251solngr2-051 4/12/05
(Open this document in 'Page Layout' view!) Graded Assignment 2
Name:
Class days and time:
Student number:
Modify the data below as follows: Add the last digit of your student number to 7 in problem 1; Add the second to last number to the 9
in problem 2.
1) For the following 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), (iv)
Compute Cov5 x  3, y  and Corr 5 x  3, y  using the formulas in section K4 of 251v2out or section C1 of 251var2. Note
y  1y  0 .
that
x
3
1
.11
5
.08
7
.16
y 3
.10
.14
.10
9
.12
.08
.11
2) For the following sample (i) Compute the sample mean and variance of
or rxy , (iii) Compute the sample mean and variance of
x , (ii) Compute Covx, y  or s xy and Corr x, y 
x  y  from the results in (i) and (ii). (iv) Compute Cov5x  3,2 y 
Corr 5x  3,2 y  using the formulas in section K4 of 251v2out or section C1 of 251var2. Note that 2 y  2 y  0 .
y
and
x
9
4
6
2
1
1
3
10
2
4
2
5
7
6
5
-3
Solution: Assume that Seymour Butz’s number is 555555. The table becomes
x
3
.11
5
.08
12
.16
y 3
.10
.14
.10
9
.12
.08
.11
1
(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 .11 and it represents Px  3   y  1 . If x and y are independent , we
would have Px  3   y  1  Px  3 P y  1  .33.37   .1221 . 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  7   y  3  .10
 Px  7 P y  3  .35.30   .1050 . Notice that the second row is not proportional to the first 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.
1
251solngr2-051 4/12/05
(ii) Compute E x  and Var x  : Looking below, we find Ex   6.93 and Varx   2.7938 .
(iii) Compute Covx, y  and Corr x, y  .
To summarize
x
2
Px   1 ,
3
5
12
P y 
yP y  y P y 


1
.11 .08
.16
.35
0.35
0.35
 x  E x  
xPx   6.93


y
3
.14
.10 
.34
1.02
3.06
2
2
 .10
Ex 
x Px   63 .75
 .12
9
.08
.11
.31
2.97
25 .11


P y   1 ,
Px 
.33  .30
 .37  1.00
4.16
28 .52
 y  Ey 
yP y   4.16
xPx  0.99  1.50  4.44  6.93

  

x 2 Px  2.97  7.50  53 .28  63 .75
and
E y2 


  y
2
P y   28 .52
 .1131  .0851  .16 12 1  0.33  0.40  1.92 
E xy  
xyPxy     .10 33  .14 53  .10 12 3   0.90  2.10  3.60   27 .97
 .12 39  .0859  .1112 9  3.24  43 .60  11 .88 
 xy  Covx, y   Exy   x  y  27.97  6.934.16  0.8588 ,

 
 
 x2  E x 2   x2  63.75  6.932  15.7251 and  y2  E y 2   y2  28.52  4.162  11.2144 . So that
 xy  Corr x, y  
 xy
 0.8588

0.8588 2

15 .7251 11 .2144 
.0041823  0.06467 .
15 .7251 11 .2144
(  x  3.9655,  y  3.3488 ) The correlation and covariance are negative, indicating a tendency of y to
 x y

2
 .0041823 hardly exists on a zero to one scale, indicating that the relationship is
fall when x rises.  xy
barely there. Note that 1   xy  1 always!
(iv) Compute Ex  y  and Var x  y  .
Ex  y   Ex  E y    x   y  6.93  4.16  11.09 and
Var x  y    x2   y2  2 xy  Varx   Var y   2Covx, y   15 .7251  11.2144  20.8588   25 .2219
To check this do the computations below.
If we run down the columns of the table:
y
x
3
3
3
5
5
5
12
12
12
1
3
9
1
3
9
1
3
9
x y
4
6
12
6
8
14
13
15
21
Px, y 
.11
.10
.12
.08
.14
.08
.16
.10
.11
2
251solngr2-051 4/12/05
Now collect probabilities that belong to the same value. For example, P6  .10  .08  .18 .
4
6
8
12
13
14
15
21
E x  y  
( x  y) Px  y 
Px  y 
x y

.11
.18
.14
.12
.16
.08
.10
.11
1.00
x  y Px, y   9.32

0.44
1.08
1.12
1.44
2.08
1.12
1.50
2.31
11.09
( x  y) 2 Px  y 
1.76
6.48
8.96
17.28
27.04
15.68
22.50
48.51
148.21

Var x  y    x2 y  E x  y 2   x2 y  148.21 11.092  25.2219
(v) Compute Cov5 x  3, y  and Corr 5x  3, y  using the formulas in section K4 of 251v2out or section
C1 of 251var2. Note that y  1y  0 .
Solution: 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  5 and c  1 .
 xy  Covx, y   0.18588 and  xy  Corrx, y   0.06467
Cov(5x  3, 1y  0)  51Cov( x, y)  50.8588   4.2940
Corr (5x  3,1y  0)  (sign(51))Corr ( x, y)  10.06467   0.06467 .
2) 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5 x  3,2 y  and Corr 5x  3,2 y  using the formulas in section
K4 of 251v2out or section C1 of 251var2. Note that 2 y  2 y  0 .
The original data
y
x
9
2
4
4
6
2
2
5
1
7
1
6
3
5
10
-3
Becomes
y
x
14
2
4
4
6
2
2
5
1
7
1
6
3
5
10
-3
3
251solngr2-051 4/12/05
The entire table is below with required computations.
Row
x
x2
xy
y
y2
1
2
3
4
5
6
7
8
sum
14
4
6
2
1
1
3
10
41
196
16
36
4
1
1
9
100
363
2
4
2
5
7
6
5
-3
28
4
16
4
25
49
36
25
9
168
28
16
12
10
7
6
15
-30
64
 x  41,  x  363 ,  y  28,  y  168 , and  xy
 x  41  5.125 and y   y  28  3.500 .
Then x 
So n  7,
2
n
(i)
s x2
x

2
8
2
n
 nx 2
n 1
 64 .
8
363  85.125 2

 21 .8393 , s 2y 
7
y
2
 ny 2
n 1

28  83.500 2
 10 .0000 .
7
( s x  21 .8393  4.67325 and s y  10.0000  3.16228 ).
(ii) s xy  Covx, y  
rxy 
s xy
sx s y
 xy  nxy  64  85.125 3.500   11.3571 and
n 1
 Corr x, y  
8 1
 11 .3531
 0.7685 . The correlation and covariance are negative,
21 .8393 10 .0000
indicating a tendency of y to fall when x rises. rxy2  .5906 is fairly large on a zero to one scale,
indicating that the relationship is moderately strong. Note that 1  rxy  1 always!
41 28 69


 8.625 .
8
8
8
(iii) x  y  x  y 
s x2 y  s x2  s 2y  2s xy  21 .8393  10 .0000  211 .3571   9.125 .
To check this do the computations below.
x
x y
y
 x  y 2
14
4
6
2
1
1
3
10
2
4
2
5
7
6
5
-3
So x  y  x  y 
 x  y 

2
16
8
8
7
8
7
8
7
69
256
64
64
49
64
49
64
49
659
69
 8.625 as above and
8

n x y

2
659  8698 
2
 9.125 . Notice how much larger the variation is in x and
n 1
7
y individually than in x  y . This is reflected in the small variance.
Computer results follow. N* means missing measurements.
s x2 y
Variable
x
y
x+y
N
8
8
8
N*
0
0
0
Mean
5.13
3.50
8.63

SE Mean
1.65
1.12
1.07
StDev
4.67
3.16
3.02
Minimum
1.00
-3.00
7.00
Q1
1.25
2.00
7.00
Median
3.50
4.50
8.00
Q3
9.00
5.75
8.00
Maximum
14.00
7.00
16.00
4
251solngr2-051 4/12/05
(iv) Compute Cov5 x  3,2 y  and Corr 5x  3,2 y  using the formulas in section K4 of 251v2out or
section C1 of 251var2. Note that 2 y  2 y  0 .
Solution: 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  5 and c  1 .
s xy  Covx, y   7.21429 and rxy  Corrx, y   .0.828475.
Cov(5x  3,  2 y  0)  52Cov( x, y)  10 11.3531   113 .531
Corr (5x  3,  2 y  0)  (sign(52))Corr ( x, y)  10.7685   0.7685 .
Appendix: Minitab Computations (This is mostly a reminder to me of how I checked my work –
but there is enough info here to run the routines and I’m happy to give them to anyone who wants them.)
Population Correlation Problem
————— 4/12/2005 9:12:27 PM ————————————————————
Welcome to Minitab, press F1 for help.
Results for: 1gr2-051aa.MTW
MTB > WSave "C:\Documents and Settings\rbove\My Documents\Minitab\1gr2051aa.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\1gr2-051aa.MTW'
MTB > echo
MTB > Execute "C:\Documents and Settings\rbove\My
Documents\Minitab\251popcorr.mtb" 1.
Executing from file: C:\Documents and Settings\rbove\My
Documents\Minitab\251popcorr.mtb
MTB > #251popcorr
MTB > # Computes population covariance and correlation
MTB > # Put x in C15, y in c17
MTB > # Put a joint probability table in c10 - c14
MTB > # Fill table with zeros to make it 5 by 5.
MTB > name k10 'varx'
MTB > name k11 'vary'
MTB > name k12 'Exy'
MTB > name k13 'covxy'
MTB > name k14 'corr'
MTB > name k15 'sdx'
MTB > name k16 'sumpx'
MTB > name k17 'sdy'
MTB > name k18 'sumpy'
MTB > name k25 'Ex'
MTB > name k26 'Ex2'
MTB > name k27 'Ey'
MTB > name k28 'Ey2'
MTB > execute 'marg973'
Executing from file: marg973.MTB
(Input was Probabilities in c10-c14, x in c15, y in c17)
MTB > #marg973.mtb #part of 251popcorr Computes marginal probabilities
MTB > let c18=c10+c11+c12+c13+c14
MTB > let c16(1)=sum(c10)
MTB > let c16(2)=sum(c11)
MTB > let c16(3)=sum(c12)
MTB > let c16(4)=sum(c13)
MTB > let c16(5)=sum(c14)
5
251solngr2-051 4/12/05
MTB > let k16=sum(c16)
MTB > let k18=sum(c18)
MTB > print c10-c18
#These are sums of x and y
#probabilities and should be 1.
Data Display
Row
1
2
3
4
5
C10
0.11
0.10
0.12
0.00
0.00
C11
0.08
0.14
0.08
0.00
0.00
C12
0.16
0.10
0.11
0.00
0.00
C13
0
0
0
0
0
C14
0
0
0
0
0
C15
3
5
12
0
0
C16
0.33
0.30
0.37
0.00
0.00
C17
1
3
9
0
0
C18
0.35
0.34
0.31
0.00
0.00
MTB > end
(End means the end of an exec and a return to Popcorr)
MTB > execute 'meany973'
Executing from file: meany973.MTB
MTB > #meany.973.mtb
part of 251popcorr
MTB > print k16,k18
Data Display
sumpx
sumpy
1.00000
1.00000
(A check to see if probabilities add to 1)
MTB > let c20=c10*c17
(c20-c24 are products for E(xy))
MTB > let c20=c20*c15(1)
MTB > let c21=c11*c17
MTB > let c21=c21*c15(2)
MTB > let c22=c12*c17
MTB > let c22=c22*c15(3)
MTB > let c23=c13*c17
MTB > let c23=c23*c15(4)
MTB > let c24=c14*c17
MTB > let c24=c24*c15(5)
MTB > let c25=c15*c16
# xP(x)
MTB > let k25=sum(c25)
# E(x)
MTB > let c27=c17*c17
MTB > let k27=sum(c27)
# E(y)?
MTB > end
MTB > execute 'meanz973'
Executing from file: meanz973.MTB
MTB > #meanz973.mtb part of 251popcorr
MTB > let c27=c17*c18
MTB > let k27=sum(c27) #E(y)
MTB > end
MTB > execute 'exysq973'
Executing from file: exysq973.MTB
MTB > #exysq973.mtb
part of 251popcorr
MTB > let c26=c15*c25 # xsqP(x)
MTB > let c28=c17*c27 # ysqP(y)
MTB > let k26=sum(c26) # E(xsq)
MTB > let k28=sum(c28) # E(ysq)
MTB > print c20-c28
Data Display
Row
1
2
3
4
5
C20
0.33
0.90
3.24
0.00
0.00
C21
0.4
2.1
3.6
0.0
0.0
C22
1.92
3.60
11.88
0.00
0.00
C23
0
0
0
0
0
C24
0
0
0
0
0
C25
0.99
1.50
4.44
0.00
0.00
C26
2.97
7.50
53.28
0.00
0.00
C27
0.35
1.02
2.79
0.00
0.00
C28
0.35
3.06
25.11
0.00
0.00
MTB > print k25-k28
6
251solngr2-051 4/12/05
Data Display
Ex
Ex2
Ey
Ey2
6.93000
63.7500
4.16000
28.5200
MTB > end
MTB > execute 'xyvar973'
Executing from file: xyvar973.MTB
MTB > #xyvar973.mtb part of popcorr
MTB > let k10=k26-k25*k25
#variance of x
MTB > let k11=k28-k27*k27
#variance of y
MTB > print k10 k11
Data Display
varx
vary
MTB
MTB
MTB
MTB
MTB
MTB
MTB
15.7251
11.2144
>
>
>
>
>
>
>
let k20=sum(c20)
(Column sums)
let k21=sum(c21)
let k22=sum(c22)
let k23=sum(c23)
let k24=sum(c24)
let k12 = k20+k21+k22+k23+k24 # E(xy)
print k20-k24, k12
Data Display
K20
K21
K22
K23
K24
Exy
4.47000
6.10000
17.4000
0
0
27.9700
MTB > end
MTB > execute 'cov973'
Executing from file: cov973.MTB
MTB > #cov973.mtb part of popcorr
MTB > let k13=k12-k25*k27
#Covariance
MTB > let k15=sqrt(k10)
#St. dev of x
MTB > let k17=sqrt(k11)
#St. dev of y
MTB > let k14=k13/k15
MTB > let k14=k14/k17
#Corr(x, y)
MTB > print k13-k18
Data Display
covxy
corr
sdx
sumpx
sdy
sumpy
-0.858800
-0.0646707
3.96549
1.00000
3.34879
1.00000
Covariance
Correlation
Std. deviation of x
Sum of x probabilities
Std. deviation of y
Sum of y probabilities
MTB > end
MTB > execute 'tb2973'
Executing from file: tb2973.MTB
MTB > #tb2973.mtb Part of 251popcorr
MTB > let k30=30
MTB > let k31=10
MTB > let k32=1
MTB > execute 'tb2s973' 5
Executing from file: tb2s973.MTB
7
251solngr2-051 4/12/05
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#tb2s973.mtb Subroutine
# Part of popcorr
let ck30=ck31
let ck30(6)=c16(k32)
let ck30(7)=c25(k32)
let ck30(8)=c26(k32)
let k30=k30+1
let k31=k31+1
let k32=k32+1
end
#tb2s973.mtb Subroutine
# Part of popcorr
let ck30=ck31
let ck30(6)=c16(k32)
let ck30(7)=c25(k32)
let ck30(8)=c26(k32)
let k30=k30+1
let k31=k31+1
let k32=k32+1
end
#tb2s973.mtb Subroutine
# Part of popcorr
let ck30=ck31
let ck30(6)=c16(k32)
let ck30(7)=c25(k32)
let ck30(8)=c26(k32)
let k30=k30+1
let k31=k31+1
let k32=k32+1
end
#tb2s973.mtb Subroutine
# Part of popcorr
let ck30=ck31
let ck30(6)=c16(k32)
let ck30(7)=c25(k32)
let ck30(8)=c26(k32)
let k30=k30+1
let k31=k31+1
let k32=k32+1
end
#tb2s973.mtb Subroutine
# Part of popcorr
let ck30=ck31
let ck30(6)=c16(k32)
let ck30(7)=c25(k32)
let ck30(8)=c26(k32)
let k30=k30+1
let k31=k31+1
let k32=k32+1
end
let c35=c18
let c35(6)=k18
let c35(7)=k25
let c35(8)=k26
let c36=c27
let c36(6)=k27
let c37=c28
let c37(6)=k28
of tb2973
of tb2973
of tb2973
of tb2973
of tb2973
8
251solngr2-051 4/12/05
MTB > print c30-c37
Data Display
Row
1
2
3
4
5
6
7
8
C30
0.11
0.10
0.12
0.00
0.00
0.33
0.99
2.97
C31
0.08
0.14
0.08
0.00
0.00
0.30
1.50
7.50
(The final table)
C32
0.16
0.10
0.11
0.00
0.00
0.37
4.44
53.28
C33
0
0
0
0
0
0
0
0
C34
0
0
0
0
0
0
0
0
C35
0.35
0.34
0.31
0.00
0.00
1.00
6.93
63.75
C36
0.35
1.02
2.79
0.00
0.00
4.16
C37
0.35
3.06
25.11
0.00
0.00
28.52
MTB > write c30-c37.
Data Display (WRITE)
0.11
0.10
0.12
0.00
0.00
0.33
0.99
2.97
0.08
0.14
0.08
0.00
0.00
0.30
1.50
7.50
0.16
0.10
0.11
0.00
0.00
0.37
4.44
53.28
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0.35
0.34
0.31
0.00
0.00
1.00
6.93
63.75
0.35
1.02
2.79
0.00
0.00
4.16
*
*
0.35
3.06
25.11
0.00
0.00
28.52
*
*
* NOTE * Column lengths not equal.
MTB > end.
MTB > execute 'tb3973'
Executing from file: tb3973.MTB
MTB > #tb3973.mtb final printout for 252popcorr
MTB >
MTB > write c20-c24;
SUBC> replace.
Data Display (WRITE)
0.33
0.90
3.24
0.00
0.00
0.4
2.1
3.6
0.0
0.0
1.92
3.60
11.88
0.00
0.00
0
0
0
0
0
(Products for E(x y) again)
0
0
0
0
0
MTB > end
MTB > end
Sample Correlation Problem
MTB > exec '251samcov'
Executing from file: 251samcov.MTB
MTB > #251samcov Computes sample variances
MTB > # and covariances using 'var973'
MTB > #Input is x column in c40, y column in c42
MTB > # Example in Covex
MTB > name c40 'x'
MTB > name c41 'xsq'
MTB > name c42 'y'
MTB > name c43 'ysq'
MTB > name c44 'xy'
MTB > name k40 'sumx'
MTB > name k41 'sumx2'
MTB > name k42 'sumy'
MTB > name k43 'sumy2'
MTB > name k44 'sumxy'
MTB > name k45 'n'
MTB > name k46 'xbar'
9
251solngr2-051 4/12/05
MTB > name k47 'ybar'
MTB > name k48 'svarx'
MTB > name k49 'svary'
MTB > name k50 'scovxy'
MTB > name k51 'sx'
MTB > name k52 'sy'
MTB > name k53 'rxy'
MTB > name k54 'rxy2'
MTB > let c1=c40
(Input for sample stats in c40, c42)
MTB > execute 'var973'
Executing from file: var973.MTB (An exec for computing variance is used for both x and y.)
MTB > #var973.mtb
MTB > #computes sample variance of data in C1.
MTB > let k1=sum(c1)
MTB > let c2=c1 * c1
MTB > let k2=sum(c2)
MTB > let k3=count(c1)
MTB > let k4=k1/k3 #mean
MTB > let k5=k3*k4*k4
MTB > let k5=k2-k5
MTB > print k5
Data Display
svar
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MTB
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MTB
152.875
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name k1 'sum'
name k2 'sumsq'
name k3 'count'
name k4 'smean'
name k5 'svar'
name k6 'sdev'
name k7 'DF'
name k8 'sterr'
let k7 = k3 - 1
let k5 = k5/k7
let k6 = sqrt(k5)
let k8 = k5/k3
let k8 = sqrt(k8)
describe c1
(DF = n – 1)
(Standard error)
Descriptive Statistics: C1
Variable
C1
N
8
N*
0
Mean
5.13
SE Mean
1.65
StDev
4.67
Minimum
1.00
Q1
1.25
Median
3.50
Q3
9.00
Maximum
14.00
MTB > print c1-c2
Data Display
Row
1
2
3
4
5
6
7
8
C1
14
4
6
2
1
1
3
10
(x and x squared)
C2
196
16
36
4
1
1
9
100
10
251solngr2-051 4/12/05
MTB > print k1-k8
Data Display
sum
sumsq
count
smean
svar
41.0000
363.000
8.00000
5.12500
21.8393
sdev
DF
sterr
4.67325
7.00000
1.65224
(A sample variance for x)
MTB > end
MTB > let C41=c2
MTB > let k40=k1
MTB > let k41=k2
MTB > let k46=k4
MTB > let k48=k5
MTB > let k51=k6
MTB > let c1=c42
(Input for sample stats in c40, c42)
MTB > execute 'var973'
Executing from file: var973.MTB (An exec for computing variance is used for both x and y.)
MTB > #var973.mtb
MTB > #computes sample variance of data in C1.
MTB > let k1=sum(c1)
MTB > let c2=c1 * c1
MTB > let k2=sum(c2)
MTB > let k3=count(c1)
MTB > let k4=k1/k3 #mean
MTB > let k5=k3*k4*k4
MTB > let k5=k2-k5
MTB > print k5
Data Display
svar
MTB
MTB
MTB
MTB
MTB
MTB
MTB
MTB
MTB
MTB
MTB
MTB
MTB
70.0000
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name k1 'sum'
name k2 'sumsq'
name k3 'count'
name k4 'smean'
name k5 'svar'
name k6 'sdev'
name k7 'DF'
name k8 'sterr'
let k7 = k3 - 1
let k5 = k5/k7
let k6 = sqrt(k5)
let k8 = k5/k3
let k8 = sqrt(k8)
MTB > describe c1
Descriptive Statistics: C1
Variable
C1
N
8
N*
0
Mean
3.50
SE Mean
1.12
StDev
3.16
Minimum
-3.00
Q1
2.00
Median
4.50
Q3
5.75
Maximum
7.00
11
251solngr2-051 4/12/05
MTB > print c1-c2
Data Display
Row
1
2
3
4
5
6
7
8
C1
2
4
2
5
7
6
5
-3
(y and y squared)
C2
4
16
4
25
49
36
25
9
MTB > print k1-k8
Data Display
sum
sumsq
count
smean
svar
sdev
DF
sterr
MTB
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28.0000
168.000
8.00000
3.50000
10.0000
3.16228
7.00000
1.11803
(A sample variance for y)
end
let C43=c2
let k42=k1
let k43=k2
let k45=k3
let k47=k4
let k49=k5
let k52=k6
let c44=c40*c42
let k44=sum(c44)
let k50=k45*k46*k47
let k50=k44-k50
let k50=k50/k7
let k53=k51*k52
let k53=k50/k53
let k54=k53*k53
Print c40-c44
Data Display
Row
1
2
3
4
5
6
7
8
x
14
4
6
2
1
1
3
10
xsq
196
16
36
4
1
1
9
100
y
2
4
2
5
7
6
5
-3
ysq
4
16
4
25
49
36
25
9
xy
28
16
12
10
7
6
15
-30
12
251solngr2-051 4/12/05
MTB > Print k40-k54
Data Display
sumx
sumx2
sumy
sumy2
sumxy
n
xbar
ybar
svarx
svary
scovxy
sx
sy
rxy
rxy2
41.0000
363.000
28.0000
168.000
64.0000
8.00000
5.12500
3.50000
21.8393
10.0000
-11.3571
4.67325
3.16228
-0.768511
0.590609
(Sum of x)
(Sum of x2)
(Sum of y)
(Sum of y2)
(Sum of xy)
(Sample size)
(Sample mean of x)
(Sample mean of y)
(Sample variance of x)
(Sample variance of y)
(Sample covariance)
(Std deviation for x)
(Std deviation of y)
(Sample correlation)
(Sample correlation squared)
13