EXAM 4 – FORM A
STAT 211
SPRING 2004
The number of pounds of steam used per month by a chemical plant is thought to be related to the average ambient
temperature in (F) for that month. The past year’s usage and temperature are shown in the following table.
Month
Jan.
Feb.
Mar.
Apr.
May
June
July
Aug.
Sept.
Oct.
No.
Dec.
Temp.
21
24
32
47
50
59
68
74
62
50
41
30
Usage/1000 185.79 214.47 288.03 424.84 454.58 539.03 621.55 675.06 562.03 452.93 369.95 273.98
Using MINITAB, true average temperature that year ( temp) is tested being 50F and the results as follows.
Test of  temp = 50 versus  temp  50
Variable
n
Mean
StDev
temp
12
46.50
17.34
Variable
temp
(
95.0% CI
35.48,
57.52)
SE Mean
5.01
T
-0.70
P
0.499
Steam usage (y) is regressed on the average monthly temperature (x). The simple linear regression model,
Y   0  1 x  e is fitted using MINITAB software. Residuals are checked at the end of the analysis and independent
residuals are found to be normally distributed with the mean zero and the constant variance. The following are the results.
Predictor
Constant
temp
S = 1.887
Coef
-5.452
9.19295
SE Coef
1.620
0.03281
R-Sq = 100.0%
Analysis of Variance
Source
DF
Regression
1
Residual Error
10
Total
11
T
-3.37
280.17
P
0.007
0.000
R-Sq(adj) = 100.0%
SS
279645
36
279680
MS
279645
4
F
78493.85
P
0.000
Unusual Observations
Obs
temp
usage
Fit
SE Fit
Residual
12
30.0
273.980
270.336
0.768
3.644
R denotes an observation with a large standardized residual
St Resid
2.11R
Use the given information to answer the following 8 questions.
1.
We claim that true average temperature that year should be 50F. Do the data support this claim ?
(a) Yes
(b) No
H0:  temp = 50 versus Ha:  temp ≠ 50
2.
Which of the following is the test statistics for testing the true standard deviation of monthly average temperature being
18 F ?
(a) 0.93
(b) 10.21
(c) 183.75
(d) 280.17
(e) 78493.85
H0:  temp = 18 versus Ha:  temp ≠ 18
3.
p-value=0.499 > 0.05 then fail to reject H0
test statistics:
2=
(n  1)  s 2
 02

11(17.34) 2
(18) 2
If the true standard deviation of monthly average temperature is known to be 18 F, Would the data support the true
average temperature that year being 50F?
(a) Yes
(b) No
_
H0:  temp = 50 versus Ha:  temp ≠ 50
test statistics:
z
x  0

__

46.40  50
18 / 12
 0.67
X
then fail to reject H0
4.
What is the estimate of the expected steam usage when the average temperature is 55F?
(a) we cannot answer this question because temperature 55 is not listed on the data
(b) we cannot answer this question because temperature 55 is not in the range of the temperature data
(c) 500.16
< -1.96
EXAM 4 – FORM A
STAT 211
SPRING 2004
(d) 517.38
(e) 524.43
55 is in the range of the temperature data then estimate of the expected steam usage is -5.452+9.19295(55)
5.
What is the estimated change in expected steam usage when the monthly average temperature changes by 1F?
(a) -5.452
(b) 9.193
(c) 17.34
(d) 46.50
It is the definition of the slope
6.
Suppose the monthly average temperature is 47F, what is the corresponding residual on the fitted model?
(a) -1.7767
(b) -1.3213
(c) 0.5574
(d) 1.3213
(e) 1.7767
e=y-yhat=424.84-[-5.452+9.19295(47)]
7.
What proportion of total variability is accounted for by the simple linear regression model?
(a) 0%
(b) 25%
(c) 50%
(d) 75%
(e) 100%
Definition of R2
8.
We believe that the change in expected steam usage when the monthly average temperature changes by 1F should be
10 pounds. Which of the following is the corresponding test statistics to test this?
(a) -24.60
(b) -3.37
(c) 10.21
(d) 280.17
(e) 78493.85
^
H0:  1 = 10 versus Ha:  1 ≠ 10
test statistics:
   0 9.1929  10
t 1

 24.60
std ( 1 )
0.03281
A paper on the Journal of the Association of Asphalt Paving Technologists describes an experiment to determine the effect
of air voids on percentage retained strength of Asphalt. For purposes of the experiment, air voids are controlled at three
levels: low (2-4%), medium (4-6%), and high (6-8%). The data are shown in the following table.
Air Voids
Retained strength (%)
Low
106 90 103 90 79 88 92 95
Medium 80
69 94
91 70 83 87 83
High
78
80 62
69 76 85 69 85
The following are the analysis results from MINITAB. The true mean of percentage retained strength is shown by low,
medium, high for the levels of low, medium, and high, respectively.
Analysis of Variance for strength
Source
DF
SS
MS
level
2
1230.3
615.1
Error
21
1555.8
74.1
Total
2786.0
Level
low
medium
high
N
8
8
8
Mean
92.875
82.125
75.500
Pooled StDev =
where the model is single factor ANOVA
F
P
8.30
0.002
StDev
8.560
9.015
8.229
8.607
Tukey's pairwise
Family error
Individual error
Critical value =
comparisons
rate = 0.0500
rate = 0.0200
3.56
Intervals for (column level mean) - (row level mean)
high
low
low
-28.21
-6.54
medium
-17.46
4.21
-0.08
21.58
Test for equal true means (low =medium)in independent samples assuming unequal true variances
STAT 211
EXAM 4 – FORM A
Estimate for difference: 10.75
95% CI for difference: (1.25, 20.25)
T-Test of difference = 0 (vs not =): T-Value = 2.45
SPRING 2004
P-Value = 0.029
DF = 13
Test for equal true means (low =medium)in independent samples assuming equal true variances
95% CI for difference: (1.32, 20.18)
T-Test of difference = 0 (vs not =): T-Value = 2.45 P-Value = 0.028 DF = 14
Both use Pooled StDev = 8.79
Test for equal true means (low =high)in independent samples assuming unequal true variances
Estimate for difference: 17.38
95% CI for difference: (8.31, 26.44)
T-Test of difference = 0 (vs not =): T-Value = 4.14 P-Value = 0.001 DF = 13
Test for equal true means (low =high)in independent samples assuming equal true variances
95% CI for difference: (8.37, 26.38)
T-Test of difference = 0 (vs not =): T-Value = 4.14 P-Value = 0.001 DF = 14
Both use Pooled StDev = 8.40
Test for equal true means(medium =high)in independent samples assuming unequal true variances
Estimate for difference: 6.63
95% CI for difference: (-2.70, 15.95)
T-Test of difference = 0 (vs not =): T-Value = 1.54 P-Value = 0.149 DF = 13
Test for equal true means(medium =high)in independent samples assuming equal true variances
95% CI for difference: (-2.63, 15.88)
T-Test of difference = 0 (vs not =): T-Value = 1.54 P-Value = 0.147 DF = 14
Both use Pooled StDev = 8.63
Testing equal variances for all levels
Bartlett's Test (normal distribution)
Test Statistic: 0.055
P-Value
: 0.973
Levene's Test (any continuous distribution)
Test Statistic: 0.018
P-Value
: 0.982
Use the given information to answer the following 8 questions.
9.
Do the different levels of air voids significantly affect mean percentage retained strength?
(a) Yes
(b) No
H0: low =medium=high
P-value=0.002 < 0.05 then reject H0
10. Which of the following Tukey’s test suggest?
(a) The different levels of air voids do not affect mean percentage retained strength
(b) Low and medium levels have different true mean percentage retained strength
(c) Low and high levels have different true mean percentage retained strength
(d) Medium and high levels have different true mean percentage retained strength
(e) Exactly two of the above
11. Is the constant variance assumption in analysis of variance satisfied for us to rely on the output MINITAB gives us?
(a) Yes. The p-value is 0.002
(b) No. The p-value is 0.002
(c) Yes. The p-value is 0.973
Bartlett’s test is used
(d) No. The p-value is 0.973
12. Assuming that the constant variance assumption is satisfied, which of the following is the point estimate for the
constant standard deviation of residuals?
(a) 8.6
(b) 24.8
(c) 74.1
(d) 615.1
Square root of MSE
EXAM 4 – FORM A
STAT 211
SPRING 2004
13. We would think the true means of percent retained strength for medium and high levels should not be different
assuming equal true variances. Do the data support this?
(a) Yes
(b) No
0 falls in the confidence interval or the p-value is larger than 0.05.
14. We would think the true means of percent retained strength for low and high levels should not be different assuming
equal true variances. We tested this using the given data. Based on our conclusion, which of the following we have
made?
(a) Possible type II error
(b) Possible type I error
(c) Correct decision
Rejected equal means when the true means were equal
15. If you were testing the equal variances of the percent retained strength for medium and high levels, which of the
following would be the corresponding test statistics?
(a) 0.018
(b) 0.055
(c) 1.096
(d) 1.200
2
2
2
2
then the test statistics: F= s medium / s high  1.2
H 0 :  medium
  high
16. I do not know if you noticed it, but the total degree of freedom is not written on the analysis of variance table. Which
of the following is the degree of freedom for total?
(a) 20
(b) 21
(c) 22
(d) 23
(e) 24
The warranty for batteries of mobile phones is set at 200 operating hours, with proper charging procedures. A study of
5000 batteries is carried out and 15 stop operating prior to 200 hours. The following is the results from MINITAB
software.
Test of p=0.002 vs p0.002 where p:true proportion of phones stop operating
hours
Exact
Sample
x
n Sample p
95.0% CI for p
P-Value
1
15
5000 0.003000 (0.001680, 0.004943)
0.150
prior to 200
Use the given information to answer the following 2 questions.
17. The claim of less than 0.2% of the company’s batteries failing during the warranty period will be tested. Which of the
following would be the corresponding null hypothesis?
(a) p 0.002
(b) p  0.002
(c) p < 0.002
(d) p  0.005
(e) p < 0.005
claim is under the Ha for the hypothesis to be valid
18. Do these experimental results support the claim that less than 0.2% of the company’s batteries will fail during the
warranty period?
(a) Yes
(b) No
The P-value on the output is twice the area above positive test statistics where test statistics is positive. We have a
lower tailed test then the P-value is 1-0.15/2=0.925 which cause failing to reject H0.
In a random sample of 200 College Station residents who drive a domestic car, 165 reported wearing their seat belt
regularly, while another sample of 250 College Station residents, who drive a foreign car, revealed 198 who regularly wore
their seat belt. The following is the results from MINITAB software.
Sample
1
2
x
165
198
n
200
250
Sample p
0.825000
0.792000
Estimate for p(1) - p(2): 0.033
95% CI for p(1) - p(2): (-0.0398310, 0.105831)
Test for p(1) - p(2) = 0 (vs not = 0): Z = 0.89
P-Value = 0.375
Where p(1) and p(2) correspond to the proportion of domestic and foreign car drivers who wore their seat belt, respectively.
Use the given information to answer the following 4 questions.
19. Which of the following is the point estimate for p(1)-p(2)?
(a) We do not have enough information to answer this question
(b) -0.033
(c) -0.375
EXAM 4 – FORM A
STAT 211
SPRING 2004
(d) 0.033
(e) 0.375
20. Are there any significant differences in seat belt usage between domestic and foreign car drivers?
(a) No
(b) Yes
0 falls in the confidence interval or the p-value is larger than 0.05 where H0:p(1)-p(2)=0
21. If all the numbers are doubled on the initial given data, does this effect your decision for the previous question?
(a) Yes (rejected H0 in the previous question and failing to reject H0 in this question or vice versa)
(b) No
(rejected or failed to reject H 0 in the previous question and rejecting or failing to reject H 0 in this
question)
Sample proportions stay the same.
22. If p(1) < p(2), it means
(a) The proportion of domestic car drivers who wore their seat belts is the same as the foreign car drivers
(b) The proportion of domestic car drivers who wore their seat belts is more than the foreign car drivers
(c) The proportion of domestic car drivers who wore their seat belts is less than the foreign car drivers
23. If the P-value is 0.0222 for the upper tailed test with the test statistics of z=2.01? Which of the following is the
corresponding p-value for the two tailed test?
(a) 0.0111
(b) 0.0222
(c) 0.0333
(d) 0.0444
24. I would like to test if the true correlation between the overall grade and the number of hours of studying is zero.
Random sample is collected from 200 students and the Pearson’s correlation is found to be 0.975. Which of the
following test statistics should be used for testing?
(a) 33.27
(b) 49.13
(c) 58.94
(d) 61.74
(e) 280.17
The test statistics is
t
r n2
1 r
2

0.975 200  2
1  0.975 2
25. Which of the following is incorrect?
(a) If the P-value is very large in any test, you have to fail to reject H 0
(b) Only if the value of the test statistics is above the critical value in two tailed test, you should reject H 0
(c) If there is a less than sign under the alternative hypothesis, you have a lower tailed test
(d) The P-value is the area below the test statistics in the lower tailed test