5/12/99 252y9943
ECO252 QBA2
Name
FINAL EXAM
Hour of Class Registered (Circle)
May 5, 1999
MWF 10 11 TR 12:30 2:00
Note: If this is the only thing you look at before taking the final, you are badly cheating yourself.
Problems like 5e, 5f and 6 appeared on the 1998 final. People who used this final and did not read
the problems carefully got very wrong answers to them.
Note: If you still think that a large p-value means that a coefficient is significant, you need a
conference with an audiologist. Further note that a p-value is a probability and can only be
compared with another probability (like the significance level).
Note: Have you reread “Things that You Should Never Do On a Statistics Exam …?” I think I could
have graded this exam by just looking for violations of these rules.
I. (16 points) Do all the following.
1.
Hand in your fourth regression problem (2 points) and answer the following questions.
a. For the regression of the number of hours of work against the number of machines, what
coefficients are significant at the 1% level? Why? What about the 5% level? (2)
b. Would you say that the regression of number of hours of work against the number of machines and
months of experience is more successful than the regression against machines alone? Why? (3)
c. What was the surprise that occurred when you did the stepwise regression? (2)
Solution:
The rule on p-value:
If the p-value is less than the significance level   reject the null
hypothesis; if the p-value is greater or equal than the significance
level, do not reject the null hypothesis.
a) part of the printout follows: (For the entire printout see 252x9943.)
MTB > print c1-c4
Data Display
Row
Hours
Number
Exper
Inter
1
2
3
4
5
6
7
8
9
10
1.0
3.1
17.0
14.0
6.0
1.8
11.5
9.3
6.0
12.2
1
3
10
8
5
1
10
5
4
10
12
8
5
2
10
1
10
2
6
18
12
24
50
16
50
1
100
10
24
180
MTB > brief 3
5/12/99 252y9943
MTB > regress c1 on 1 c2 'resid''pred'
Regression Analysis
The regression equation is
Hours = 0.10 + 1.42 Number
Predictor
Constant
Number
Coef
0.101
1.4192
s = 2.056
Stdev
1.267
0.1908
R-sq = 87.4%
t-ratio
0.08
7.44
p
0.939
0.000
R-sq(adj) = 85.8%
Analysis of Variance
SOURCE
Regression
Error
Total
Obs.
1
2
3
4
5
6
7
8
9
10
DF
1
8
9
Number
1.0
3.0
10.0
8.0
5.0
1.0
10.0
5.0
4.0
10.0
SS
233.84
33.83
267.67
Hours
1.000
3.100
17.000
14.000
6.000
1.800
11.500
9.300
6.000
12.200
MS
233.84
4.23
Fit
1.520
4.358
14.293
11.454
7.197
1.520
14.293
7.197
5.777
14.293
F
55.30
Stdev.Fit
1.108
0.830
1.047
0.785
0.664
1.108
1.047
0.664
0.727
1.047
p
0.000
Residual
-0.520
-1.258
2.707
2.546
-1.197
0.280
-2.793
2.103
0.223
-2.093
St.Resid
-0.30
-0.67
1.53
1.34
-0.61
0.16
-1.58
1.08
0.12
-1.18
Since the p-value column has .936 as the p-value for 0.101, the coefficient ‘constant,’ and .939 is
above the significance levels of .01 and .05, we can say that the constant is not significant at these levels. If
you look up the values of t for 8 degrees of freedom and significance levels of .005 and .025, you will find
that the t-ratio of 0.08 is less than the table value.
Likewise, since the p-value column has zero as the p-value for 0.1.4192, the coefficient of
‘Number,’ and zero is below the significance levels of .01 and .05, we can say that the constant is
significant at these levels. If you look up the values of t for 8 degrees of freedom and significance levels of
.005 and .025, you will find that the t-ratio of 0.08 is more than the table value.
b) part of the printout follows:
MTB > regress c1 on 2 c2 c3 'resid''pred'
Regression Analysis
The regression equation is
Hours = 1.62 + 1.53 Number - 0.293 Exper
Predictor
Constant
Number
Exper
Coef
1.6191
1.5333
-0.29311
s = 1.388
Stdev
0.9746
0.1335
0.09019
R-sq = 95.0%
t-ratio
1.66
11.49
-3.25
p
0.141
0.000
0.014
R-sq(adj) = 93.5%
Analysis of Variance
SOURCE
Regression
Error
Total
DF
2
7
9
SS
254.19
13.48
267.67
MS
127.09
1.93
F
65.99
p
0.000
2
5/12/99 252y9943
SOURCE
Number
Exper
Obs.
1
2
3
4
5
6
7
8
9
10
DF
1
1
Number
1.0
3.0
10.0
8.0
5.0
1.0
10.0
5.0
4.0
10.0
SEQ SS
233.84
20.34
Hours
1.000
3.100
17.000
14.000
6.000
1.800
11.500
9.300
6.000
12.200
Fit
-0.365
3.874
15.487
13.299
6.355
2.859
14.021
8.699
5.994
11.676
Stdev.Fit
0.946
0.579
0.796
0.776
0.518
0.854
0.712
0.644
0.495
1.071
Residual
1.365
-0.774
1.513
0.701
-0.355
-1.059
-2.521
0.601
0.006
0.524
St.Resid
1.34
-0.61
1.33
0.61
-0.28
-0.97
-2.12R
0.49
0.00
0.59
R denotes an obs. with a large st. resid.
This is one of the few really good ones. R-squared and R-squared adjusted went up, and , equally important,
the coefficient of ‘Number’ remained significant while the coefficient of ‘Exper’ was significant at the 5%
level.
c) part of the printout follows:
Stepwise Regression
F-to-Enter:
Response is
Step
Constant
Number
T-Ratio
Inter
T-Ratio
4.00
Hours
on
F-to-Remove:
4.00
3 predictors, with N =
1
0.1005
2
-0.4758
1.42
7.44
1.85
10.81
10
-0.040
-3.56
S
2.06
1.31
R-Sq
87.36
95.51
More? (Yes, No, Subcommand, or Help)
SUBC> y
No variables entered or removed
More? (Yes, No, Subcommand, or Help)
SUBC> n
MTB > print c1-c6
This was a surprise! Though the coefficients in column 1 are the same as those in the regression of ‘Hours’
against ‘Number’ above, the variable it brought in in column 2 is the interaction variable. I had assumed
that it would only help the other two variables to explain ‘Number,’ but the computer refused to bring in
‘Exper.’
3
5/12/99 252y9943
2.
The following pages show the regression of the variable 'mins', the winning time in minutes in a
triathlon, against some of the following independent variables:
'female' A dummy variable that is 1 if the contestant is female.
'swim' Number of miles of swimming
'bike'
Number of miles of biking
'run'
Number of miles of running
c6
‘swim’ multiplied by ‘female’
c7
‘bike’ multiplied by ‘female’
c8
‘run’ multiplied by ‘female’
c9
‘swim’ squared
c10
‘bike’ squared
c11
‘run’ squared
a. In the regression of ‘mins’ against ‘female’, ‘swim’, ‘bike’ and ‘run’, which coefficients have
signs that look wrong? Why? Which coefficients are not significant at the 99% confidence level?
(3)
b. Look at the regression of ‘mins’ against ‘run‘, c8 and c11 and the regression of ‘mins’ against
‘run’, and c8. Use   .10 . Does either seem to be an improvement over the regression of ‘mins’
against ‘run’ alone? Why?(2)
c. Explain the meaning of the F test in the regression of ‘mins’ against ‘female’, ‘swim’, ‘bike’ and
‘run’ . What is being tested and what are the conclusions? (2)
d. The printout concludes with a printout of the data and of a correlation matrix. What does this
suggest about the problems that are occurring with these regressions? (2)
Solution: The printout enclosed with the exam follows:
Worksheet size: 100000 cells
MTB > RETR 'C:\MINITAB\LR13-49.MTW'.
Retrieving worksheet from file: C:\MINITAB\LR13-49.MTW
Worksheet was saved on 5/ 3/1999
MTB > regress c1 on 4 c2 c3 c4 c5
Regression Analysis
The regression equation is
mins = - 24.6 + 35.5 female - 25.0 swim + 7.13 bike - 6.37 run
Predictor
Constant
female
swim
bike
run
s = 33.02
Coef
-24.57
35.47
-25.01
7.130
-6.372
Stdev
20.13
14.77
45.75
1.331
5.384
R-sq = 98.0%
t-ratio
-1.22
2.40
-0.55
5.36
-1.18
p
0.241
0.030
0.593
0.000
0.255
R-sq(adj) = 97.4%
a) There is no reason to expect the constant to be positive or negative. However in this equation, a negative
coefficient for ‘swim’, ‘bike’ or ‘run’ would lead us to believe that an extra mile of swimming, biking or
running would lead to a faster time. The positive coefficient for ‘female’ is expected, because, at least at
present, women’s times in speed events have been longer than men’s. (However, at the rate that women’s
times in athletic events are falling this may not always be true!)
The constant and the coefficients of ‘swim’ and ‘run’ have p-values above .01. If the confidence level is
99%, the significance level is .01, so these coefficients are not significant.
4
5/12/99 252y9943
Analysis of Variance
SOURCE
Regression
Error
Total
DF
4
15
19
SS
786104
16351
802455
MS
196526
1090
F
180.29
p
0.000
c) The F test here is a test of the null hypothesis that the independent variables do not explain the dependent
variable. The low p-value means that we reject the null hypothesis.
SOURCE
female
swim
bike
run
DF
1
1
1
1
SEQ SS
6291
726098
52189
1526
Unusual Observations
Obs.
female
mins
1
0.00
489.25
18
1.00
660.48
Fit
547.00
582.47
Stdev.Fit
17.48
17.48
Residual
-57.75
78.01
St.Resid
-2.06R
2.79R
R denotes an obs. with a large st. resid.
MTB > regress c1 on 1 c5
Regression Analysis
The regression equation is
mins = - 19.2 + 23.6 run
Predictor
Constant
run
Coef
-19.25
23.615
s = 57.74
Stdev
23.19
1.582
R-sq = 92.5%
t-ratio
-0.83
14.92
p
0.417
0.000
R-sq(adj) = 92.1%
Analysis of Variance
SOURCE
Regression
Error
Total
DF
1
18
19
SS
742445
60011
802455
Unusual Observations
Obs.
run
mins
1
26.2
489.2
12
18.6
589.1
MS
742445
3334
Fit
599.5
420.0
F
222.69
Stdev.Fit
25.7
16.4
p
0.000
Residual
-110.2
169.1
St.Resid
-2.13R
3.05R
R denotes an obs. with a large st. resid.
MTB > regress c1 on 2 c5 c8
Regression Analysis
The regression equation is
mins = - 19.2 + 22.1 run + 3.02 C8
Predictor
Constant
run
C8
s = 54.36
Coef
-19.25
22.106
3.017
Stdev
21.83
1.705
1.659
R-sq = 93.7%
t-ratio
-0.88
12.96
1.82
p
0.390
0.000
0.087
R-sq(adj) = 93.0%
5
5/12/99 252y9943
Analysis of Variance
SOURCE
Regression
Error
Total
DF
2
17
19
SS
752216
50240
802455
SOURCE
run
C8
DF
1
1
SEQ SS
742445
9771
Unusual Observations
Obs.
run
mins
2
18.6
505.1
11
26.2
540.9
12
18.6
589.1
MS
376108
2955
Fit
391.9
639.0
448.0
F
127.27
Stdev.Fit
21.9
32.5
21.9
p
0.000
Residual
113.2
-98.1
141.0
St.Resid
2.27R
-2.25R
2.83R
R denotes an obs. with a large st. resid.
MTB > regress c1 on 2 c5 c11
Regression Analysis
The regression equation is
mins = - 102 + 39.6 run - 0.519 C11
Predictor
Constant
run
C11
Coef
-101.71
39.550
-0.5192
s = 54.11
Stdev
49.18
8.654
0.2778
R-sq = 93.8%
t-ratio
-2.07
4.57
-1.87
p
0.054
0.000
0.079
R-sq(adj) = 93.1%
Analysis of Variance
SOURCE
Regression
Error
Total
DF
2
17
19
SS
752675
49780
802455
SOURCE
run
C11
DF
1
1
SEQ SS
742445
10230
Unusual Observations
Obs.
run
mins
12
18.6
589.1
MS
376337
2928
Fit
454.3
F
128.52
Stdev.Fit
24.0
p
0.000
Residual
134.8
St.Resid
2.78R
R denotes an obs. with a large st. resid.
MTB > regress c1 on 3 c5 c8 c11
Regression Analysis
The regression equation is
mins = - 102 + 38.0 run + 3.02 C8 - 0.519 C11
Predictor
Constant
run
C8
C11
s = 50.01
Coef
-101.71
38.042
3.017
-0.5192
Stdev
45.45
8.033
1.526
0.2567
R-sq = 95.0%
t-ratio
-2.24
4.74
1.98
-2.02
p
0.040
0.000
0.066
0.060
R-sq(adj) = 94.1%
b) Some people have all the luck! Unlike most choices I gave these are really good. Not only do R-squared
and R-squared adjusted go up, but the p-values are all below 10% and the signs of the coefficients are
reasonable. So yes, these are probably improvements.
6
5/12/99 252y9943
Analysis of Variance
SOURCE
Regression
Error
Total
DF
3
16
19
SS
762446
40009
802455
SOURCE
run
C8
C11
DF
1
1
1
SEQ SS
742445
9771
10230
Unusual Observations
Obs.
run
mins
12
18.6
589.1
MS
254149
2501
Fit
482.3
F
101.64
Stdev.Fit
26.3
p
0.000
Residual
106.7
St.Resid
2.51R
R denotes an obs. with a large st. resid.
MTB > print c1-c11
Data Display
Row
mins
female
swim
bike
run
C6
C7
C8
C9
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
489.250
505.150
245.500
204.400
114.533
108.267
79.417
566.500
74.983
116.117
540.933
589.067
280.100
235.033
127.167
120.750
90.317
660.483
83.150
131.817
0
0
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
2.40
2.00
1.20
1.50
0.93
0.93
0.50
2.40
0.50
0.60
2.40
2.00
1.20
1.50
0.93
0.93
0.50
2.40
0.50
0.60
112.0
100.0
55.3
48.0
24.8
24.8
18.0
112.0
20.0
25.0
112.0
100.0
55.3
48.0
24.8
24.8
18.0
112.0
20.0
25.0
26.2
18.6
13.1
10.0
6.2
6.2
5.0
26.2
4.0
6.2
26.2
18.6
13.1
10.0
6.2
6.2
5.0
26.2
4.0
6.2
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
2.40
2.00
1.20
1.50
0.93
0.93
0.50
2.40
0.50
0.60
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
112.0
100.0
55.3
48.0
24.8
24.8
18.0
112.0
20.0
25.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
26.2
18.6
13.1
10.0
6.2
6.2
5.0
26.2
4.0
6.2
5.7600
4.0000
1.4400
2.2500
0.8649
0.8649
0.2500
5.7600
0.2500
0.3600
5.7600
4.0000
1.4400
2.2500
0.8649
0.8649
0.2500
5.7600
0.2500
0.3600
Row
C10
C11
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
12544.0
10000.0
3058.1
2304.0
615.0
615.0
324.0
12544.0
400.0
625.0
12544.0
10000.0
3058.1
2304.0
615.0
615.0
324.0
12544.0
400.0
625.0
686.44
345.96
171.61
100.00
38.44
38.44
25.00
686.44
16.00
38.44
686.44
345.96
171.61
100.00
38.44
38.44
25.00
686.44
16.00
38.44
7
5/12/99 252y9943
MTB > Correlation c1 c2 c3 c4 c5 c6 c7 c8 c9 c10 c11.
Correlations (Pearson)
mins
0.089
0.951
0.984
0.962
0.510
0.584
0.564
0.956
0.975
0.928
female
swim
bike
run
C6
C7
C8
female
swim
bike
run
C6
C7
C8
C9
C10
C11
0.000
0.000
0.000
0.792
0.716
0.726
0.000
0.000
0.000
0.973
0.965
0.432
0.480
0.470
0.985
0.954
0.932
0.985
0.420
0.494
0.479
0.979
0.989
0.954
0.417
0.487
0.487
0.982
0.983
0.985
0.982
0.980
0.426
0.412
0.403
0.993
0.483
0.488
0.471
0.478
0.478
0.479
C9
0.982
0.974
C10
C10
C11
0.975
d) There are many correlations above .9 here. These indicate that collinearity is a problem and that many
regressions will give coefficients that are insignificant or have unreasonable signs.
8
5/12/99 252y9943
II. Do at least 4 of the following 7 Problems (at least 15 each) (or do sections adding to at least 60 points Anything extra you do helps, and grades wrap around) . Show your work! State H 0 and H1 where
applicable. Use a significance level of 5% unless noted otherwise.
Note: These problems involve comparing population means, variances, proportions or medians. To
do this you use sample means, variances or proportions. If you look at a problem and tell me that the
sample means, variances, proportions or medians differ without incorporating them in a test, you are
wasting both your time and mine, and it is possible that my annoyance will affect how I grade the rest
of your exam, since I will now suspect that you have no idea what a significant difference is or what
we mean by a statistical test!
1. a. Premiums on a group of 11closed end mutual funds were as follows. (These are in per cent, but that
shouldn’t affect your analysis.)Test the hypothesis that the mean is 3 per cent using (i) Either a test ratio or a
critical value and (ii) A confidence interval. (6)
+4.7 -0.7 +5.3 +9.2 -0.3 -0.3 +5.0 +0.4 -1.9 +0.5 -3.1
b. Test that the following data (i) has a Poisson distribution (6) and (ii)has a Poisson distribution with a
mean of 4.5 (6). If you do both parts do only one with a chi-square method.
x 0 1 2 3 4 5 6 7
O 23 19 42 60 89 79 48 40
Solution: a) From Table 3 of the Syllabus Supplement:
Interval for
Confidence
Hypotheses
Interval
Mean (
  x  t 2 s x
H0 :   0
unknown)
H1 :    0
DF  n  1
Test Ratio
t
H0 :   3
H1 :   3
11
 0  3, DF  n  1  10,   .05, tn 1  t .025
 2.228
2
x  1.7091 and
sx 
s
14 .1989 3.7681

 1.2908  1.1361
11
11

n
.
(i) Test Ratio:
x   0 1.7091  3
t

 1.136 . This is in the
sx
1.1361
‘accept’ region between 2.228 , so do not
reject H 0 .
Critical Value: Since this is a 2-sided test,
xcv  0  t sx
2
 3  2.228 1.1361   3  2.5312 or 0.469 to
Critical Value
xcv   0  t 2 s x
x  0
sx
x
x2
+4.7
22.09
-0.7
0.49
+5.3
28.09
+9.2
84.64
-0.3
0.09
-0.3
0.09
+5.0
25.00
+0.4
0.16
-1.9
3.61
+0.5
0.25
-3.1
9.61
18.8 174.12
x 2  174 .12 , we
x  18 .8 and


 x  18.8  1.7091
find x 
Since
s2 
x
n
2
11
 nx 2
n 1

and
174 .12  111.7091 2
 14 .1989
10
2.531. This means that we reject H 0 if the
sample mean is above 2.531 or below 0.469.
Since x  1.7091 is between these critical values,
do not reject H 0 .
9
5/12/99 252y9943
(ii) Confidence Interval: Since this is a two-sided test,
  x  t s x  1.7091  2.228 1.1361   1.7091  2.5312 or 0.822    4.240 . This does not
2
contradict H 0 :   3 , because 3 is between –0.822 and 4.240, so do not reject H 0 .
b) (i) H 0 : Poisson Since the parameter is unknown, the chi-squared method is the only possible method.
To find the mean, sum xO and divide by n , which is the sum of O . The actual comparison can be done by
E  O  2
O2
and subtracting n .
E
E
x
0
1
2
3
4
5
6
7
xO 1602
so  
O 23 19
42
60
89
79
48
40

 4.0005 .
400
O
xO 0  19  84  180  356  395  288  280  1602
We look up probabilities in the Poisson table (these are in the column labeled p ) and multiply them by n
to get E .
E  O  2
O2
x
O
E O
E  O2
p
E
E
E
0
23
7.3264 -15.6736 245.662 33.5310
72.205 0.018316
1
19 29.3052 10.3052 106.197
3.6238
12.319 0.073263
2
42 58.6100 16.6100 275.892
4.7073
30.097 0.146525
3
60 78.1468 18.1468 329.306
4.2139
46.067 0.195367
4
89 78.1468 -10.8532 117.792
1.5073 101.361 0.195367
5
79 62.5172 -16.4828 271.683
4.3457
99.829 0.156293
6
48 41.6784 -6.3216
39.963
0.9588
55.280 0.104196
7
40 23.8160 -16.1840 261.922 10.9977
67.182 0.059540
8+
0 20.4520 20.4520 418.284 20.4520
0.000 0.051130
400 399.9998
0.0012
84.3388 484.339
either summing
or by summing


E  O2
O2
 n  484 .339  400  84 .339 Since there are 9 items on
E
E
2 7 
the comparison and we have used the data to estimate 1 parameter df  9  2  7 and  .05  14 .0671 , we
So  2 

 84.3377 or  2 

reject H 0 .
(ii) H 0 : Poisson4.5
We use the Kolmogorov-Smirnov method, though the Chi-squared method could
also be used if you did not do part i. The column Fe is the cumulative distribution from the Poisson table
with a mean of 4.5.
x
O Cumulative O
Fo
Fe
D
0
23
23
.05750
.01111
.0464
1
19
42
.10500
.06110
.0439
2
42
84
.21000
.17358
.0364
3
60
144
.36000
.34230
.0177
4
89
233
.58250
.53260
.0504
5
79
312
.78000
.70293
.0771
6
48
360
.90000
.83105
.0789
7
40
400
1.00000
.91341
.0876
8+
0
400
1.00000 1.00000
.0000
400
1.36
 .0680 . Since the
From the Kolmogorov-Smirnov, the critical value for a 95% confidence level is
400
largest number in D is above this value, we reject H 0 .
Exam continues in 252z9943.
10