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Brase 5e Basic IRG part3 Answers

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Instructor’s Resource Guide Understanding Basic Statistics, 5th Edition
Answers to Sample Chapter Tests
All calculations performed using Minitab 15 or a TI-83 Plus. Answers given as provided by the technology.
CHAPTER 1, FORM A
1.
2.
3.
4.
5.
6.
7.
(a) Length of time to earn a bachelor’s degree
(b) Quantitative
(c) All Colorado residents who earned (or will
earn) a bachelor’s degree
Explanations will vary.
(a) Nominal
(b) Nominal
(c) Ratio
(d) Interval
(e) Nominal
(f) Ordinal
(g) Ordinal
(a) Census
(b) Experiment
(c) Sampling
(d) Simulation
Answers will vary.
The outcomes are the number of dots on the face,
1 through 6. Consider single digits in the random
number table. Select a starting place at random.
Record the first five digits you encounter that are
between (and including) 1 and 6. The first five
outcomes are 3, 6, 1, 5, and 6.
(a) Systematic
(b) Cluster
(c) Stratified
(d) Convenience
(e) Simple random
(a) Parameter
(b) Parameter
(c) Statistic
3.
4.
5.
6.
7.
CHAPTER 1, FORM C
1.
2.
3.
CHAPTER 1, FORM B
1.
2.
(a) Observed book purchase (mystery or not
mystery)
(b) Qualitative
(c) All current customers of the bookstore
Explanations will vary.
(a) Nominal
(b) Ratio
(c) Interval
(d) Ordinal
(e) Ratio
(a) Census
(b) Sampling
(c) Simulation
(d) Experiment
Answers will vary.
Assign each of the 736 employees a number
between 1 and 736. Select a starting place in the
random number table at random. Use groups of
three digits. Use the first 30 distinct groups of
three digits that correspond to employees
numbers. For this problem, use employees 622,
413, 055, 401, and 334.
(a) Simple random
(b) Stratified
(c) Cluster
(d) Convenience
(e) Systematic
(a) Statistic
(b) Parameter
(c) Statistic
4.
5.
6.
(b)
A.
B.
C.
D.
E.
A.
B.
C.
D.
(e)
(d)
A.
B.
C.
D.
Copyright © Houghton Mifflin Harcourt Company. All rights reserved.
(b)
(a)
(d)
(c)
(a)
(b)
(c)
(a)
(d)
(e)
(d)
(a)
(c)
Part III: Answers to Sample Chapter Tests
E. (b)
4.
Time Series Plot of Air Quality Index
5.5
CHAPTER 2, FORM A
5.0
1.
Air Quality Index
4.5
Pareto Chart of Majors at Hendrix College
700
600
4.0
3.5
3.0
2.5
Number of Students
500
2.0
400
1.5
Sunday
Monday
Tuesday
300
Wednesday
Day
Thursday
Friday
Saturday
200
100
0
5.
Business Administration
Social Science
Natural Science
Major
Humanities
Philosophy
Circle Graph of Noise Preference
Background
8.0%
2. (a) The class width is 6
Quiet
38.0%
Frequency and Relative Frequency Table
Stereo
34.0%
Class
Rel.
Class Cum.
Boundaries Freq. Frequency Midpoint Freq.
2.5  8.5
8.5  14.5
14.5  20.5
20.5  26.5
26.5  32.5
10
9
6
2
3
0.3333
0.3000
0.2000
0.0667
0.1000
5.5
11.5
17.5
23.5
29.5
10
19
25
27
30
TV
20.0%
6.
Circle Graph of Age of Shoppers
Men 60+
20.0%
(b)
Women under 60
35.0%
Relative Frequency Histogram of Student Visits
35
Relative Frequency
30
25
Men under 60
25.0%
20
15
Women 60+
20.0%
10
5
0
2.5
8.5
14.5
20.5
Weekly Student Visits
26.5
32.5
7.
Dotplot of Student Visits
3.
Minitab output:
Stem-and-leaf of Dollars N = 20
Leaf Unit = 1.0
0
1
2
3
4
5
5789
25679
124
357
269
17
4
8
12
16
20
Student Visits
24
28
32
Copyright © Houghton Mifflin Harcourt Company. All rights reserved.
Instructor’s Resource Guide Understanding Basic Statistics, 5th Edition
Relative Frequency Histogram of Incomplete Grades
35
30
Relative Frequency
8.
9.
The dotplot and the histogram show the same
information, except the dotplot retains the
original values in the data set.
Eleven motorists were speeding.
Minitab output:
Stem-and-leaf of Age of Runner N = 22
Leaf Unit = 1.0
25
20
15
10
5
0
1
2
3
4
5
8
089
2445889
12568
0237
05
0
3.
The distribution is fairly symmetrical with a
center near 30.
1.
33.5
48.5
63.5
Number of Incomplete Grades
78.5
93.5
Minitab output:
Stem-and-leaf of Study Abroad Students
N = 20
Leaf Unit = 1.0
2
3
4
5
6
CHAPTER 2, FORM B
18.5
169
3468
2457
012689
348
Pareto Chart of Sales
1400
4.
Time Series Plot of Price of Gold
1000
325
800
320
600
315
Price of Gold ($)
Number of Books Sold
1200
400
200
0
Nonfiction
Fiction
Children's
Electronic Media
Genre
310
305
300
295
290
1
2. (a) The class width is 15
2
3
4
5
6
7
Week
8
9
10
Frequency and Relative Frequency Table
Class
Rel.
Class Cum.
Boundaries Freq. Frequency Midpoint Freq.
18.5  33.5
33.5  48.5
48.5  63.5
63.5  78.5
78.5  93.5
3
7
8
4
2
0.1250
0.2917
0.3333
0.1667
0.0833
26
41
56
71
86
5.
Circle Graph of Chosen Major
Undecided
8.0%
3
10
18
22
24
Social Science
16.0%
Engineering
18.0%
Science
8.0%
Computers
28.0%
6.
(b)
Copyright © Houghton Mifflin Harcourt Company. All rights reserved.
Humanities
22.0%
11
12
Part III: Answers to Sample Chapter Tests
8.
9.
Circle Graph of Bus Riders
Junior
12.0%
Senior
2.0%
(a)
(c)
CHAPTER 3, FORM A
1.
Sophomore
25.0%
Freshman
61.0%
2.
3.
7.
(c) s 2  81.67
Dotplot of Incomplete Grade
4.
5.
20
30
40
50
60
Incomplete Grade
70
x = 16.61; median = 13; mode = 12
$42.19 (thousand); $35.76 (thousand); The
trimmed mean because it does not include the
extreme value. Note that all the salaries are
below $42.19 (thousand) except one.
(a) Range = 25
(b) x = 33
80
(d)
(a)
(b)
(a)
90
s = 9.04
CV = 20.9%
$6.04 to $14.72
Low value = 15; Q1 = 23; median = 33; Q3 =
49; High value = 72
(b)
Boxplot of Age of Skier
80
70
8.
9.
Age of Skier
60
The dotplot and the histogram show the same
information, except the dotplot retains the
original values in the data set.
6 days
Minitab output:
Stem-and-leaf of Diameter N = 16
Leaf Unit = 0.10
0
1
2
3
4
9
278
345568
01677
1
The distribution is fairly symmetrical with a
center near 2.6 mm.
(c)
(b)
(e)
(a)
(d)
(e)
(b)
40
30
20
10
6.
7.
8.
(c) Interquartile range = 26
Weighted average = 85.65
163.31 lb
79% below; 21% above
CHAPTER 3, FORM B
1.
x = 10.55; median = 11; mode = 11
2.
37.8 inches; 31.63 inches; The trimmed mean
because it does not include the extreme value.
Note that all but three values are below 37.8
inches.
(a) Range = 14
(b) x = 12.17
CHAPTER 2, FORM C
1.
2.
3.
4.
5.
6.
7.
50
3.
(c) s 2  25.37
4.
(d) s = 5.04
(a) CV = 26.5%
(b) 15.3 to 49.7
Copyright © Houghton Mifflin Harcourt Company. All rights reserved.
Instructor’s Resource Guide Understanding Basic Statistics, 5th Edition
5.
(a) Low value = 27; Q1 = 35.25; median = 42.5;
Q3 = 57.5; High value = 68
(b)
2.
(b)
 x  167;  y  53;  x2  6069;
 y 2  789;  xy  2103
Boxplot of Faculty Age
70
(c) b = 0.678; a = 12.0; yˆ  0.678x  12.0
60
Age
y  10.6
(a) x  33.4;
50
3.
40
4.
5.
30
(d) See problem 1.
r = 0.996; r2 = 0.992; 99.2% of the variation in y
can be explained by the least squares line using x
as the predicting variable.
For x = 32 inches, y = 9.696 pounds.
Scatter Diagram of Sales ($) versus TV Ads ($)
900
800
Sales ($ Thousands)
6.
7.
8.
(c) Interquartile range = 22.25
Weighted average = 84.75
$44.40
19% above; 81% below
CHAPTER 3, FORM C
1.
2.
3.
4.
5.
6.
7.
8.
(b)
(d)
A.
B.
C.
D.
A.
B.
(b)
(a)
(d)
(e)
600
500
400
300
200
100
0
(e)
(a)
(c)
(d)
(b)
(a)
6.
10
20
30
40
50
60
TV Ads ($ Thousands)
70
80
90
The linear correlation coefficient appears to be
positive since as x increases, y tends to increase.
(a) x = 35.2; y = 358
(b)
 x  211;  y  2150;  x2  12143;
 y 2  1124798;  xy  114639
(c) b = 8.26; a = 67.7; yˆ  8.26 x  67.7
7.
CHAPTER 4, FORM A
1.
8.
(d) See problem 5.
r = 0.954; r2 = 0.910; 91.0% of the variation in
y can be explained by the least squares line using
x as the predicting variable.
For x = $37, y = $373.32 (in thousands).
Scatter Diagram of Pounds versus Inches for Northern Pike
20
15
Pounds
700
10
5
0
20
25
30
35
40
45
Inches
The linear correlation coefficient appears to be
positive since as x increases, y tends to increase.
Copyright © Houghton Mifflin Harcourt Company. All rights reserved.
Part III: Answers to Sample Chapter Tests
7.
CHAPTER 4, FORM B
1.
8.
r = 0.994; r2 = 0.988; 98.8% of the variation in y
can be explained by the least squares line using x
as the predicting variable.
For x = $35, y = $100.40.
Scatter Diagram of Revenue versus Salary
26
CHAPTER 4, FORM C
Revenue ($ Billions)
24
22
20
18
16
14
12
10
1.0
2.
1.5
Salary ($ Millions)
2.0
2.5
The linear correlation coefficient appears to be
positive since as x increases, y tends to increase.
(a) x  1.38; y  17.8
(b)
4.
5.
Scatter Diagram of Price ($) versus Cost ($)
175
Price ($)
150
125
100
75
50
20
30
40
50
Cost ($)
60
2.
3.
4.
5.
6.
200
6.
1.
 x  6.9;  y  89;  x  11.03;
 y 2  1703;  xy  134.6
(d) See problem 1.
r = 0.880; r2 = 0.775; 77.5% of the variation in y
can be explained by the least squares line using x
as the predicting variable.
For x = $1.5 million, y = $18.735 billion.
70
(a)
(d)
(b)
(a)
(c)
(e)
(d)
(c)
CHAPTER 5, FORM A
2
(c) b = 7.81; a = 7.02; yˆ  7.81x  7.02
3.
1.
2.
3.
4.
5.
6.
7.
8.
(a) Relative frequency; 19/317 = 0.0599 or
5.99%
(b) 1  0.0599 = 0.9401 = 94.01%
(c) Defective, not defective; yes
1/3
(a) 1/12 (b) 1/12 (c) 1/4
(a) 21/144 = 0.149
(b) 21/132 = 0.159
P(Approval on written and interview) =
P(written)P(interview | written) =
(0.63)(0.85) = 0.5355 = 53.55%
(a) 35/360
(b) 21/360
(c) 5/149
(d) 149/360
(e) 48/149
(f) 31/360
(g) No; P(referred) = 149/360 ≠
P(referred | satisfied) = 59/138.
The linear correlation coefficient appears to be
positive since as x increases, y tends to increase.
(a) x = 44.83; y = 126.3
(b)
 x  269;  y  758;  x2  13659;
 y 2  107100;  xy  38216
(c) b = 2.6471; a = 7.653; yˆ  2.65x  7.65
(d) See problem 5.
Copyright © Houghton Mifflin Harcourt Company. All rights reserved.
Instructor’s Resource Guide Understanding Basic Statistics, 5th Edition
7.
There are 8 packages to display.
7.
He can walk 6 different routes.
8.
9.
P16,5 = 524,160
C24,4 = 10,626
CHAPTER 5, FORM C
8.
9.
P40,4 = 2,193,360
C11, 3 = 165
CHAPTER 5, FORM B
1.
2.
3.
4.
5.
6.
(a) P(woman) = 4/10 = 0.4 = 40%
(b) P(man) = 6/10 = 0.6 = 60%
(c) P(President) = 1/10 = 0.1 = 10%
2/3
(a) 1/6 (b) 1/6 (c) 1/9
(a) (6/17)(8/17) = 0.166 = 16.6%
(b) (6/17)(8/16) = 0.176 = 17.6%
P(female and computer science major) =
P(female)P(computer science major | female) =
(0.64)(0.12) = 0.0768 = 7.68%
(a) 180/417
(b) 65/115
(c) 98/417
(d) 61/417
(e) 61/101
(f) 65/180
(g) No; P(neutral) = 166/417 ≠
P(neutral | freshman) = 61/101
1.
2.
3.
4.
5.
6.
7.
8.
9.
A.
B.
(a)
A.
B.
C.
A.
B.
(b)
A.
B.
C.
D.
E.
F.
(d)
(e)
(a)
(d)
(b)
(c)
(e)
(b)
(a)
(d)
(e)
(c)
(a)
(d)
(c)
(b)
CHAPTER 6, FORM A
1.
(a) P(0) = 0.038; P(1) = 0.300; P(2) = 0.263;
P(3) = 0.171; P(4) = 0.117; P(5) = 0.058;
P(6) = 0.033; P(7) = 0.021
Copyright © Houghton Mifflin Harcourt Company. All rights reserved.
Part III: Answers to Sample Chapter Tests
(b)
(b)
Probability Distribution of x
0.25
Probability Distribution of x
0.20
0.30
Probability
Probability
0.25
0.20
0.15
0.10
0.15
0.05
0.10
0.00
0.05
1
2
3
4
5
6
7
8
9
10
x
0.00
0
1
2
3
4
5
6
7
x
(c) 0.641
(d) 0.101
(e)  = 4.40
(c) P(2  x  5) = 0.609
(d) P(x < 3) = 0.601
(e)  = 2.44
(f) σ = 1.828
2.
3.
2.
3.
(f)  = 1.574
P(4) = 0.122
(a) P(11) = 0.0088
(b) P(r  8) = 0.050
(c) P(r  4) = 0.515
(b) P(r  9) = 0.200
4.
5.
(c) P(r  3) = 0.012
4.
5.
(f)  = 2.08
P(3) = 0.251
(a) P(15) = 0.000 (to three digits)
(d) P(5  r  10) = 0.484
51
(a)
(d) P(9  r  11) = 0.200
6
(a)
Distribution Plot
Binomial, n=7, p=0.55
0.30
0.25
Probability
Distribution Plot
Binomial, n=5, p=0.75
0.4
0.20
0.15
0.10
Probability
0.3
0.05
0.00
0.2
0
1
2
3
4
5
6
7
8
r
0.1
0.0
0
1
2
3
4
5
(b)  = 3.85
6
(c)  = 1.316
r
6.
6.
(a)  = 1.2
(b)  = 3.75
(b)  = 0.980
(c)  = 0.968
(a) µ = 1.05
(c) Yes, since 5 > µ + 2.5 =
1.2 + (2.5)(0.980) = 3.65.
(b)  = 0.945
(c) Yes, since 4 > µ + 2.5 =
1.05 + (2.5)(0.945) = 3.4125.
CHAPTER 6, FORM C
1.
CHAPTER 6, FORM B
1.
(a) P(1) = 0.066; P(2) = 0.092; P(3) = 0.202;
P(4) = 0.224; P(5) = 0.184; P(6) = 0.079;
P(7) = 0.053; P(8) = 0.044; P(9) = 0.035;
P(10) = 0.022
2.
3.
A.
B.
C.
D.
(e)
A.
(c)
(e)
(b)
(d)
(b)
Copyright © Houghton Mifflin Harcourt Company. All rights reserved.
Instructor’s Resource Guide Understanding Basic Statistics, 5th Edition
4.
5.
6.
B.
C.
D.
(b)
A.
B.
(c)
(a)
(d)
(c)
CHAPTER 7, FORM B
(e)
(c)
2.
3.
(a) The tails of a normal curve must get closer
and closer to the x-axis.
(b) A normal curve must be symmetrical.
68%
1.82 ≈ 2 babies
4.
(a) z  0.33
1.
(b) 0.86  z  0.57
CHAPTER 7, FORM A
(c) z  1.33
1.
(d) x  2.6
2.
3.
(a) A normal curve is bell-shaped with one
peak. This curve has two peaks.
(b) A normal curve never crosses the horizontal
axis.
99.7%
0.91 ≈ 1 boy
4.
(a) z  1.8
(e) 6.8  x  11
5.
(b) 1.2  z  0.8
(c) z  2.2
6.
(d) x  3.1
(e) 4.1  x  4.6
5.
6.
(f) x  4.35
(a) No, look at z values to compare different
populations.
(b) For Joel, z = 1.51. For John, z = 2.77.
Relative to the district, John is a better
salesman.
(a) P(x  5) = 0.2198
(b) P(x  10) = 0.0668
7.
8.
9.
7.
8.
(c) P(5  x  10) = 0.7134
29.90 minutes or 30 minutes
(a) Here, np = (57)(.87) > 5 and
nq = (57)(.13) > 5.
Yes, the assumptions are satisfied.
(b) P(r ≤ 47) ≈ 0.2053
(c) P(47 ≤ r ≤ 55) ≈ 0.8781
9.
(f) x  13.1
(a) Generator II is hotter; see the z values.
(b) z = 0.72 for Generator I
z = 2.75 for Generator II
Generator II could be near a melt down
since it is hotter.
(a) P(x < 6) = 0.0912
(b) P(x > 7) = 0.0228
(c) P(6  x  7) = 0.8860
3 years
(a) Here, np = (88)(.71) > 5 and
nq = (88)(.29) > 5.
Yes, the assumptions are satisfied.
(b) P(r ≥ 60) ≈ 0.6791
(c) P(60 ≤ r ≤ 70) ≈ 0.7283
(a) P  5.2  x  = 0.9664
(b) P  x  7.1 = 0.0916
(c) P 5.2  x  7.1 = 0.8748
10. (a) P 8  x  = 0.9845
(b) P  x  9 = 0.2949
(c) P 8  x  9 = 0.6896
(a) P  x  3 = 0.3036
(b) P  x  3.5 = 0.2203
(c) P  3.1  x  3.3 = 0.2029
10. (a) P  x  23 = 0.0472
(b) P  x  28 = 0.1129
(c) P  23  x  28 = 0.8399
CHAPTER 7, FORM C
1.
2.
3.
4.
5.
(c)
(b)
(e)
A.
B.
A.
B.
C.
Copyright © Houghton Mifflin Harcourt Company. All rights reserved.
(d)
(a)
(d)
(b)
(a)
Part III: Answers to Sample Chapter Tests
6.
A.
B.
C.
7. (c)
8. A.
B.
9. A.
B.
C.
D.
10. A.
B.
C.
11. A.
B.
C.
(c)
(e)
(b)
3.
4.
(a)
(b)
(c)
(b)
(a)
(a)
(b)
(d)
(c)
(a)
(d)
(e)
5.
6.
7.
2.
3.
4.
5.
6.
7.
(a) p̂ = 41/56 or 0.732
(b)
(c)
(d)
20
(a)
(b)
0.62 to 0.85
np > 5 and nq > 5; yes
246 more
271
167
CHAPTER 8, FORM C
1.
2.
CHAPTER 8, FORM A
1.
(b) The distribution is mound-shaped
symmetrical and  is known.
(c) $120 to $156
10.90 to 18.50 pages
89.0 to 97.4 minutes; We are 99% confident that
the interval 89.0 to 97.4 contains the population
mean repair time at Computer Depot.
27.70 to 31.90 inches
We are 95% confident that the interval 27.70 to
31.90 contains the population mean Blue Spruce
circumference.
(a) $33.04 to $36.36
(b) The distribution is mound-shaped
symmetrical and  is unknown.
(c) $826.00 to $909.00
27.98 to 38.78 minutes
10.05 to 13.75 inches; We are 99% confident
that the interval 10.05 to 13.75 contains the
population mean rainbow trout length.
3.
4.
5.
6.
7.
(b)
A.
B.
(a)
(b)
A.
B.
C.
(d)
A.
B.
(c)
(e)
(d)
(a)
(e)
(a)
(d)
(a) p̂ = 59/78  0.756
(b) 0.68 to 0.84
(c) np > 5 and nq > 5; yes
(d) 122 more
147
(a) 1692
(b) 715
CHAPTER 8, FORM B
1.
2.
1.88 to 2.92; We are 95% confident that the
interval 1.88 to 2.92 contains the population
mean number of calls received each night at the
O’Sullivan household.
(a) $6.00 to $7.80
Copyright © Houghton Mifflin Harcourt Company. All rights reserved.
Instructor’s Resource Guide Understanding Basic Statistics, 5th Edition
For all hypothesis testing questions in chapters 9 – 11, the P-values will be given as determined by a TI-84 Plus.
Sketches of the sampling distributions were generated using Minitab 15. As such, there may be slight rounding
discrepancies in the P-values between the technologies.
(e) Do not reject H0. At the 5% level of
significance, there is not enough evidence to
CHAPTER 9, FORM A
conclude that the waiting time has
1. (a)   0.01
decreased.
3. (a)   0.01
H0:  = 24,819
H0: p = 0.75
H1:  > 24,819
H1: p < 0.75
Right-tailed
Left-tailed
(b) Standard normal; z = 3.60
(b) Standard normal; z = 2.81
(c) P-value = 0.0024
(c) P-value = 0.00016
Distribution Plot
Normal, Mean=0, StDev=1
Distribution Plot
Normal, Mean=0, StDev=1
0.4
0.4
0.3
Density
Density
0.3
0.2
0.2
0.1
0.00248
0.1
0.0
0.000159
0.0
2.
0
z
(d) z0 = 2.33
(e) Reject H0. At the 1% level of significance,
there is evidence that the population mean
daily sales has increased.
(a)   0.05
H0:  = 18.3
H1:  < 18.3
Left-tailed
(b) Student’s t; d.f. = 4; t = 1.154
(c) P-value = 0.1564
-2.81
0
4.
(d) z0 = 2.33
(e) Reject H0. At the 1% level of significance,
there is evidence that the proportion of
rainbow trout is less than 75%.
(a)   0.05
H0:  = 71
H1:   71
Two-tailed
(b) Standard normal; z = 6.236
(c) P-value ≈ 0.0000
Distribution Plot
Distribution Plot
T, df=4
Normal, Mean=0, StDev=1
0.4
0.4
0.3
0.3
Density
Density
z
3.6
0.2
0.1
0.2
0.1
0.156
0.0
2.2445E-10
-1.154
0
t
(d) t0 = -2.132
Copyright © Houghton Mifflin Harcourt Company. All rights reserved.
0.0
-6.24
2.2445E-10
z
0
6.236
Part III: Answers to Sample Chapter Tests
5.
(d) z0 = 1.96
(e) Reject H0. At the 5% level of significance,
there is evidence that the population mean
height is different from 71 inches.
(a)   0.05
H0:  = 30
(e) Do not reject H0. At the 5% level of
significance, there is not enough evidence
to conclude that the proportion is less than
60%.
CHAPTER 9, FORM B
H1:  > 30
Right-tailed
(b) Student’s t; d.f. = 13; t = 0.364
(c) P-value = 0.361
1.
(a)   0.01
H0:  = 5.3
H1:  < 5.3
Left-tailed
(b) Standard normal; z = 3.28
(c) P-value = 0.00053
Distribution Plot
T, df=13
Distribution Plot
Normal, Mean=0, StDev=1
0.4
0.4
0.3
Density
Density
0.3
0.2
0.1
0.361
0.1
0.000519
0.0
t
0.364
0.0
(d) t0 = 1.771
(e) Do not reject H0. At the 5% level of
significance, there is not enough evidence to
conclude that the delivery time is greater
than 30 minutes.
(a)   0.05
H0: p = 0.6
H1: p < 0.6
Left-tailed
(b) Standard normal; z = 1.14
(c) P-value = 0.1269
2.
-3.28
z
0
(d) z0 = 2.33
(e) Reject H0. At the 1% level of significance,
there is evidence that the average recovery
time is less as an outpatient.
(a)   0.05
H0:  = 4.19
H1:  > 4.19
Right-tailed
(b) Student’s t; d.f. = 15; t = 3.2
(c) P-value = 0.0030
Distribution Plot
T, df=15
0.4
Distribution Plot
Normal, Mean=0, StDev=1
0.4
Density
0.3
0.3
Density
6.
0.2
0.2
0.2
0.1
0.00298
0.1
0.0
0.127
0.0
(d) z0 = 1.645
0
t
-1.14
3.2
0
z
(d) t0 = 1.753
(e) Reject H0. At the 5% level of significance,
there is evidence that the average yield of
Arizona municipal bonds is higher than the
national average yield.
Copyright © Houghton Mifflin Harcourt Company. All rights reserved.
Instructor’s Resource Guide Understanding Basic Statistics, 5th Edition
3.
(a)   0.01
Right-tailed
(b) Student’s t; d.f. = 15; t = 2.38
(c) P-value = 0.0155
H0: p = 0.37
H1: p  0.37
Two-tailed
(b) Standard normal; z = 0.467
(c) P-value = 0.6408
Distribution Plot
T, df=15
0.4
Distribution Plot
0.3
Normal, Mean=0, StDev=1
Density
0.4
0.2
0.3
Density
0.1
0.2
0.0155
0.0
0.1
0.320
0.0
2.38
t
0.320
-0.467 0 0.467
z
4.
0
(d) z0 = 2.576
(e) Do not reject H0. At the 1% level of
significance, there is no evidence to
conclude that the proportion of freshmen on
the Dean’s List is different than 37%.
(a)   0.05
6.
(d) t0 = 1.753
(e) Reject H0. At the 5% level of significance,
there is evidence that the average age is
greater than 30 years.
(a)   0.05
H0: p = 0.27
H1: p > 0.27
Right-tailed
(b) Standard normal; z = 2.746
(c) P-value = 0.0030
H0:  = 2.8
H1:   2.8
Two-tailed
Distribution Plot
Normal, Mean=0, StDev=1
(b) Standard normal; z = 2.598
(c) P-value = 0.0094
0.4
Density
0.3
0.2
Distribution Plot
Normal, Mean=0, StDev=1
0.1
0.4
0.00302
0.0
Density
2.746
0.2
(d) z0 = 1.645
(e) Reject H0. At the 5% level of significance,
there is sufficient evidence to conclude that
the proportion is greater than 27%.
0.1
0.00469
0.0
-2.598
0.00469
0
z
2.60
CHAPTER 9, FORM C
5.
0
z
0.3
(d) z0 = 1.96
(e) Reject H0. At the 5% level of significance,
there is evidence that the population mean
time is different from 2.8 hours.
(a)   0.05
H0:  = 30
H1:  > 30
1.
2.
A.
B.
C.
D.
E.
A.
Copyright © Houghton Mifflin Harcourt Company. All rights reserved.
(b)
(b)
(b)
(d)
(b)
(e)
Part III: Answers to Sample Chapter Tests
4.
5.
6.
(a)
(b)
(d)
(b)
(d)
(e)
(c)
(a)
(a)
(b)
(c)
(e)
(c)
(a)
(c)
(b)
(d)
(c)
(a)
(c)
(e)
(d)
(a)
(a)
Distribution Plot
Normal, Mean=0, StDev=1
0.4
Density
0.3
0.115
0.0
5.
z
1.2
(d) z0 = 1.96
(e) Do not reject H0. At the 5% level of
significantly, there is not enough evidence to
conclude that there is difference in
population proportions of success between
the two different sales pitches.
(a)   0.05
(b) Student’s t; t = 2.263
(c) P-value = 0.0365
Distribution Plot
T, df=5
0.4
0.3
8.29 to 1.71 with d.f. = 15.14 using a TI-83;
Since the interval contains numbers that are all
negative it, appears that the population mean
duration of the first drug is less than that of the
second drug.
3.
0.189 to 0.130; Since the interval contains both
positive and negative values, there is no evidence
of a difference in the population proportions of
successful portfolios managed by Kendra
compared to those managed by Lisa.
(a)   0.05
0.2
0.1
(a) 0.61 to 2.79
(b) Since the interval contains all positive
numbers, it seems that battery I has a larger
population mean.
H 0: p 1 = p 2
H 1: p 1  p 2
Two-tailed
(b) Standard normal; z = 1.20
(c) P-value = 0.2304
0
H1: d < 0
left-tailed
2.
4.
0.115
-1.2
H0: d = 0
CHAPTER 10, FORM A
1.
0.2
0.1
Density
3.
B.
C.
D.
E.
A.
B.
C.
D.
E.
A.
B.
C.
D.
E.
A.
B.
C.
D.
E.
A.
B.
C.
D.
E.
0.0
6.
0.0365
-2.263
t
0
(d) t0 = 2.015
(e) Reject H0. At the 5% level of significance,
there is evidence to conclude that the new
process increases the mean number of items
processed per shift.
(a)   0.01
H0: 1 = 2
H1: 1 > 2
Right-tailed
(b) Student’s t; d.f. = 8; t = 1.480
(c) P-value = 0.0886
Copyright © Houghton Mifflin Harcourt Company. All rights reserved.
Instructor’s Resource Guide Understanding Basic Statistics, 5th Edition
Distribution Plot
T, df=8
0.4
Density
0.3
0.2
3.
0.1
0.0886
0.0
7.
0
X
1.48
(d) t0 = 2.896
(e) Do not reject H0. At the 1% level of
significance, there is not sufficient evidence
to show that the population mean profit per
employee in computer stores is higher than
those for building supply stores.
(a)   0.05
4.
(b) n1p1, n1q1, n2p2, n2q2 are all greater than 5;
yes
(c) The interval contains all positive values and
shows that at the 90% confidence level, the
population proportion of defects is greater
on shift I.
(a) 15.49 to 46.51
(b) The interval contains values that are all
positive. At the 90% confidence level, the
population mean stopping distance of the old
tread design is greater than that for the new.
(a)   0.05
H0: p1 = p2
H1: p1 < p2
Left-tailed
(b) Standard normal; z = 0.894
(c) P-value = 0.1858
H0: 1 = 2
H1: 1  2
Two-tailed
Distribution Plot
Normal, Mean=0, StDev=1
(b) Standard normal; z = 1.52
(c) P-value = 0.1292
0.4
Density
0.3
Distribution Plot
Normal, Mean=0, StDev=1
0.2
0.4
0.1
0.186
Density
0.3
0.0
0.2
-0.894
0
z
0.1
0.0643
0.0
0.0643
-1.52
0
z
1.52
(d) z0 = 1.96
(e) Do not reject H0. At the 5% level of
significance, there is not enough evidence to
say that the mean weight of tomatoes grown
organically is different than that of other
tomatoes.
CHAPTER 10, FORM B
1.
2.
(a) -$103.60 to $461.60 with d.f. = 29.29 using
a TI-83.
(b) Since the interval contains both positive and
negative values, it does not appear that the
population mean daily sales of the printers
differ.
(a) 0.018 to 0.093
5.
(d) z0 = 1.645
(e) Do not reject H0. At the 5% level of
significance, there is not enough evidence to
say that the proportion of seniors is higher
than the proportion of juniors.
(a)   0.05
H0: d = 0
H1: d < 0
Left-tailed
(b) Student’s t; d.f. = 4; t = 2.566
(c) P-value = 0.0311
Copyright © Houghton Mifflin Harcourt Company. All rights reserved.
Part III: Answers to Sample Chapter Tests
Distribution Plot
Normal, Mean=0, StDev=1
Distribution Plot
0.4
T, df=4
0.4
0.3
Density
Density
0.3
0.2
0.2
0.1
0.000200
0.1
0.0
0.0
6.
0.0311
-2.566
(d) z0 = 1.645
(e) Reject H0. At a 5% level of significance, the
evidence indicates that the average egg
production of range free chickens is higher.
CHAPTER 10, FORM C
1.
2.
3.
4.
H1: 1 > 2
Right-tailed
(b) Student’s t; d.f. = 8; t = 1.744
(c) P value = 0.0597
Distribution Plot
T, df=8
0.4
5.
0.3
Density
3.54
t
(d) t0 = 2.132
(e) Reject H0. At the 5% level of significance,
there is evidence that the average time for
runners at higher elevation is longer.
(a)   0.01
0.2
0.1
0.0597
0.0
z
0
H0: 1 = 2
0
t
7.
0
1.744
(d) t0 = 2.896
(e) Do not reject H0. At the 1% level of
significance, there is not enough evidence to
conclude that the average mileage for the
Pacer is greater.
(a)   0.05
6.
7.
H0: 1 = 2
H1: 1 > 2
Right-tailed
(b) Standard normal; z = 3.54
(c) P-value = 0.0002
(b)
(e)
(d)
A.
B.
C.
D.
E.
A.
B.
C.
D.
E.
A.
B.
C.
D.
E.
A.
B.
C.
D.
E.
(e)
(b)
(c)
(d)
(a)
(d)
(c)
(c)
(e)
(a)
(b)
(b)
(b)
(a)
(a)
(c)
(e)
(c)
(e)
(b)
CHAPTER 11, FORM A
1.
Copyright © Houghton Mifflin Harcourt Company. All rights reserved.
Instructor’s Resource Guide Understanding Basic Statistics, 5th Edition
Scatter Diagram of Pounds versus Inches for Northern Pike
(c) P-value = 0.0046
20
Distribution Plot
Chi-Square, df=4
0.20
0.15
10
Density
Pounds
15
0.10
5
0.05
0.00464
0
20
25
30
35
40
45
0.00
Inches
2.
The linear correlation coefficient appears to be
positive since as x increases, y tends to increase.
(a) x  33.4; y  10.6
(b)
 x  167;  y  53;  x2  6069;
 y 2  789;  xy  2103
(c) b = 0.678; a = 12.0; yˆ  0.678x  12.0
4.
5.
(d) See problem 1.
r = 0.996; r2 = 0.992; 99.2% of the variation in y
can be explained by the least squares line using x
as the predicting variable.
Se = 0.757
For x = 32 inches, y = 9.65 pounds
6.
7.70  y  11.61 pounds
7.
 = 0.01
H0:  = 0
H1:   0
3.
9.
 = 0.05
H0:  = 0
H1:  > 0
t = 19.31; d.f. = 3; P-value = 0.00015
Reject H0; At the 5% level of significance, there
is sufficient evidence that the slope is positive.
Chi-square goodness of fit
(a)  = 0.05
H0: The distribution of ages of college
students is the same in the Western
Association as in the nation.
H1: The distribution of ages of college
students is different in the Western
Association than it is in the nation.
(b) Chi-Square; d.f. = 4;  2 = 15.03
15.03
X
(d) 02  9.49
(e) Reject H0. At the 5% level of significance,
there is evidence that the distribution of ages
is different.
10. Chi-square test of independence
(a)  = 0.05
H0: Teacher evaluations are independent of
midterm grades
H1: Teacher evaluations are not independent
of midterm grades
2
(b) Chi-Square; d.f. = 4;   12.36
(c) P-value = 0.0149
Distribution Plot
Chi-Square, df=4
0.20
0.15
Density
t = 19.31; d.f. = 3; P-value = 0.000304
Reject H0; There is evidence that the population
correlation coefficient is not equal to 0 at the 1%
significance level.
8.
0
0.10
0.05
0.00
0.0149
0
x
12.36
(d) 02  9.49
(e) Reject H0. There is evidence to say that at
the 5% level of significance teacher
evaluations are not independent of midterm
grades.
11. Chi-square test for 
(a)  = 0.01
H0:  = 4.1
H1:  > 4.1
(b) Chi-Square; d.f. = 24;  2 = 34.28
(c) P-value = 0.0798
Copyright © Houghton Mifflin Harcourt Company. All rights reserved.
Part III: Answers to Sample Chapter Tests
Distribution Plot
Chi-Square, df=24
5.
For a CEO salary of $1.5 million, the model
predicts an annual company revenue of $18.74
billion.
6.
$11.00  y  $26.43 billion
7.
 = 0.01
H0:  = 0
H1:   0
0.06
0.05
Density
0.04
0.03
0.02
0.01
0.0798
0.00
x
t = 3.21; d.f. = 3; P-value = 0.0490
Do not reject H0; There is not enough evidence
that the population correlation coefficient is not
equal to 0 at the 1% significance level.
34.28
(d) 02  42.98
(e) Do not reject H0. At the 1% significance
level, there is not enough evidence to reject
 = 4.1.
8.
9.
 = 0.05
H0:  = 0
H1:  > 0
t = 3.21; d.f. = 3; P-value = 0.0245
Reject H0; At the 5% level of significance, there
is sufficient evidence that the slope is positive.
Chi-square goodness of fit
(a)  = 0.05
H0: The distribution of fish fits the initial
stocking distribution
H1: The distribution of fish after six years
does not fit the initial stocking distribution
CHAPTER 11, FORM B
1.
(b) Chi-Square; d.f. = 4;  2 = 19.0
Scatter Diagram of Revenue versus Salary
(c) P-value = 0.00079
26
Distribution Plot
Chi-Square, df=4
22
0.20
20
18
0.15
16
Density
Revenue ($ Billions)
24
14
0.10
12
0.05
10
1.0
2.
1.5
Salary ($ Millions)
2.0
2.5
The linear correlation coefficient appears to be
positive since as x increases, y tends to increase.
(a) x  1.38; y  17.8
(b)
 x  6.9;  y  89;  x2  11.03;
 y 2  1703;  xy  134.6
(c) b = 7.81; a = 7.02; yˆ  7.81x  7.02
3.
4.
0.000786
0.00
(d) See problem 1.
r = 0.880; r2 = 0.775; 77.5% of the variation in y
can be explained by the least squares line using x
as the predicting variable.
Se = 2.988
0
x
19
(d) 02  9.49
(e) Reject H0. At the 5% level of significance,
there is enough evidence to conclude that
the fish distribution has changed.
10. Chi-square test of independence
(a)  = 0.01
H0: The choice of college major is
independent of grade average
H1: The choice of college major is not
independent of college average
Copyright © Houghton Mifflin Harcourt Company. All rights reserved.
Instructor’s Resource Guide Understanding Basic Statistics, 5th Edition
6.
7.
8.
9.
(c)
(b)
(c)
A.
B.
C.
D.
10. A.
B.
C.
D.
11. A.
B.
2
(b) Chi-Square; d.f. = 4;  = 2.64
(c) P-value = 0.6198
Distribution Plot
Chi-Square, df=4
0.20
Density
0.15
0.10
0.05
0.620
0.00
0
2.64
x
(d) 02  13.28
(e) Do not reject H0. At the 1% level of
significance, there is not enough evidence to
conclude that college major is not
independent of grade average.
11. Chi-square test for 
(a)  = 0.01
H 0:  = 6
H 1:  ≠ 6
(b) Chi-Square; d.f. = 21;  2 = 30.24
(c) P-value = 0.1745
Distribution Plot
Chi-Square, df=21
0.07
0.06
Density
0.05
0.04
0.03
0.02
0.01
0.00
0.0873
0
z
30.24
2  41.40
(d)  L2  8.03;  H
(e) Do not reject H0. At the 1% significance
level, there is not enough evidence to reject
 = 60.
CHAPTER 11, FORM C
1.
2.
3.
4.
5.
(a)
(d)
(b)
(a)
(a)
Copyright © Houghton Mifflin Harcourt Company. All rights reserved.
(c)
(e)
(d)
(a)
(d)
(c)
(a)
(b)
(e)
(b)
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