08
02/07
02/06
y
x2
y2
xy
203.32 496.60 41339.0 246612 100969
195.62 489.97 38267.2 240071
95848
202.98 482.76 41200.9 233057
97991
196.58 477.50 38643.7 228006
93867
202.84 492.40 41144.1 242458
99878
197.84 496.04 39140.7 246056
98137
193.13 504.63 37299.2 254651
97459
188.12 500.78 35389.1 250781
94207
189.94 497.54 36077.2 247546
94503
196.56 491.82 38635.8 241887
96672
189.56 499.95 35933.0 249950
94771
186.10 499.85 34633.2 249850
93022
177.32 500.41 31442.4 250410
88733
176.87 498.79 31283.0 248791
88221
179.43 504.05 32195.1 254066
90442
176.65 502.08 31205.2 252084
88692
182.06 497.90 33145.8 247904
90648
181.25 494.39 32851.6 244421
89608
174.88 496.96 30583.0 246969
86908
129.98 498.66 16894.8 248662
64816
3721.03 9923.08 697304
4924233 1845390
251grass2-081
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 x  3721 .03,  y  9923 .08,
 x  3721 .03  186 .0515 and
xy  1845390 . Thus x 
To summarize the results of these computations n  20 ,
 x  697304 ,  y  4924233 and 
 y  9923 .08  496 .1540 . s   x
y
2
2
n
2
x
20
n
2
 nx
2
n 1

20
697304  20 186 .0515 
5000 .787

 263 .199 .
19
19
2
Minitab says 263.201. s x  263 .199  16 .2234 Minitab says 16.2235.
Though the mean and variance of y were not requested, we will need them.
s 2y 
y
2
 ny 2
n 1

4924233  20 496 .1540 2 857 .166

 45 .1140 Minitab says 45.1342.
19
19
s y  45.1140  6.7167 Minitab says 6.7167.
(ii) Compute Covx, y  or s xy and Corr x, y  or rxy .
s xy 
 x  x  y  y    xy  nx y
n 1
Minitab says -42.8130
rxy 
s xy
sx s y

n 1
42 .8378
263 .199 45 .1140


1845390  20 186 .0515 496 .1540 
813 .9190

 42 .8378 .
19
19
42.8378 2

263 .199 45.1140 
(iii) Compute the sample mean and variance of
0.1545  .3931 . Minitab says -.3928.
x  y  from the results in (i) and (ii).
We have x  186 .0515 , y  496.1540 , s x2  263.199 , s 2y  45 .1140 and s xy  42 .8378 .
So x  y  x  y  186.0515  496.1540  682.2055 and s x2 y  s x2  s 2y  2 s xy
 263 .199  45.1140  242.8378  = 222.6374 .
(iv) The coefficient of variation is computed by dividing the standard deviation by the mean. Compute a
coefficient of variation for x , y and x  y and compare the relative safety of investing in precious metal
stocks, investing in utilities and doing both.
s x  16 .2234 , s y  6.7167 and s x  y  222.6374  14.9210 .
s x y
sy
sx
16 .2234
14 .9210
6.7167

 .0872 , C y 

 .0219 . This

 .0135 and C x  y 
x 186 .0515
682
.2055
y 496 .154
x y
seems to show that investing in precious metals is much more (over 6 times as) risky than either utilities or
a 50-50 strategy of doing both. However, because of the negative covariance, the 50-50 strategy is only
about 62% riskier than utilities alone.
Cx 
(v) Just for practice, compute Cov6 x  3, y  and Corr 6 x  3, y  using the formulas in section K4 of
251v2out or section C1 of 251var2. Note that y  1y  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  6 and c  1. From the work above we have s xy  Covx, y 
 42 .8378 and rxy  Corrx, y   .3931 . This means that Cov(6 x  3, 1y  0)  61Cov( x, y)
 642.8378   42.8378 and Corr (6 x  3, 1y  0)  (sign(61))Corr ( x, y)  1.3931   .3931 .
251grass2-081
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Summary of solutions of individualized sample problems.
The following chart includes the means, medians and standard deviations of all the data sets you could have
used including the original set.
Descriptive Statistics: PHLXGS, DJUT, PHLXGS_1, DJUT_1, PHLXGS_2, DJUT_2,
etc. Note that N* is the number of missing numbers and is always zero.
Variable
PHLXGS
DJUT
PHLXGS_1
DJUT_1
PHLXGS_2
DJUT_2
PHLXGS_3
DJUT_3
PHLXGS_4
DJUT_4
PHLXGS_5
DJUT_5
PHLXGS_6
DJUT_6
PHLXGS_7
DJUT_7
PHLXGS_8
DJUT_8
PHLXGS_9
DJUT_9
PHLXGS_10
DJUT_10
n
20
20
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
N*
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Mean
186.05
496.15
196.04
493.64
195.73
493.63
194.93
493.68
194.89
493.53
195.12
494.01
194.87
493.83
195.36
493.45
195.29
493.13
194.71
493.36
190.63
493.52
SE Mean StDev
3.63 16.22
1.50
6.72
1.66
5.49
2.41
7.99
1.80
5.98
2.41
7.98
2.33
7.72
2.42
8.03
2.36
7.83
2.38
7.91
2.19
7.26
2.53
8.40
2.38
7.88
2.47
8.18
2.02
6.71
2.37
7.85
2.07
6.87
2.33
7.72
2.50
8.30
2.35
7.80
6.25 20.74
2.38
7.90
Minimum
129.98
477.50
188.12
477.50
186.10
477.50
177.32
477.50
176.87
477.50
179.43
477.50
176.65
477.50
182.06
477.50
181.25
477.50
174.88
477.50
129.98
477.50
Q1
177.85
492.90
189.94
489.97
189.94
489.97
189.94
489.97
189.94
489.97
189.94
489.97
189.94
489.97
189.94
489.97
189.94
489.97
189.94
489.97
189.94
489.97
Median
188.84
497.72
196.56
496.04
196.56
496.04
196.56
496.04
196.56
496.04
196.56
496.04
196.56
496.04
196.56
496.04
196.56
494.39
196.56
496.04
196.56
496.04
Q3
196.57
500.30
202.84
499.95
202.84
499.85
202.84
500.41
202.84
498.79
202.84
500.78
202.84
500.78
202.84
497.90
202.84
497.54
202.84
497.54
202.84
498.66
The First Three Data Sets
Row
1
2
3
4
5
6
7
8
9
10
11
PHLXGS_1
203.32
195.62
202.98
196.58
202.84
197.84
193.13
188.12
189.94
196.56
189.56
DJUT_1
496.60
489.97
482.76
477.50
492.40
496.04
504.63
500.78
497.54
491.82
499.95
PHLXGS_2
203.32
195.62
202.98
196.58
202.84
197.84
193.13
188.12
189.94
196.56
186.10
DJUT_2
496.60
489.97
482.76
477.50
492.40
496.04
504.63
500.78
497.54
491.82
499.85
PHLXGS_3
203.32
195.62
202.98
196.58
202.84
197.84
193.13
188.12
189.94
196.56
177.32
DJUT_3
496.60
489.97
482.76
477.50
492.40
496.04
504.63
500.78
497.54
491.82
500.41
PHLXGS_4
203.32
195.62
202.98
196.58
202.84
197.84
193.13
188.12
189.94
196.56
176.87
DJUT_4
496.60
489.97
482.76
477.50
492.40
496.04
504.63
500.78
497.54
491.82
498.79
251grass2-081
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The First data set. The p-value represents the probability that the correlation is exactly zero. Because the
data set is relatively small, most of these are disturbingly high. We cover this in ECO 252. The covariance
table gives the two variances (the positive numbers) and the covariance. This is followed by the sum of the
x column, the sum of the x squared column, the sum of the y column, the sum of the y squared column, the
contents of the xy column and the sum of the xy column.
Correlations: PHLXGS_1, DJUT_1
Pearson correlation of PHLXGS_1 and DJUT_1 = -0.500
P-Value = 0.117
Covariances: PHLXGS_1, DJUT_1
PHLXGS_1
DJUT_1
PHLXGS_1
30.1775
-21.9399
DJUT_1
63.8500
Sum of PHLXGS_1
Sum of PHLXGS_1 = 2156.49
Sum of Squares of PHLXGS_1
Sum of squares (uncorrected) of PHLXGS_1 = 423070
Sum of DJUT_1
Sum of DJUT_1 = 5429.99
Sum of Squares of DJUT_1
Sum of squares (uncorrected) of DJUT_1 = 2681074
Xy Column
100969
94503
95848
96672
97991
94771
93867
99878
98137
97459
94207
Sum of xy column
Sum of xy = 1064301
The Second data set. The p-value represents the probability that the correlation is exactly zero. Because
the data set is relatively small, most of these are disturbingly high. We cover this in ECO 252. The
covariance table gives the two variances (the positive numbers) and the covariance.
Correlations: PHLXGS_2, DJUT_2
Pearson correlation of PHLXGS_2 and DJUT_2 = -0.503
P-Value = 0.114
Covariances: PHLXGS_2, DJUT_2
PHLXGS_2
PHLXGS_2
35.7531
DJUT_2
-24.0285
MTB > sum c7
DJUT_2
63.7246
Sum of PHLXGS_2
Sum of PHLXGS_2 = 2153.03
Sum of Squares of PHLXGS_2
Sum of squares (uncorrected) of PHLXGS_2 = 421770
Sum of DJUT_2
Sum of DJUT_2 = 5429.89
Sum of Squares of DJUT_2
Sum of squares (uncorrected) of DJUT_2 = 2680974
Xy Column
100969
94503
95848
96672
Sum of xy column
Sum of xy = 1062552
97991
93022
93867
99878
98137
97459
94207
251grass2-081
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The Third data set. The p-value represents the probability that the correlation is exactly zero. Because the
data set is relatively small, most of these are disturbingly high. We cover this in ECO 252. The covariance
table gives the two variances (the positive numbers) and the covariance. This is followed by the sum of the
x column, the sum of the x squared column, the sum of the y column, the sum of the y squared column, the
contents of the xy column and the sum of the xy column.
Correlations: PHLXGS_3, DJUT_3
Pearson correlation of PHLXGS_3 and DJUT_3 = -0.491
P-Value = 0.125
Covariances: PHLXGS_3, DJUT_3
PHLXGS_3
PHLXGS_3
59.6715
DJUT_3
-30.4791
MTB > sum c9
DJUT_3
64.4502
Sum of PHLXGS_3
Sum of PHLXGS_3 = 2144.25
Sum of Squares of PHLXGS_3
Sum of squares (uncorrected) of PHLXGS_3 = 418579
Sum of DJUT_3
Sum of DJUT_3 = 5430.45
Sum of Squares of DJUT_3
Sum of squares (uncorrected) of DJUT_3 = 2681534
Xy Column
100969
94503
95848
96672
97991
88733
93867
99878
98137
97459
94207
Sum of xy column
Sum of xy = 1058263
The Fourth, Fifth, Sixth and Seventh Data Sets
Row
1
2
3
4
5
6
7
8
9
10
11
PHLXGS_4
203.32
195.62
202.98
196.58
202.84
197.84
193.13
188.12
189.94
196.56
176.87
DJUT_4
496.60
489.97
482.76
477.50
492.40
496.04
504.63
500.78
497.54
491.82
498.79
PHLXGS_5
203.32
195.62
202.98
196.58
202.84
197.84
193.13
188.12
189.94
196.56
179.43
DJUT_5
496.60
489.97
482.76
477.50
492.40
496.04
504.63
500.78
497.54
491.82
504.05
PHLXGS_6
203.32
195.62
202.98
196.58
202.84
197.84
193.13
188.12
189.94
196.56
176.65
DJUT_6
496.60
489.97
482.76
477.50
492.40
496.04
504.63
500.78
497.54
491.82
502.08
PHLXGS_7
203.32
195.62
202.98
196.58
202.84
197.84
193.13
188.12
189.94
196.56
182.06
DJUT_7
496.60
489.97
482.76
477.50
492.40
496.04
504.63
500.78
497.54
491.82
497.90
251grass2-081
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The Fourth data set. The p-value represents the probability that the correlation is exactly zero. Because
the data set is relatively small, most of these are disturbingly high. We cover this in ECO 252. The
covariance table gives the two variances (the positive numbers) and the covariance. This is followed by the
sum of the x column, the sum of the x squared column, the sum of the y column, the sum of the y squared
column, the contents of the xy column and the sum of the xy column.
Correlations: PHLXGS_4, DJUT_4
Pearson correlation of PHLXGS_4 and DJUT_4 = -0.450
P-Value = 0.165
Covariances: PHLXGS_4, DJUT_4
PHLXGS_4
DJUT_4
PHLXGS_4
61.2749
-27.8627
DJUT_4
62.5074
Sum of PHLXGS_4
Sum of PHLXGS_4 = 2143.8
Sum of Squares of PHLXGS_4
Sum of squares (uncorrected) of PHLXGS_4 = 418420
Sum of DJUT_4
Sum of DJUT_4 = 5428.83
Sum of Squares of DJUT_4
Sum of squares (uncorrected) of DJUT_4 = 2679916
Xy Column
100969
94503
95848
96672
97991
88221
93867
99878
98137
97459
94207
Sum of xy column
Sum of xy = 1057751
The Fifth data set. The p-value represents the probability that the correlation is exactly zero. Because the
data set is relatively small, most of these are disturbingly high. We cover this in ECO 252. The covariance
table gives the two variances (the positive numbers) and the covariance. This is followed by the sum of the
x column, the sum of the x squared column, the sum of the y column, the sum of the y squared column, the
contents of the xy column and the sum of the xy column.
Correlations: PHLXGS_5, DJUT_5
Pearson correlation of PHLXGS_5 and DJUT_5 = -0.571
P-Value = 0.067
Covariances: PHLXGS_5, DJUT_5
PHLXGS_5
DJUT_5
PHLXGS_5
52.6440
-34.7710
DJUT_5
70.5561
Sum of PHLXGS_5
Sum of PHLXGS_5 = 2146.36
Sum of Squares of PHLXGS_5
Sum of squares (uncorrected) of PHLXGS_5 = 419332
Sum of DJUT_5
Sum of DJUT_5 = 5434.09
Sum of Squares of DJUT_5
Sum of squares (uncorrected) of DJUT_5 = 2685190
Xy Column
100969
94503
95848
96672
Sum of xy column
Sum of xy = 1059972
97991
90442
93867
99878
98137
97459
94207
251grass2-081
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The Sixth data set. The p-value represents the probability that the correlation is exactly zero. Because the
data set is relatively small, most of these are disturbingly high. We cover this in ECO 252. The covariance
table gives the two variances (the positive numbers) and the covariance. This is followed by the sum of the
x column, the sum of the x squared column, the sum of the y column, the sum of the y squared column, the
contents of the xy column and the sum of the xy column.
Correlations: PHLXGS_6, DJUT_6
Pearson correlation of PHLXGS_6 and DJUT_6 = -0.527
P-Value = 0.096
Covariances: PHLXGS_6, DJUT_6
PHLXGS_6
PHLXGS_6
62.0722
DJUT_6
-33.9731
MTB > sum c15
DJUT_6
66.9524
Sum of PHLXGS_6
Sum of PHLXGS_6 = 2143.58
Sum of Squares of PHLXGS_6
Sum of squares (uncorrected) of PHLXGS_6 = 418342
Sum of DJUT_6
Sum of DJUT_6 = 5432.12
Sum of Squares of DJUT_6
Sum of squares (uncorrected) of DJUT_6 = 2683208
Xy Column
100969
94503
95848
96672
97991
88692
93867
99878
98137
97459
94207
Sum of xy column
Sum of xy = 1058222
The Seventh data set. The p-value represents the probability that the correlation is exactly zero. Because
the data set is relatively small, most of these are disturbingly high. We cover this in ECO 252. The
covariance table gives the two variances (the positive numbers) and the covariance. This is followed by the
sum of the x column, the sum of the x squared column, the sum of the y column, the sum of the y squared
column, the contents of the xy column and the sum of the xy column.
Correlations: PHLXGS_7, DJUT_7
Pearson correlation of PHLXGS_7 and DJUT_7 = -0.455
P-Value = 0.160
Covariances: PHLXGS_7, DJUT_7
PHLXGS_7
DJUT_7
PHLXGS_7
45.0180
-23.9488
DJUT_7
61.6431
Sum of PHLXGS_7
Sum of PHLXGS_7 = 2148.99
Sum of Squares of PHLXGS_7
Sum of squares (uncorrected) of PHLXGS_7 = 420283
Sum of DJUT_7
Sum of DJUT_7 = 5427.94
Sum of Squares of DJUT_7
Sum of squares (uncorrected) of DJUT_7 = 2679028
Xy Column
100969
94503
95848
96672
Sum of xy column
Sum of xy = 1060178
97991
90648
93867
99878
98137
97459
94207
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The Eighth, Ninth, and Tenth Data Sets
Row
1
2
3
4
5
6
7
8
9
10
11
PHLXGS_8
203.32
195.62
202.98
196.58
202.84
197.84
193.13
188.12
189.94
196.56
181.25
DJUT_8
496.60
489.97
482.76
477.50
492.40
496.04
504.63
500.78
497.54
491.82
494.39
PHLXGS_9
203.32
195.62
202.98
196.58
202.84
197.84
193.13
188.12
189.94
196.56
174.88
DJUT_9
496.60
489.97
482.76
477.50
492.40
496.04
504.63
500.78
497.54
491.82
496.96
PHLXGS_10
203.32
195.62
202.98
196.58
202.84
197.84
193.13
188.12
189.94
196.56
129.98
DJUT_10
496.60
489.97
482.76
477.50
492.40
496.04
504.63
500.78
497.54
491.82
498.66
The Eighth data set. The p-value represents the probability that the correlation is exactly zero. Because
the data set is relatively small, most of these are disturbingly high. We cover this in ECO 252. The
covariance table gives the two variances (the positive numbers) and the covariance. This is followed by the
sum of the x column, the sum of the x squared column, the sum of the y column, the sum of the y squared
column, the contents of the xy column and the sum of the xy column.
Correlations: PHLXGS_8, DJUT_8
Pearson correlation of PHLXGS_8 and DJUT_8 = -0.365
P-Value = 0.269
Covariances: PHLXGS_8, DJUT_8
PHLXGS_8
DJUT_8
PHLXGS_8
47.2326
-19.3816
DJUT_8
59.6386
Sum of PHLXGS_8
Sum of PHLXGS_8 = 2148.18
Sum of Squares of PHLXGS_8
Sum of squares (uncorrected) of PHLXGS_8 = 419988
Sum of DJUT_8
Sum of DJUT_8 = 5424.43
Sum of Squares of DJUT_8
Sum of squares (uncorrected) of DJUT_8 = 2675546
Xy Column
100969
94503
95848
96672
Sum of xy column
Sum of xy = 1059138
97991
89608
93867
99878
98137
97459
94207
251grass2-081
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The Ninth data set. The p-value represents the probability that the correlation is exactly zero. Because the
data set is relatively small, most of these are disturbingly high. We cover this in ECO 252. The covariance
table gives the two variances (the positive numbers) and the covariance. This is followed by the sum of the
x column, the sum of the x squared column, the sum of the y column, the sum of the y squared column, the
contents of the xy column and the sum of the xy column.
Correlations: PHLXGS_9, DJUT_9
Pearson correlation of PHLXGS_9 and DJUT_9 = -0.391
P-Value = 0.235
Covariances: PHLXGS_9, DJUT_9
PHLXGS_9
PHLXGS_9
68.8073
DJUT_9
-25.2805
MTB > sum c21
DJUT_9
60.8866
Sum of PHLXGS_9
Sum of PHLXGS_9 = 2141.81
Sum of Squares of PHLXGS_9
Sum of squares (uncorrected) of PHLXGS_9 = 417720
Sum of DJUT_9
Sum of DJUT_9 = 5427
Sum of Squares of DJUT_9
Sum of squares (uncorrected) of DJUT_9 = 2678093
Xy Column
100969
94503
95848
96672
97991
86908
93867
99878
98137
97459
94207
Sum of xy column
Sum of xy = 1056438
The Tenth data set. The p-value represents the probability that the correlation is exactly zero. Because the
data set is relatively small, most of these are disturbingly high. We cover this in ECO 252. The covariance
table gives the two variances (the positive numbers) and the covariance. This is followed by the sum of the
x column, the sum of the x squared column, the sum of the y column, the sum of the y squared column, the
contents of the xy column and the sum of the xy column.
Correlations: PHLXGS_10, DJUT_10
Pearson correlation of PHLXGS_10 and DJUT_10 = -0.316
P-Value = 0.344
Covariances: PHLXGS_10, DJUT_10
PHLXGS_10
DJUT_10
PHLXGS_10
430.1543
-51.7384
DJUT_10
62.3721
Sum of PHLXGS_10
Sum of PHLXGS_10 = 2096.91
Sum of Squares of PHLXGS_10
Sum of squares (uncorrected) of PHLXGS_10 = 404032
Sum of DJUT_10
Sum of DJUT_10 = 5428.7
Sum of Squares of DJUT_10
Sum of squares (uncorrected) of DJUT_10 = 2679786
Xy Column
100969
94503
95848
96672
Sum of xy column
Sum of xy = 1034346
97991
64816
93867
99878
98137
97459
94207