QMB 3250 **
Spring 2011
Exam 1
Form Code: D
**
Feb. 9, 2011
Page 2 of 8
-- FOR THE QUESTIONS ON THIS PAGE, USE THE CONDOMINIUM DATA ON PAGE 6 -A condominium-management company owns three large developments: Windy Meadows
(290 total condo units), Observatory Hill (300 units) and Spanish Oaks (280
units). They recently surveyed properties in all three developments and
gathered information on monthly mortgage payments. The table on page 6 displays
the number of condos sampled, average payment, standard deviation in payments
and the number of condos that had one or more late payments in the past year.
Answer the questions below for the Observatory Hill development.
1.
If you were estimating the average monthly mortgage payment for the
condominiums, you would compute the standard error of the mean then multiply
it by the Finite Population Correction Factor. What is this corrected
standard error?
33.0
34.0
35.0
36.0
Choose one answer:
<-------+-------+-------+-------+------->
A
B
C
D
E
FPC=sqrt[N-n/N-1]=sqrt[300-34/300-1]=sqrt[.8896]=.9432
se=[s/sqrt(n)]*FPC=[211/sqrt(34)].9432=34.131
2. Make a two-sided 95% interval estimate of the average monthly mortgage
payment for the condominiums. What is the upper endpoint of this interval?
1419
1431
1465
1524
<-------+-------+-------+-------+------->
A
B
C
D
E
n=34 use 33 df so t= 2.040
1454 -+ t*se = 1454 -+ 2.040(34.131) = 1454 -+ 69.627 upper is 1523.63
3. Now make a two-sided 95% interval estimate of the total monthly mortgage
payments for all condominiums in the development. What is the lower
endpoint of this interval?
368000
383000
398000
413000
Choose one answer: <--------+--------+--------+--------+-------->
A
B
C
D
E
N(1454) -+ N(69.627) = 300(1454) -+ 300(69.627) = 436200 -+ 20888
lower endpoint is 415312
4. Now make a one-sided 95% interval estimate for the upper bound on the
proportion of condominiums in the development that had one or more late
payments in the past year. What is this estimated upper bound?
Choose one answer:
Choose one answer:
.3000
.3350
.3550
.3750
<--------+--------+--------+--------+-------->
A
B
C
D
E
p=9/34 = .26471
p + Z sqrt[p(1-p)/n]*FPC = .26471 + 1.645 sqrt[.005725](.9432)
= .26471 + 1.645(.07566)(.9432) = .26471 + .11739 = .38210
5. Suppose your answer to question 4 above was an upper bound estimate of
.3186. How many condominiums in the development does this imply had one or
more late payments in the past year?
90.7
92.7
94.7
96.7
Choose one answer: <--------+--------+--------+--------+-------->
A
B
C
D
E
N*p = 300*.3186 = 95.58
6. Please carefully check the UFID on your answer sheet. Make sure you have the ID
written in correctly, and that all the bubbles are filled.
doing this. Did you do this correctly? Hope so
You get 3 points for
QMB 3250 **
Spring 2011
Exam 1
Form Code: D
**
Feb. 9, 2011
Page 3 of 8
-- FOR THE QUESTIONS ON THIS PAGE, USE THE COMPLAINT DATA ON PAGE 6 -–
A large furniture company has stores all over Florida and warehouses in
Jacksonville, Orlando and Tampa. Their customer service technicians operate out
of offices at these warehouses. On page 6 of the exam is data on customer
complaints about the furniture delivered. The variable analyzed is how many
days it took until those complaints were resolved.
Answer the questions below using the data for Orlando.
7. If you did an outlier check for this data, what is the upper fence?
Choose one:
24.0
26.0
28.0
30.0
<-------+-------+-------+-------+------->
A
B
C
D
E
IQR=q3-q1=22-16.5=5.5
UpFence=q3+1.5IQR = 22+8.25 = 30.25
8. How many points would be considered potential outliers?
A. 0 or 1
B. 2
C. 3
D. 4
E. 5 or more
The lower fence is 8.25 and two points are below. Two points are above 30.25
9. If you were going to do an approximately 95% confidence interval estimate for
the median complaint resolution time at this warehouse, what is the value of r
you would use for the interval?
8.5
9.5
10.5
11.5
<-------+-------+-------+-------+------->
A
B
C
D
E
r=.4n-2 = .4*25-2 = 10-2 = 8
Choose one:
10. What is the lower endpoint of this interval estimate for the median
complaint resolution time at this warehouse?
12.5
13.5
14.5
15.5
<-------+-------+-------+-------+------->
A
B
C
D
E
smallest is 17, 8th largest is 22
Choose one:
8th
11. Now consider the more-standard 95% confidence interval for the average complaint
resolution time.
What is the lower endpoint of this interval estimate?
13.7
14.7
15.7
16.7
<-------+-------+-------+-------+------->
A
B
C
D
E
Xbar -+ t(24) s/Sqrt(n) = 19.400 -+2.080 6.831/sqrt(25) = 19.400 -+ 2.8417
Lower endpoint is 16.158
Choose one:
12. Now compare the two interval estimates.
If you judge by the interval width, which
estimate is worse?
median has width 5, mean has width 5.683
a. The median interval is wider by at least .30.
b. The two are about equal (within .30 of each other in width).
c. The interval for the average is wider by at least .30.
13. This is test form code “D”.
You earn 2 points for correctly bubbling in this code.
Did you do this correctly?
A. Yes
B. No, I screwed it up.
QMB 3250 **
Spring 2011
Exam 1
Form Code: D
**
Feb. 9, 2011
Page 4 of 8
Displayed in the table at the right
is an analysis of the service times
for 44 customers at four offices of
the license bureau. Use this
information for all questions on this
page.
14. Suppose you were going to compare the average service time on the North office to
the West office. If you assume that the variances in the two offices’ times were
about the same, what is your estimate of this “pooled” sample variance?
82
88
94
100
<-------+-------+-------+-------+------->
A
B
C
D
E
s2pool = [(11-1)4.5732 + (9-1)3.5282 ] /(11+9-2) = 17.1499
It looks like I messed up the answer scales here, and I did not intend that. But, A
is the right answer for the numbers above.
15. For the comparison in question 14, suppose you compute a 90% confidence interval for
the difference. What is the margin of error for the comparison (t-value times the
standard error)?
Choose one:
7.29
7.39
7.49
7.59
<-------+-------+-------+-------+------->
A
B
C
D
E
df=18, t value is 1.746.
ME = t*sqrt(17.15(1/11 +1/9)) = 1.746*1.8613 = 3.2499 Same comment on answer scale
Choose one:
16. For the comparison in questions 14 and 15, what is the best conclusion reached about
the service times?
diff=20.909-12.444=8.465 and this is larger than the ME
a. We can conclude the average times at the North office are significantly longer.
b. We can conclude the average times at the North office are significantly shorter.
c. We cannot conclude either office has faster average service times.
d. We can conclude the average service times are significantly different.
17. Now suppose we are going to test the hypotheses that the average service time at the
East office is the same as the average time at the Central office. This time we are
NOT willing to assume that the variances in the service times are about equal. If
you ignore the  sign, what is the value of the TCALC value for this test?
0.0
2.7
3.2
3.7
4.2
+-------+-------+-------+-------+------->
A
B
C
D
E
std error = sqrt(4.3902/11 + 6.2262/13) = 2.176 t=(19.273-38.769)/1.7268 = 8.96
18. For the test outlined in question 17 above, the test statistic has a t-distribution
with degrees of freedom that must be computed from a formula. Although you were not
required to use this formula, I did give you guidelines for what it would produce.
What is the least amount for the degrees of freedom that the formula could produce?
Choose one:
A. Up to 7
B. 8
C. 9
D. 10
E. 11
minimum of n1-1 and n2-1 = min 11-1 and 13-1 = 10
19. For the hypothesis test in questions 17 and 18, what is the best conclusion reached
about the service times?
a.
b.
c.
d.
We
We
We
We
can conclude the average times at the Central office are
can conclude the average times at the Central office are
cannot conclude either office has faster average service
can conclude the average service times are significantly
significantly longer.
significantly shorter.
times.
different.
QMB 3250 **
Spring 2011
Exam 1
Form Code: D
**
Feb. 9, 2011
Page 5 of 8
An unethical accountant has joined forces with a software engineer to create a
new income tax program, AutoDeduct. This program will not only search for
legitimate tax deductions but will also claim those in “grey areas” if it
determines the Internal Revenue Service is unlikely to contest them. The user
has a choice in how “aggressive” the program is, so it is possible to make it
commit outright fraud. Using the program, many taxpayers would pay fewer taxes,
assuming they don’t get caught.
The firm that the accountant works for filed joint tax returns last year for
9543 clients. Without the firm’s permission, he ran the AutoDeduct program on a
sample of 69 of these returns. Six returns would have paid more taxes under
AutoDeduct, 14 would have paid up to $1000 less and the rest of the 69 had no
real change in taxes paid.
The total “savings” in the sample of 69 tax returns was $4787. This information
(analyzed below in PhStat) was projected to the entire year’s worth of joint tax
returns. Use it to answer the questions below.
20. For the tax returns in the sample, 49 of them had no difference in taxes
using AutoDeduct versus the original preparation method.
A. True
B. False
21. For the tax returns in the sample, 20.29% of them would have paid fewer
taxes using AutoDeduct.
A. True
B. False
14 would pay less so proportion is 14/69 = .202898
22. For the tax returns in the sample, the average difference would have been a
“savings” of $68.38 if AutoDeduct was used to prepare the returns.
No, it was 69.38
A. True
B. False
23. If the AutoDeduct program had been used on all 9543 joint tax returns that
the firm prepared, the total “savings” could have been over a million
dollars.
The upper limit is 1.2 million so this could be true
A. True
B. False
QMB 3250 **
Spring 2011
Exam 1
Form Code: D
**
Feb. 9, 2011
Page 6 of 8
-- CONDO DATA FOR QUESTIONS ON PAGE 2 --
-- COMPLAINT DATA FOR QUESTIONS ON PAGE 3 -Office
Days Elapsed Until Complaint Was Resolved
Jax
1
14
16
19
10
14
17
20
11
15
17
22
11
15
17
28
12
16
17
30
13
16
17
14
16
18
14
16
19
Orlando
2
17
21
37
7
18
22
14
19
22
15
20
22
15
20
22
16
20
23
17
21
24
17
21
33
Tampa
6
12
15
18
34
6
13
16
18
6
13
17
18
10
13
17
18
10
14
17
19
11
14
18
20
11
14
18
27
11
15
18
31
Summary Statistics
Mean
Std. Dev.
Minimum
First Quartile
Median
Third Quartile
Maximum
Jax
16.034
5.254
Orlando
19.400
6.831
Tampa
15.697
6.192
1
14
16
17.5
30
2
16.5
20
22
37
6
11.5
15
18
34
QMB 3250 **
Spring 2011
Exam 1
Form Code: D
**
Feb. 9, 2011
Page 7 of 8
FORMULA SHEET
--- Outlier check --IQR = Q3 – Q1
Fences are 1.5*IQR away from quartiles
--- Interval for population median --compute .4n-2
round this to r = nearest integer
interval is rth smallest to rth largest data values
--- Interval estimate of population mean (finite population) --df = n-1
with no population information,
do not use the term involving N
X t
s
n
N n
N 1
--- Interval estimate of population total --df = n-1
N X  Nt
s
n
N n
N 1
--- One-sided interval estimate of population proportion --Could be adapted for two-sided interval
with  Z
pZ
p(1  p)
n
N n
N 1
--- Two-sample interval or hypothesis test, variances assumed equal --S p2 
(n1  1) S12  (n 2  1) S 22
(n1  n 2  2)
1
1 
Std Error  s p2   
 n1 n2 
df  n1  n2  2
Interval is (Xbar1 – Xbar2)  t StdError
TCALC = (Xbar1 – Xbar2)/StdError
--- Two-sample hypothesis test, variances are not considered equal --df computed by formula (not required)
but is bounded below by what it would be in
a one-sample analysis
TCALC = (Xbar1 – Xbar2)/StdError
Std Error 
s12 s22

n1 n2
QMB 3250 **
Spring 2011
Exam 1
Form Code: D
**
Feb. 9, 2011
Page 8 of 8
Scratch paper Sheet is intentionally blank
Except to remind you that there are a total of 23 questions on this test