Download Final Exam Questions

1
KING FAHD UNIVERSITY OF PETROLEUM & MINERALS
DEPARTMENT OF MATHEMATICS AND STATISTICS
DHAHRAN, SAUDI ARABIA
STAT 211: BUSINESS STATISTICS I
Final Exam, Semester-072 (2008)
Time: 7:00 pm to 9:30 pm. Wednesday, June 11, 2008.
Tick () the box below corresponding to your Class Section, Time, and Instructor:
 Section
Time
Instructor
1
9.00-9.50am Dr. Mohammad H. Omar
2
10.00-10.50am Dr. Mohammad H. Omar
4
1:10-2.00pm Dr. Mohammad H. Omar
Student Name:
ID#
Serial #
______________________________________________________________________
1) You are allowed to use any scientific/electronic calculator.
2) Mobiles are NOT allowed. And maybe considered grounds for cheating. Turn mobiles off
and put under your seats.
3) Answer all questions.
4) On problem solving questions, set-up the problem and show key important steps to
maximize your scores. For example,
For the following data, compute the mean.
12
14
6
8
10
The following answer gives the necessary and sufficient steps to the solution.
ANSWER:
Mean = Total/size = (12+…+10)/5 = 50/5 = 10.
Notice that you can abbreviate (shorten) the steps to save time.
Question No
1-6
7-15
Marks
6
18
(2 points each)
Total
16
5
17
7
18
3
19
13
20
8
21
13
73
Marks Obtained
Comment
2
Part I. For the following questions, state either True or False.
(1point each =6 points total).
1. A study of 4000 Dell PC customers listed among other variables, the age of the customer.
The youngest customer was 14 years old and the oldest customer was 84 years old. If you
wish to develop a frequency distribution with 7 classes, the class width should be 10 years.
Answer: True
2. When cluster random sampling is employed, the population must first be divided into
homogeneous subgroups. Then either a systematic or a simple random sampling is done in
each group to obtain the cluster sample.
Answer: True
3. The Al-Zamil Construction Company has the opportunity to enter into a contract to build a
stretch of highway. The following table shows the probability distribution for the profit that
could occur if they take the contract:
Profit (in thousands SAR)
Probability
30
0.15
50
0.20
70
0.30
100
0.35
Based on this information, the expected profit for the company if they take the contract is
70.5 thousand riyals.
Answer: True
4. Hypothesis testing and estimation are two statistical tools that are used to draw inferences
about a large total data set based on a subset of the data.
Answer: True
5. A warehouse contains five parts made by Al-Salim Company and eight parts made by the
Al-Ataa Company. If an employee selects three of the parts from the warehouse at random,
the probability that none of the three parts is from Al-Salem Company is best approximated
by the hypergeometric distribution.
Answer: True
6. If the covariance between cost and revenue is -200, the correlation between these variable
must be positive.
Answer: False
3
Part II. Circle the best answer to the questions below.
(2 points each =18 total).
7. On the basis of labor data supplied by the Saudi Bureau of Statistics, SAMA has recently
decided to employ and provide further training to students majoring in mathematics or
finance. For SAMA, these labor data are examples of:
a.
b.
c.
d.
census data.
electronic data.
primary data.
secondary data. ()
8. Let A1 and A2 be two mutually exclusive and collectively exhaustive events. Which of the
following probability, if not directly given, makes it necessary to use the Bayes theorem to
evaluate P(A1|B)?
a. P(A1)
b. P(B)
()
c. P(A1 or B)
d. P(A2 or B) (
e. P(A1 or B) + P(A2 or B) )
9. The number of finance students graduating from KFUPM can be described as what type and
measurement level of data?
a. discrete data and measured as interval level.
b. continuous data and measured as interval level. ()
c. discrete data and measured as ratio level.
d. continuous data and measured as ratio level.
10. FedEx delivery service claims that no more than 6 percent of all packages arrive at the
destination address late. Assuming that the conditions for the binomial hold, if a sample of
size 15 packages is randomly selected and the 6 percent rate holds, what is the probability
that at least 2 packages will be delivered late?
a. 0.2262 ()
b. 0.3784
c. 0.3953
d. 0.7738.
4
11. You are given the following sample data:
46
68
22
The standard deviation is approximately
80
50
94
a. 1680
b. 320
c. 280
d. 25.92 ()
e. 23.66
12. The probability distribution for the number of traffic accidents that occur in Jubail industrial
city each day is given as follows:
Accidents
Probability
0
0.25
1
0.20
2
0.20
3
0.20
4
0.15
Which of the following is the probability that at least 1 accidents occur?
a. 0.20
b. 0.25
c. 0.40
d. 0.45 (
e. 0.75 ())
13. Customers are serviced at the counters of SAUDIA airline at the rate of 3 per 30 minutes.
What is the average amount of time a customer must wait before he can expect service?
a. 3 minutes
b. 5 minutes
c. 10 minutes (())
d. 12 minutes
e.
30 minutes
5
14. A Riyadh Bank Financial Risk manager wishes to estimate the difference between two
potentially lucrative population mean stock options with 95% confidence. He doesn’t know
the population variances for these stocks and has reason to believe the variances cannot be
equal. He only has access to 15 randomly sampled days for the closing price of stock A and
29 randomly sampled days for stock B. Which of the following critical values should he use
to construct the interval estimate?
a. z = 1.645
b. z = 1.96
c. t0.025 value with df =42
d. t0.025 value with df =43
e. t0.025
( s12 /15  s22 / 29) 2
value with df = 2
()
( s1 /15) 2 ( s22 / 29) 2

(14)
(28)
15. Suppose it is known that the ages of all employees working for Saudi Hollandi Bank is
normally distributed with a mean of 35.3 and standard deviation of 6.5 years. Which of the
following describes what the sampling distribution for x looks like? x will be distributed
a. normally with mean x = 35.3 and standard error  x = 6.5 / n ()
b. normally with mean x = 35.3 and standard deviation  x = 6.5
c. approximately normally with mean x = 35.3 and standard error  x = 6.5 / n
d. approximately normally with mean x = 35.3 and standard deviation  x = 6.5
6
Part III. Solving Questions. You MUST show the key steps to a problem to maximize credit
for your answers. You may shorten some steps to save time as long as the main steps are
shown (see the example on the first page).
(3+2 = 5 points).
16. The following data shows the Stem-and-Leaf plot for price-earnings (P/E) ratios for the
shares of 30 different companies.
Stem-and-leaf of P/E Ratio n = 30
Stem Unit = 1 (Leaf Unit = 0.1)
Stem Leaf
Frequency
5
46
2
6
7
3
1
8
5
1
9
689
3
10
11
78
2
12
6
1
13
14
66
2
15
348
3
16
3799
4
17
39
2
18
46
2
19
1
1
20
23
2
21
22
23
34
2
24
35
2
In addition, the following summary information are also calculated for the above data:
Median= 16.05
P75= 18.675
a. Calculate the 25th percentile of price-earning (P/E) ratios.
Data position is at 25/100(30+1) = 7.75.9.9 + 0.75(11.7 - 9.9) = 11.25
b. Based on (a) and the available information, describe the shape of the distribution of the
P/E ratio data.
Unimodal (1pt) Skewed Left (2pts) distribution
7
(3+4 = 7 points)
17. Fortune magazine July 20, 1998 issue reported the results of a survey on executive training.
The survey showed that 75 percent of 300 randomly selected CEOs believe that companies
should have fast-track training programs for developing managerial talent.
a. Describe the sampling distribution of the sample proportion p .
Approximately Normal with
E( p ) = .76 and
 p =0 .0214
b. What is the probability of finding a sample proportion of 0.70 or more?
.8384
P (0.03  p  p  0.03)  P (
0.03011

.76(1  .76) / 400
pp
0.03011

)
p (1  p ) / n
.76(1  .76) / 400
 P(
0.03011

.76(1  .76) / 400
pp
0.03011

)
p (1  p ) / n
.76(1  .76) / 400
0.03011
0.03011
Z 
)
0.021354
0.021354
 P (1.41  Z  1.41)
 P(
 2P (0  Z  1.41)  2(.4192)  0.8384
(3 points). H
18. The population standard deviation of D-E (debt-to-equity) ratio for financial institutions in
Saudi Arabia is  = 2. Supposed that we are interested in estimating the population mean
D-E ratio for all Saudi Arabian financial institutions with 94.26% confidence. How many
companies should be included in the sample if we want a margin error of 3?
 z    1.9(7.8) 
n    /2   
  54.9081 55
2 
 e  
2
2
8
(2+4+6+1 = 13 points)
19. Suppose a random sample of seven Saudi Arabian banks is selected and that their customer
loan bad debt ratios (in percents) for these banks are 7, 4, 6, 7, 5, 4 and 9. Assuming that the
debt ratios are approximately normally distributed, answer the following:
n
(Hint: To speed up your calculations, you can use:
X
i 1
2
i
 272,
n
X
i 1
i
 42, sx =1.8257)
a. Calculate the standard error of the mean bad debt ratio?
 x   / n  5.00 / 49  .7143
b. Construct the 95% confidence interval for the population mean bad debt ratio.
c. Supposed the population standard deviation of bad debt ratio for all Saudi Arabian banks is
known to be 1.8. Construct a 95% confidence interval for the population mean bad debt
ratio.
d. Interpret the interval you constructed in (c).
(33.40 to 36.20)
34.80 1.4 or
9
(2+5+1 = 8 points)
20. A marketing research firm wishes to compare the prices charged by two supermarket chains,
Panda and Farms. Using a standardized one-week shopping plan (grocery list), the research
firm makes identical purchases at each chain store. The stores for each chain are randomly
selected, and all purchases are made during a single week. The shopping expenses obtained
at the two supermarket chains are as follows:
Panda
Farms
119.25
123.71
121.32
121.72
122.34
122.42
120.14
123.63
122.19
122.44
111.99
116.62
114.88
115.38
115.11
114.40
117.02
113.91
116.89
111.87
Descriptive analyses by MINITAB for the data are also provided below.
n Mean StDev
Panda 10 121.9
1.4
Farms 10 114.8 1.84
a) Provide a point estimate of the difference between mean prices in the Panda and Farms
supermarket chain stores?
$300-$270=30
300-270 = 30
b) Assuming equal variances, develop a 90% confidence interval for the difference between mean
prices in the Panda and Farms supermarket chain stores?
df  n1  n 2  2  15  10  2  23 and   1  .90  0.10, t  / 2  1.7139
(15  1)162  (10  1)202
sp 
 17.6734
15  10  2
1 1
30  1.7139(17.6734)

90%CI:
15 10
 30  12.366
= (17.634,42.366)
()
c) Interpret your results in part b.$30270=30
df  n1  n 2  2  15  10  2  23 and   1  .90  0.10, t  / 2  1.7139
(15  1)162  (10  1)202
sp 
 17.6734
15  10  2
10
(6+6+1= 13 points)
21. User profile survey data randomly collected at the ESPN Sports Zone Web site showed that
65% of the 400 registered users were single men while 40% of 300 other users are
married men (USA Today, January 21, 1998). Based on this sample information,
a) develop the desired 99 percent confidence interval estimate for single men.
The parameter of interest in this situation is the difference in population proportions. The general format for
any
b) develop the desired 99 percent confidence interval estimate for the proportion difference
between single men and married men.
confidence interval estimate is:
Point estimate + critical value (standard error).
We are given the following sample information:
p1 
x 1 11
x
15

 .11 and p 2  2 
 .15 .
n 2 100
n1 100
In this case, the point estimate will be the difference between the two sample proportions which is:
p1  p 2  .11  .15  .04 . The standard error for the difference in proportions is:
p1 (1  p1 ) p 2 (1  p 2 )
=

n1
n2
.11(1  .11) .15(1  .15)
= 0.0475.

100
100
For a 99 percent confidence level, the critical value is z = 2.575.
The confidence interval estimate is found as:
p1  p 2  z
p1 (1  p1 ) p 2 (1  p 2 )

n1
n2
.04  2.575
.11(1  .11) .15(1  .15)
= .04  0.122313.

100
100
Thus, the 95 percent confidence interval is:
-0.162313 -------------------- +0.082313.
c)
interpret your results in part (b).
We are 99% confident that the true difference in proportion of redeemed coupons to be
between -0.162313 and +0.082313.
Thus, the range of possibilities for the difference in population proportions ranges from over
0.16 more for coupon A versus coupon B all the way to over 0.08 more for coupon B versus
coupon A. Therefore, since the interval crosses zero, we are unable to conclude that a d.