Test Bank For Abnormal Psychology 18th Edition By James N Butcher, Jill M Hooley, Matthew Nock, Susan Mineka All Chapters LATEST

Test Bank For Probability And Statistics For Engineering
vv vv vv vv vv vv vv




And The Sciences 8th Ed by Jay L. Devore. vv vv vv vv vv vv vv vv vv




Chapter 1 – Overview and Descriptive Statistics
vv vv vv vv vv vv




SHORT ANSWER
vv




1. Give one possible sample of size 4 from each of the following populations:
vv vv vv vv vv vv vv vv vv vv vv vv

a. All daily newspapers published in the United States
vv vv vv vv vv vv vv

b. All companies listed on the New York Stock Exchange
vv vv vv vv vv vv vv vv

c. All students at your college or university
vv vv vv vv vv vv

d. All grade point averages of students at your college or university
vv vv vv vv vv vv vv vv vv vv




ANS:

a. Houston Chronicle, Des Moines Register, Chicago Tribune, Washington Post
vv vv vv vv vv vv vv vv

b. Capital One, Campbell Soup, Merrill Lynch, Pulitzer
vv vv vv vv vv vv

c. John Anderson, Emily Black, Bill Carter, Kay Davis
vv vv vv vv v v vv vv

d. 2.58. 2.96, 3.51, 3.69
vv vv vv




PTS: v v v v 1

2. A Southern State University system consists of 23 campuses. An administrator wishes to make an
vv vv vv vv vv vv vv vv vv vv vv vv vv vv

inference about the average distance between the hometowns of students and their campuses. Describe and
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

discuss several different sampling methods that might be employed. Would this be an enumerative or an
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

analytic study? Explain your reasoning.
vv vv vv vv vv




ANS:
One could take a simple random sample of students from all students in the California State University
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

system and ask each student in the sample to report the distance from their hometown to campus.
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

Alternatively, the sample could be generated by taking a stratified random sample by taking a simple
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

random sample from each of the 23 campuses and again asking each student in the sample to report the
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

distance from their hometown to campus.
vv vv vv vv vv vv

Certain problems might arise with self reporting of distances, such as recording error or poor recall. This
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

study is enumerative because there exists a finite, identifiable population of objects from which to sample.
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv




PTS: v v v v 1

3. A Michigan city divides naturally into ten district neighborhoods. How might a real estate appraiser select
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

a sample of single-family homes that could be used as a basis for developing an equation to predict
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

appraised value from characteristics such as age, size, number of bathrooms, and distance to the nearest
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

school, and so on? Is the study enumerative or analytic?
vv vv vv vv v v vv vv vv vv vv




ANS:

One could generate a simple random sample of all single family homes in the city or a stratified random
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

sample by taking a simple random sample from each of the 10 district neighborhoods. From each of the
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

homes in the sample the necessary variables would be collected. This would be an enumerative study
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

because there exists a finite, identifiable population of objects from which to sample.
vv vv vv vv vv vv vv vv vv vv vv vv vv

, PTS: v v v v 1

4. An experiment was carried out to study how flow rate through a solenoid valve in an automobile‘s
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

pollution-control system depended on three factors: armature lengths, spring load, and bobbin depth.
vv vv vv vv vv vv v v vv vv vv vv vv vv

Two different levels (low and high) of each factor were chosen, and a single observation on flow was
v v vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

made for each combination of levels.
vv vv vv vv vv vv




a. The resulting data set consisted of how many observations?
vv vv vv vv vv vv vv vv

b. Is this an enumerative or analytic study? Explain your reasoning.
vv vv vv vv vv vv vv vv vv




ANS:
a. Number observations equal 2 2 2=8 vv vv vv vv v v vv v v

b. This could be called an analytic study because the data would be collected on an existing
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

process. There is no sampling frame.
vv vv vv vv vv vv




PTS: v v v v 1

5. The accompanying data specific gravity values for various wood types used in construction .
vv vv vv vv vv vv vv vv vv vv vv vv vv




.41 .41 .42 .42. .42 .42 .42 .43 .44
.54 .55 .58 .62 .66 .66 .67 .68 .75
.31 .35 .36 .36 .37 .38 .40 .40 .40
.45 .46 .46 .47 .48 .48 .48 .51 .54

Construct a stem-and-leaf display using repeated stems and comment on any interesting features of the display.
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv




ANS:
One method of denoting the pairs of stems having equal values is to denote the stem by L, for ‗low‘ and
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

the second stem by H, for ‗high‘. Using this notation, the stem-and-leaf display would appear as follows:
vv vv vv vv vv vv vv v v vv vv vv vv vv vv vv vv vv




3L 1 stem: tenths vv

3H 56678 leaf: v v hundredths
4L 000112222234
5L 144
5H 58
6L 2
6H 6678
7L
7H 5

The stem-and-leaf display on the previous page shows that .45 is a good representative value for the data.
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

In addition, the display is not symmetric and appears to be positively skewed. The spread of the data is
v v vv vv vv vv vv vv vv vv vv vv vv vv v v vv vv vv vv vv

.75 - .31 = .44, which is .44/.45 = .978 or about 98% of the typical value of .45. This constitutes a
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv v v vv vv

reasonably large amount of variation in the data. The data value .75 is a possible outlier.
vv vv vv vv vv vv vv vv v v vv vv vv vv vv vv vv




PTS: v v v v 1

6. Temperature transducers of a certain type are shipped in batches of 50.
vv A sample of 60 batches was
vv vv vv vv vv vv vv vv vv vv v v vv vv vv vv vv

selected, and the number of transducers in each batch not conforming to design specifications was
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

determined, resulting in the following data:
vv vv vv vv vv vv




0 4 v v v v 2 v v 1 1 3 4 1 2 3 2 2 8 4 5 1 3 1
v v 1 v v v v 3
2 1 v v v v 2 v v 1 3 2 0 5 3 3 1 3 2 4 7 0 2 3
v v 4 v v v v 0
5 0 v v v v 2 v v 1 0 6 4 2 1 6 0 3 3 3 6 1 2 3
v v 3 v v v v 2

, a. Determine frequencies and relative frequencies for the observed values of x = number of
vv vv vv vv vv vv vv vv vv vv vv vv vv

nonconforming transducers in a batch. vv vv vv vv

b. What proportion of batches in the sample has at most four nonconforming transducers? What
vv vv vv vv vv vv vv vv vv vv vv vv vv

proportion has fewer than four? What proportion has at least four nonconforming units?
vv vv vv vv vv vv vv vv vv vv vv vv vv




ANS:

a.
Number Nonconforming vv Relative Frequency Frequency vv


0 0.117 7
1 0.200 12
2 0.217 13
3 0.233 14
4 0.100 6
5 0.050 3
6 0.050 3
7 0.017 1
8 0.017 1
1.001
The relative vv v v frequencies don’t add up exactly to 1because they have been rounded
vv vv vv vv vv vv vv vv vv vv




b. The number of batches with at most 4 nonconforming items is 7+12+13+14+6=52, which is a proportion of
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

52/60=.867. The proportion of batches with (strictly) fewer than 4 nonconforming items is 46/60=.767.
vv vv vv vv vv vv vv vv vv vv vv vv vv vv




PTS: v v v v 1

7. The number of contaminating particles on a silicon wafer prior to a certain rinsing process was determined
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

for each wafer in a sample size 100, resulting in the following frequencies:
vv vv vv vv vv vv vv vv vv vv vv vv vv




Number of particles vv v v Frequency Number of particles vv v v Frequency
0 1 8 12
1 2 9 4
2 3 10 5
3 12 11 3
4 11 12 1
5 15 13 2
6 18 14 1
7 10


a. What proportion of the sampled wafers had at least two particles? At least six particles?
vv vv vv vv vv vv vv vv vv vv vv vv vv vv

b. What proportion of the sampled wafers had between four and nine particles, inclusive? Strictly between
vv vv vv vv vv vv vv vv vv vv vv vv vv vv

four and nine particles?
vv vv vv vv




ANS:
a. From this frequency distribution, the proportion of wafers that contained at least two particles is (100-1-
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

2)/100 = vv

.97, or 97%. In a similar fashion, the proportion containing at least 6 particles is (100 – 1-2-3-12-11-
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

15)/100 = 56/100 = .56, or 56%. vv vv vv vv vv vv

b. The proportion containing between 4 and 9 particles inclusive is (11+15+18+10+12+4)/100 = 70/100
vv vv vv vv vv vv vv vv vv vv vv vv

= .70, or 70%. The proportion that contain strictly between 4 and 9 (meaning strictly more than 4
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

and strictly less than 9) is (15+ 18+10+12)/100= 55/100 = .55, or 55%.
vv vv vv vv vv vv vv vv vv vv vv vv vv

, PTS: v v v v 1

8. The cumulative frequency and cumulative relative frequency for a particular class interval are the sum
vv vv vv vv vv vv vv vv vv vv vv vv vv vv

of frequencies and relative frequencies, respectively, for that interval and all intervals lying below it.
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

Compute the cumulative frequencies and cumulative relative frequencies for the following data:
vv vv vv vv vv vv vv vv vv vv vv vv




75 89 80 93 64 67 72 70 66 85
89 81 81 71 74 82 85 63 72 81
81 95 84 81 80 70 69 66 60 83
85 98 84 68 90 82 69 72 87 88

ANS:

Class Frequency Relative Cumulative Cumulative
Frequency
vv Frequency
vv vv Relative
vv Frequency
60 vv – vv under vv65 3 .075 3 .075
65 vv – vv under vv70 6 .15 9 .225
70 vv – vv under vv75 7 .175 16 .40
75 vv – vv under vv80 1 .025 17 .425
80 vv – vv under vv85 12 .30 29 .725
85 vv – vv under vv90 7 .175 36 .90
90 vv – vv under vv95 2 .05 38 .95
95 vv – vv under vv100 2 .05 40 1.0

PTS: v v v v 1

9. Consider the following observations on shear strength of a joint bonded in a particular
vv vv vv vv vv vv vv vv vv vv vv vv vv

vv manner:

30.0 4.4 33.1 66.7 81.5 22.2 40.4 16.4 73.7 36.6 109.9

a. Determine the value of the sample mean.
vv vv vv vv vv vv

b. Determine the value of the sample median. Why is it so different from the mean?
vv vv vv vv vv vv vv vv vv vv vv vv vv vv

c. Calculate a trimmed mean by deleting the smallest and largest observations. What is the
vv vv vv vv vv vv vv vv vv vv vv vv vv


corresponding trimming percentage? How does the value of this
vv vv vv vv vv vv vv vv vv v v vv compare to the mean vv vv vv


vv and median?
vv




ANS:
a. The sum of the n = 11 data points is 514.90, so
vv vv vv = 514.90/11 = 46.81.
vv vv vv vv vv vv vv vv v v v v vv vv vv

b. The sample size (n = 11) is odd, so there will be a middle value. Sorting from smallest to largest: 4.4
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

16.4 22.2
v v v v

30.0 33.1 36.6 40.4 66.7 73.7 81.5 109.9. The sixth value, 36.6 is the middle, or median,
v v v v v v v v v v v v v v vv vv vv vv vv vv vv vv vv

value. The mean differs from the median because the largest sample observations are much further from
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

the median than are the smallest values.
vv vv vv vv vv vv vv

c. Deleting the smallest (x = 4.4) and largest (x = 109.9) values, the sum of the remaining 9
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv


observations is 400.6. The trimmed mean
vv is 400.6/9 = 44.51. The trimming percentage is
vv vv vv vv vv v v v v vv vv vv vv vv vv vv


100(1/11) = 9.1%.
vv lies between the mean and median.
vv vv v v v v vv vv vv vv vv




PTS: v v v v 1

10. A sample of 26 offshore oil workers took part in a simulated escape exercise, resulting in the
vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv vv

accompanying data on time (sec) to complete the escape:
vv vv vv vv vv vv vv vv vv




373 370 364 366 364 325 339 393
356 359 363 375 424 325 394 402