Statistics 2080B: Practice Problems - Matrices
Practice P

A2.
 A2.	 MATRIX A	MATRIX B      MATRIX C
3   1   3 	1   0   2	3
2   0   4 			0
-4   1   2 			2


a)  MATRIX AC
15
14
-8

b)  MATRIX BC
7

c) No, it is not possible to find the product AB since the number of columns in A(3) does not equal the number of rows in B(1).

A3.

a) AB is a 3 x 4 matrix
b) No, it is not possible to find the product BA because the number of columns in B(4) does not equal the number of rows in A(3).

A4.

a) BC has dimensions

b) CB has dimensions

c)
	     MATRIX CB
3   0   6
0   0   0
2   0   4


A7.


A =   3   0   2
-1   1   4

a) The identity matrix used to get the product IA is .

		I = 1  0
0  1


c) The identity matrix used to get the product AI is

			1  0  0
I = 	0  1  0
0  0  1


A8.

Since AB = I and BA = I then B is the inverse of A


AB = BA = I =   1   0   0
0   1   0
0   0   1


A18.

a) Y has dimensions

X has dimensions

b)

        X'X [X'X]-1 = I  I = 1  0  0
0  1  0
0  0  1


c) A matrix A is symmetric if aij = aji for all i and j.
d)

 MTB > read 10 x 3 m20  M20 = X  X has dimensions 10 x 3
DATA> 1 -2 4
DATA> 1 -2 4
DATA> 1 -1 1
DATA> 1 -1 1
DATA> 1 0 0
DATA> 1 0 0
DATA> 1 1 1
DATA> 1 1 1
DATA> 1 2 4
DATA> 1 2 4
MTB > trans m20 m21
MTB > print m21
MATRIX M21       M21 = X'  X' has dimensions 3 x 10

1   1   1   1   1   1   1   1   1   1
-2  -2  -1  -1   0   0   1   1   2   2
4   4   1   1   0   0   1   1   4   4

MTB > mult m21 m20 m22
MTB > print m22

MATRIX M22	  M22 = X'X	X'X has dimensions  3 x 3

10    0   20
0   20    0
20    0   68

MTB > invert m22 m23
MTB > print m23

MATRIX M23     M23 = [X'X]-1    [X'X] -1 has dimensions 3x3

0.242857  0.000000 -0.071429
0.000000  0.050000  0.000000
-0.071429  0.000000  0.035714

MTB > read 10 x 1 m24  M24 = Y  Y has dimensions 10 x 1
DATA> 1.1
DATA> 1.3
DATA> 2.0
DATA> 2.1
DATA> 2.7
DATA> 2.8
DATA> 3.4
DATA> 3.6
DATA> 4.1
DATA> 4.0

MTB > mult m21 m24 m25
MTB > print m25
MATRIX M25    M25 = X'Y  X'Y has dimensions 3 x 1

27.1
14.3
53.1

MTB > mult m23 m25 m26
MTB > print m26
MATRIX M26	M26 = [X'X]-1X'Y	[X'X]-1X'Y has dimensions 3 x 1


The least squares equation is

A21.

MTB > mult m20 m26 m27     M20 = X  M26= beta (hat) so M27 = Y(hat)
MTB > print m27
MATRIX M27

1.20143
1.20143
2.03429
2.03429
2.78857
2.78857
3.46429
3.46429
4.06143
4.06143

MTB > copy m24 c1		c1 = yi
MTB > copy m27 c2		c2 = yi(hat)
MTB > let c3 = c1 - c2 	c3 = ei = residuals
MTB > let c4 = c3**2
MTB > sum c4
SUM     =    0.061286  SSE = .061286


when x = 1

V22 = 0.357

t has 7 df

0.05 < p < 0.10

Reject the null hypothesis at since the p-value is less than . The data indicates curvature.

A.26

  MATRIX M30	M30 = l

1
1
1

MTB > trans m30 m31
MTB > mult m31 m23 m32
MTB > mult m32 m30 m33

MTB > copy m20 c10 c11 c12
MTB > print c10-c12

ROW   C10   C11   C12

1     1    -2     4
2     1    -2     4
3     1    -1     1
4     1    -1     1
5     1     0     0
6     1     0     0
7     1     1     1
8     1     1     1
9     1     2     4
10     1     2     4

MTB > regress c1 on 2 in c11 c12;
SUBC> predict 1 1.

The regression equation is
C1 = 2.79 + 0.715 C11 - 0.0393 C12

Predictor       Coef       Stdev    t-ratio        p
Constant     2.78857     0.04611      60.48    0.000
C11          0.71500     0.02092      34.17    0.000
C12         -0.03929     0.01768      -2.22    0.062

s = 0.09357     R-sq = 99.4%     R-sq(adj) = 99.2%

Analysis of Variance

SOURCE       DF          SS          MS         F        p
Regression    2     10.2677      5.1339    586.39    0.000
Error         7      0.0613      0.0088
Total         9     10.3290

SOURCE       DF      SEQ SS
C11           1     10.2245
C12           1      0.0432

Fit  Stdev.Fit         95% C.I.         95% P.I.
3.4643     0.0403   ( 3.3689, 3.5597)  ( 3.2233, 3.7053)


The confidence interval for E[y] when x=1, , s = 0.09357, and l'(x'x)-1l = 0.1857.

t has 7 df, t=1.895

(3.36,3.51)

A.27

prediction interval for a new value of y when x=1

the same as in A.26

(3.27,3.66)

A.28

a)

s = 1.1

b) confidence interval for E[y] when x=1.10, x2 = 1.21

l'(x'x)-1l = 0.1077

t has 22 df, t = 2.074

(8.60,10.10)

We are confident that the mean number of items produced per hour per worker when the piece rate, , is between 8.6 and 10.10 items/hour/worker.

c) prediction interval for individual worker who is paid per piece.

(6.945,11.747)

We are confident that if a worker is paid a piecework rate of per piece his productivity will be between 6,95 and 11.75.

d) SSE = 26.26

of the valuability is explained by the regression model.

Jonathan Payne
1999-03-11