QTM7010 - Statistics
Spring Semester 2001
Midterm Exam
Name______Date 3/14/01
Directions: Show all work! You may use Minitab wherever applicable.
1) Typically, the more attractive a corporate common stock is to an investor, the higher the stock's price-earnings (P/E) ratio. For example, if investors expect the stock's future earnings per share to increase, the price of the stock will be bid up and a high P/E ratio will result. Thus, the level of a stock's P/E ratio is a function of both the current financial performance of the firm and an investor's expectation of future performance. The table contains samples of P/E ratios from manufacturing firms and holding companies for the last day of March 1996.
MANUFACTURING FIRMS HOLDING COMPANIES
Company P/E ratio Company P/E ratio
Block Drug 9 EMC Ins. Group 8
Daig Corp. 38 First Essex Bancorp 9
Modtech Inc. 31 MLG Bancorp 14
Guest Supply 16 State Auto Financial 10
Astro Systems Inc. 36 Boston Bancorp 6
Fischer Imaging 33 Cellular Communications 48
Casino Data Systems 74 Anderson Group 5
Data Key Inc. 69 Provident Bancorp 12
Network Peripherals 23 Pubco Corp. 4
Brenco, Inc. 12 Condor Services 16
Day Runner 16 Eselco Inc. 16
Safeskin Corp. 16 Mesaba 2
Marisa Christina 14 Keystone Financial 13
Merix Corp. 17 JSB Financial 16
Cognex Corp. 39 Argonaut Group 14
FLIR Systems 17 ONBANCorp 12
Stant Corp. 11 Great Amer. Mgmt. Invst. 3
Grief Bros. Corp. 13 CPB, Inc. 12
Computer Identics 37 GBC Bancorp 18
PRI Automation 14 California Bancshares 16
a) Compare the P/E ratio distributions of manufacturing and holding firms using separate stem-and-leaf displays. (10 points)
Stem-and-leaf of C1 C2 = 1 N = 20
Leaf Unit = 1.0
1 0 9
6 1 12344
(5) 1 66677
9 2 3
8 2
8 3 13
6 3 6789
2 4
2 4
2 5
2 5
2 6
2 6 9
1 7 4
Stem-and-leaf of C1 C2 = 2 N = 20
Leaf Unit = 1.0
3 0 234
7 0 5689
(7) 1 0222344
6 1 66668
1 2
1 2
1 3
1 3
1 4
1 4 8
b) What do your stem-and-leaf displays suggest about the level of the P/E ratios of firms in the manufacturing business as compared to firms in the holding business? Explain. (4 points)
Higher in mfg
To assist you in answering the following questions, the above P/E ratios for the 20 manufacturing firms are displayed in order from the smallest P/E ratio to the largest P/E ratio:
9 / 11 / 12 / 13 / 14 / 14 / 16 / 16 / 16 / 1717 / 23 / 31 / 33 / 36 / 37 / 38 / 39 / 69 / 74
Descriptive Statistics: C1 by C2
Variable C2 N Mean Median TrMean StDev
C1 1 20 26.75 17.00 25.11 18.34
2 20 12.70 12.00 11.33 9.61
Variable C2 SE Mean Minimum Maximum Q1 Q3
C1 1 4.10 9.00 74.00 14.00 36.75
2 2.15 2.00 48.00 6.50 16.00
c) Find the median of these P/E ratios. (4 points)
17
d) Find the first quartile of these P/E ratios. (4 points)
14.00
e) Find the third quartile of these P/E ratios. (4 points)
36.75
f) Find the interquartile range. (2 points)
IQR = 36.75-14.00 = 22.75
g) Draw a boxplot for these P/E ratios. (10 points)
h) Are there any unusual values, i.e., outliers? If so, identify them. (4 points)
74
2) Travelers who have no intention of showing up often fail to cancel their hotel reservations in a timely manner. These travelers are known, in the parlance of the hospitality trade, as "no-shows." To protect against no-shows and late cancellations, hotels invariably overbook rooms. A recent study reported in the Journal of Travel Research examined the problems of overbooking rooms in the hotel industry. The following data, extracted from the study, represent daily numbers of late cancellations and no shows for a random sample of 10 days at a large (500-room) hotel:
18 10 15 13 17 15 12 15 18 17
Descriptive Statistics: C3
Variable N Mean Median TrMean StDev SE Mean
C3 10 15.000 15.000 15.250 2.667 0.843
Variable Minimum Maximum Q1 Q3
C3 10.000 18.000 12.750 17.250
a) Calculate the mean of this data set. (4 points)
15
b) Calculate the range of this data set. (4 points)
range = 18-10=8
c) Calculate the variance of this data set. (5 points)
(2.667)^2 =
d) Calculate the standard deviation of this data set. (5 points)
s = 2.667
3) If you randomly select a jail inmate convicted of DWI (driving while intoxicated), the probability distribution for the number x of prior DWI sentences is as described in the accompanying table (based on data from the U.S. Department of Justice).
x P(x)
------
0 0.50
1 0.30
2 0.15
3 0.05
x / P(x) / x*P(x) / x^2*P(x)0 / 0.5 / 0 / 0
1 / 0.3 / 0.3 / 0.3
2 / 0.15 / 0.3 / 0.6
3 / 0.05 / 0.15 / 0.45
1 / 0.75 / 1.35
mean / 0.75
variance / 0.7875
std dev / 0.887412
a) Find the mean. (5 points)
.75
b) Find the variance. (5 points)
.7875
c) Find the standard deviation. (5 points)
.8874
4) A restaurant has found that 73 percent of its dinner parties request seating in the non-smoking section. Find the probability that exactly three of the next five dinner parties will request the non-smoking section. (5 points)
n = 5
p = .73
q = .27
x = 3
Probability Density Function
Binomial with n = 5 and p = 0.730000
x P( X = x )
3.00 0.2836
5) Many personnel executives feel that the limitations imposed on salary increases by the Council on Wage and Price Stability (COWPS) are unfair. Furthermore, many say that compliance regulations are often conflicting and unclear. In fact, a study by Thomsen suggests that 80% of the personnel officers of America's largest corporations have difficulty with the acceptance and interpretation of COWPS guidelines. Suppose 15 corporate personnel officers are randomly selected from among major publicly listed firms to record their reaction to COWPS guidelines.
x P(x)
0 0.00000
1 0.00000
2 0.00000
3 0.00000
4 0.00001
5 0.00011
6 0.00078
7 0.00424
8 0.01806
9 0.06105
10 0.16423
11 0.35184
12 0.60198
13 0.83287
14 0.96482
15 1.00000
a) Find the probability that at least 9 of the 15 express dissatisfaction with the guidelines. (5 points)
n = 15
p = .80
q = .20
P(x >= 9) = 1 – P(x<=8) = 1 - .01806 = .98194
b) Find the probability that no more than 10 express dissatisfaction with the guidelines. (5 points)
P(x<=10) = .16423
c) Find the probability that exactly 12 express dissatisfaction with the guidelines. (5 points)
P(x = 12) = P(x<=12) – P(x<=11) = 0.60198 - .35184 = .25014
6) Individual banks that issue Visa and MasterCard are concerned about the cost of maintaining small accounts. A particular bank wishes to analyze these costs for accounts under $200. If this bank had a large number of Visa accounts with an average billing for 1993 of $785 and a standard deviation of $363, what proportion of these accounts are $200 or less in annual billing? (Assume that the annual billings for 1993 are approximately normally distributed.) (10 points)
mean = 785
std dev = 363
P(x < = 200)
Cumulative Distribution Function
Normal with mean = 785.000 and standard deviation = 363.000
x P( X <= x )
200.0000 0.0535
7) Most manufacturers of machinery or appliances offer an optional warranty to protect buyers against repair costs for product breakdowns during the early stages of product life. In doing so, the manufacturer faces one difficult issue: to price the warranty so that it covers expected maintenance costs without raising the product price to the point of dissuading buyers. A manufacturer of nonintelligent computer terminals has found that, in continuous operation, the time before required repair of the terminals is normally distributed, with a mean of 324 days and a standard deviation of 47 days. How long should the warranty time be for the terminals if the manufacturer is willing to cover the repair costs of at most 10% of the terminals under warranty? (10 points)
mean = 324 days
std dev = 47 days
P(x<=?) =.10
Inverse Cumulative Distribution Function
Normal with mean = 324.000 and standard deviation = 47.0000
P( X <= x ) x
0.1000 263.767
8) A telephone company has determined that during nonholidays the number of phone calls that pass through the main branch office each hour has a mean m of 80,000 and a standard deviation s of 32,000. Find the probability that the mean number of incoming phone calls per hour for a random sample of 64 nonholiday hours will be larger than 85,000. (10 points)
pop
mean = 80,000 hrs
std dev = 32,000 hrs
random sample
n = 64
P(xbar > 85,000) = 1 - .8944 = .1056
Cumulative Distribution Function
Normal with mean = 80000.0 and standard deviation = 4000.00
x P( X <= x )
8.50E+04 0.8944
9) A company is interested in estimating the mean number of days of sick leave taken by all its employees. From previous studies, it is known that the population standard deviation, s, is in the neighborhood of 10 days. The firm's statistician selects at random 100 personnel files and notes the number of sick days taken by each employee. She calculates a sample mean of 12.2 days.
a) Estimate the population mean using a 90% confidence interval. (10 points)
12.2 +– 1.645(10/sqrt(100))
10.555 <= mu <= 13.845
b) How many personnel files would the statistician have to select in order to estimate the population mean to within 2 days with 99% confidence? (10 points)
n = [(2.575*10)/2]^2 = 166
10) Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS) of their patients. In 1993, the average LOS for men in the United States was 6.5 days and the average for women was 5.6 days (Statistical Abstract of the United States: 1995). A random sample of 20 hospitals in one state had a mean LOS for women in 1996 of 3.8 days and a standard deviation of 1.2 days. Use a 95% confidence interval to estimate the population mean LOS for women for the state's hospitals in 1996. (10 points)
3.8 +- 2.093(1.2/sqrt(20))
3.8 +- .5616
3.2384 <= mu <= 4.3616
11) Family-owned companies are notorious for having difficulties in transferring control from one generation to the next. Part of this problem can be traced to lack of a well-documented strategic business plan. In a survey of 3,900 privately held family firms with revenues exceeding $1,000,000 a year. Arthur Andersen, the international accounting and consulting firm, found that 1,911 had no strategic business plan (Minneapolis Star Tribune, Sept. 4, 1995). Assume the 3,900 firms were randomly sampled from the population. Use a 90% confidence interval to estimate the proportion of family-owned companies without strategic business plans. (10 points)
Test and CI for One Proportion
Test of p = 0.5 vs p not = 0.5
Sample X N Sample p 90.0% CI Z-Value P-Value
1 1911 3900 0.490000 (0.476833, 0.503167) -1.25 0.212
12) In a survey conducted for Money Magazine by the ICR Survey Research Group, 26% of parents with college-bound high school children reported not having saved any money for college. The poll had a "… margin of error of plus or minus 4 percentage points" (Newark Star-Ledger, Aug. 16, 1996).
a) Assume that random sampling was used in conducting the survey and that the researchers wanted to have 95% confidence in their results. Estimate the sample size used in the survey. (10 points)
n = (1.96/.04)^2* (.26)(.74) = 462
b) Repeat part a but this time assume the researchers wanted to be 99% confident. (5 points)
n = (2.575/.04)^2 * (.26)(.74) = 798
Directions for 13 and 14: Data files are given in Excel format. Integrate appropriate portions of Minitab (or other statistical package) printouts with your answers.
13) Variable description:
ServiceTime: Time that it takes for a photocopy repair person to respond to a call about a malfunctioning photocopier.
A supplier of photocopying machines is interested in the time that it takes for a repair person to respond to a request for service when a photocopier is malfunctioning. A random sample of 35 calls is selected, and times in minutes are recorded for the repair person to respond. The variable ServiceTime is the label for the number of minutes to respond. Find a 99% confidence interval for the mean of ServiceTime and interpret your results. (10 points)
One-Sample T: ServiceTime
Variable N Mean StDev SE Mean 99.0% CI
ServiceTime 35 54.89 17.24 2.91 ( 46.93, 62.84)
14) Variable description:
Error: Coded values of 0s and 1s for no error and error found, respectively.
An internal auditor at a manufacturing plant needs to estimate the proportion of invoices that are in error. A random sample of 100 invoices is selected with an error being recorded as a 1, and a 0 indicating no error. Compute a 95% confidence interval for the proportion of invoices in error and interpret your results. (10 points)
Test and CI for One Proportion:
Test of p = 0.5 vs p not = 0.5
Success = 1
Variable X N Sample p 95.0% CI Z-Value P-Value
Error 66 100 0.660000 (0.567155, 0.752845) 3.20 0.001 (assuming normal approx)
Error 66 100 0.660000 (0.558467, 0.751776) 0.002 (binomial)