Solutions Manual – Statistics for Business and Economics, 11th Edition – Anderson – modified 2/16/ – Studocu
modified 2/16/
Mục Lục
EXCERPTS FROM:
Solutions Manual to Accompany
Statistics for Business
and Economics
Eleventh Edition
David R. Anderson
University of Cincinnati
Dennis J. Sweeney
University of Cincinnati
Thomas A. Williams
Rochester Institute of Technology
The material from which this was excerpted is copyrighted by
SOUTH-WESTERN
CENGAGE Learning
TM
Contents
-
- Data and Statistics
-
- Descriptive Statistics: Tabular and Graphical Methods
-
- Descriptive Statistics: Numerical Methods
-
- Introduction to Probability
-
- Discrete Probability Distributions…………………………………………………………………………………
-
- Continuous Probability Distributions
-
- Sampling and Sampling Distributions
-
- Interval Estimation
-
- Hypothesis Testing……………………………………………………………………………………………………..
-
- Statistical Inference about Means and Proportions with Two populations
-
- Simple Linear regression
-
- Multiple Regression
-
- Regression Analysis: Model Building
-
- Decision Analysis
2. Descriptive Statistics: Tabular and Graphical Methods
- a/b.
Waiting Time Frequency Relative Frequency
0 – 4 4 0.
5 – 9 8 0.
10 – 14 5 0.
15 – 19 2 0.
20 – 24 1 0.
Totals 20 1.
c/d.
Waiting Time Cumulative Frequency Cumulative Relative Frequency
Less than or equal to 4 4 0.
Less than or equal to 9 12 0.
Less than or equal to 14 17 0.
Less than or equal to 19 19 0.
Less than or equal to 24 20 1.
e. 12/20 = 0.
- a.
y
x
A
B
C
5
11
2
0
2
10
18 12
5
13
12
30
12 Total
Total
b.
y
x
A
B
C
100.
84.
16.
1
0.
15.
83.
2
100.
100.
100.
Total
c.
y
x
A
B
C
27.
61.
11.
100.
0.
16.
83.
100.
12
Total
d. Category A values for x are always associated with category 1 values for y. Category B values for x
are usually associated with category 1 values for y. Category C values for x are usually associated
with category 2 values for y.
- a.
Fuel Type
Year Constructed Elec Nat. Gas Oil Propane Other Total
1973 or before 40 183 12 5 7 247
1974-1979 24 26 2 2 0 54
1980-1986 37 38 1 0 6 82
1987-1991 48 70 2 0 1 121
Total 149 317 17 7 14 504
b.
Year Constructed Frequency Fuel Type Frequency
1973 or before 247 Electricity 149
1974-1979 54 Nat. Gas 317
1980-1986 82 Oil 17
1987-1991 121 Propane 7
Total 504 Other 14
Total 504
c. Crosstabulation of Column Percentages
Fuel Type
Year Constructed Elec Nat. Gas Oil Propane Other
1973 or before 26 57 70 71 50.
1974-1979 16 8 11 28 0.
1980-1986 24 12 5 0 42.
1987-1991 32 22 11 0 7.
Total 100 100 100 100 100.
d. Crosstabulation of row percentages.
Fuel Type
Year Constructed Elec Nat. Gas Oil Propane Other Total
1973 or before 16 74 4 2 2 100.
1974-1979 44 48 3 3 0 100.
1980-1986 45 46 1 0 7 100.
1987-1991 39 57 1 0 0 100.
3. Descriptive Statistics: Numerical Methods
- a.
3181
$
20
i
x
x
n
Σ
== =
b. Median 10th $160 Los Angeles
11th $162 Seattle
Median =
160 162
$
2
+
=
c. Mode = $167 San Francisco and New Orleans
d.
25
20 5
100
i
⎛⎞
==
⎜⎟
⎝⎠
5th $
6th $
1
134 139
$136.
2
Q
+
==
e.
75
20 15
100
i
⎛⎞
==
⎜⎟
⎝⎠
15th $
16th $
3
167 173
$
2
Q
+
==
- a. Range = 60 – 28 = 32
IQR = Q 3 – Q 1 = 55 – 45 = 10
b. x ==
435
9
48 33.
2
Σ− =( xxi ) 742
2
2 () 742
92.
18
xxi
s
n
Σ−
===
−
s ==92 9.
c. The average air quality is about the same. But, the variability is greater in Anaheim.
- a.
765
76.
10
xi
x
n
Σ
===
2
()442.
7
1101
xxi
s
n
Σ−
==
−−
=
b.
84 76.
1.
7
xx
z
s
−−
== =
Approximately one standard deviation above the mean. Approximately 68% of the scores are within
one standard deviation. Thus, half of (100-68), or 16%, of the games should have a winning score of
84 or more points.
90 76.
1.
7
xx
z
s
−−
== =
Approximately two standard deviations above the mean. Approximately 95% of the scores are
within two standard deviations. Thus, half of (100-95), or 2%, of the games should have a winning
score of more than 90 points.
c.
122
12.
10
xi
x
n
Σ
===
2
()559.
7.
1101
xxi
s
n
Σ−
==
−−
=
Largest margin 24:
24 12.
1.
7.
xx
z
s
−−
== =. No outliers.
- a.
b.
1.
.
9
xi
x
n
Σ
== =
1.
.
9
xi
y
n
Σ
== =
–
-0.
0
0.
1
S&P 500 -1 -1 -0 0 0 1 1.
DJIA
i
x
i
y ()
i
x − x ()
i
yy −
2
()
i
x − x
2
()
i
yy − () xii − xy y (− )
0 0 0 0 0 0 0.
0 0 0 0 0 0 0.
-0 -0 -1 -1 1 1 1.
0 0 -0 -0 0 0 0.
-0 -0 -0 -0 0 0 0.
1 0 0 0 0 0 0.
0 0 0 0 0 0 0.
0 0 0 0 0 0 0.
-0 -0 -0 -0 0 0 0.
Total 2 2 2.
4. Introduction to Probability
- a.
H
T
H
T
H
T
H T H T H T H T
(H,H,H)
(H,H,T)
(H,T,H)
(H,T,T)
(T,H,H)
(T,H,T)
(T,T,H)
(T,T,T)
1st Toss 2nd Toss 3rd Toss
b. Let: H be head and T be tail
(H,H,H) (T,H,H)
(H,H,T) (T,H,T)
(H,T,H) (T,T,H)
(H,T,T) (T,T,T)
c. The outcomes are equally likely, so the probability of each outcomes is 1/8.
- No. Requirement (4) is not satisfied; the probabilities do not sum to 1. P (E 1 ) + P (E 2 ) + P (E 3 ) +
P (E 4 ) = .10 + .15 + .40 + .20 =.
- a. Use the relative frequency method. Divide by the total adult population of 227 million.
Age Number Probability
18 to 24 29 0.
25 to 34 40 0.
35 to 44 43 0.
45 to 54 43 0.
55 to 64 32 0.
65 and over 37 0.
Total 227 1.
b. P (18 to 24) =.
c. P (18 to 34) = .1309 + .1757 =.
d. P (45 or older) = .1929 + .1437 + .1661 =.
- a. Let D = Domestic Equity Fund
P ( D ) = 16/25 =.
b. Let A = 4- or 5-star rating
13 funds were rated 3-star of less; thus, 25 – 13 = 12 funds must be 4-star or 5-star.
P ( A ) = 12/25 =.
c. 7 Domestic Equity funds were rated 4-star and 2 were rated 5-star. Thus, 9 funds were Domestic
Equity funds and were rated 4-star or 5-star
P ( D ∩ A ) = 9/25 =.
d. P ( D ∪ A ) = P ( D ) + P ( A ) – P ( D ∩ A )
= .64 + .48 – .36 =.
- Let: B = rented a car for business reasons
P = rented a car for personal reasons
a. P (B ∪ P) = P (B) + P (P) – P (B ∩ P)
= .54 + .458 – .30 =.
b. P (Neither) = 1 – .698 =.
- a. P (A ∩ B) = 0
b.
(A B) 0
(A B) 0
(B).
P
P
P
∩
===
c. No. P (A | B) ≠ P (A); ∴ the events, although mutually exclusive, are not independent.
d. Mutually exclusive events are dependent.
- a. Let O = flight arrives on time
O
c
= flight arrives late
S = Southwest flight
U = US Airways flight
J = JetBlue flight
Given: P (O | S) = .834 P (O | U) = .751 P (O | J) =.
P (S) = .40 P (U) = .35 P (J) =.
P (O | S) =
(O S)
(S)
P
P
∩
∴ P (O ∩ S) = P (O | S) P (S) = (.834)(.4) =.
Similarly
P (O ∩ U) = P (O | U) P (U) = (.751)(.35) =.
P (O ∩ J) = P (O | J) P (J) = (.701)(.25) =.
Joint probability table
On time Late Total
Southwest .3336 .0664.
US Airways .2629 .0871.
JetBlue .1753 .0747.
Total: .7718 .2282 1.
b. Southwest Airlines; P (S) =.
c. P (O) = P (S ∩ O) + P (U ∩ O) + P (J ∩ O) = .3336 + .2629 + .1753 =.
d.
c
c
c
(S O ).
(S O ).
(O ).
P
P
P
∩
===
Similarly,
c.
(U O ).
.
P ==
c.
(J O ).
.
P ==
Most likely airline is US Airways; least likely is Southwest
- M = missed payment
D 1 = customer defaults
D 2 = customer does not default
P(D 1 ) = .05 P(D 2 ) = .95 P(M | D 2 ) = .2 P(M | D 1 ) = 1
a.
11
1
112 2
P(D )P(M D ) (.05)(1).
P(D M).
P(D )P(M D ) P(D )P(M D ) (.05)(1) (.95)(.2).
==
++
==
b. Yes, the probability of default is greater than .20.
5. Discrete Probability Distributions…………………………………………………………………………………
- a. Let x = time (in minutes) to assemble the product.
b. It may assume any positive value: x > 0.
c. Continuous
- a. f (200) = 1 – f (-100) – f (0) – f (50) – f (100) – f (150)
= 1 – .95 =.
This is the probability MRA will have a $200,000 profit.
b. P(Profit) = f (50) + f (100) + f (150) + f (200)
= .30 + .25 + .10 + .05 =.
c. P(at least 100) = f (100) + f (150) + f (200)
= .25 + .10 +.05 =.
- a. E ( x ) = Σ x f ( x ) = 0 (.56) + 2 (.44) =.
b. E ( x ) = Σ x f ( x ) = 0 (.66) + 3 (.34) = 1.
c. The expected value of a 3 – point shot is higher. So, if these probabilities hold up, the team will
make more points in the long run with the 3 – point shot.
- a. Medium E ( x ) = Σ x f ( x ) = 50 (.20) + 150 (.50) + 200 (.30) = 145
Large: E ( x ) = Σ x f ( x ) = 0 (.20) + 100 (.50) + 300 (.30) = 140
Medium preferred.
b. Medium
x f ( x ) x – μ ( x – μ)
2
( x – μ)
2
f (x )
50 .20 -95 9025 1805.
150 .50 5 25 12.
200 .30 55 3025 907.
σ
2
= 2725.
Large
y f ( y ) y – μ ( y – μ)
2
( y – μ)
2
f (y )
0 .20 -140 19600 3920
100 .50 -40 1600 800
300 .30 160 25600 7680
σ
2
= 12,
Medium preferred due to less variance.
- a. f (0) =.
b. f (2) =.
c. P ( x ≤ 2) = f (0) + f (1) + f (2) = .3487 + .3874 + .1937 =.
d. P ( x ≥ 1) = 1 – f (0) = 1 – .3487 =.
e. E ( x ) = n p = 10 (.1) = 1
f. Var ( x ) = n p (1 – p ) = 10 (.1) (.9) = .9, σ = .9=.
- a. () ( )(1 )
x nx
n
fx p p
x
⎛⎞ −
=−
⎜⎟
⎝⎠
10! 31
(3) (.30) (1 .30)
3!(10 3)!
f
0 − 3
=−
−
10(9)(8) 37
(3) (.30) (1 .30).
3(2)(1)
f =−=
b. P ( x > 3) = 1 – f (0) – f (1) – f (2)
10! 010
(0) (.30) (1 .30).
0!(10)!
f =−=
10! 19
(1) (.30) (1 .30).
1!(9)!
f =−=
10! 28
(2) (.30) (1 .30).
2!(8)!
f =−=
P ( x > 3) = 1 – .0282 – .1211 – .2335 =.
- a.
2
2
()
!
x
e
fx
x
−
=
b. μ = 6 for 3 time periods
c.
6
6
()
!
x
e
fx
x
−
=
d.
22
2 4(.1353)
(2).
2! 2
e
f
−
== =
e.
66
6
(6).
6!
e
f
−
==
f.
54
4
(5).
5!
e
f
−
==
- Since the shipment is large we can assume that the probabilities do not change from trial to trial and
use the binomial probability distribution.
a. n = 5
505
(0) (0) (0) 0.
0
f
⎛⎞
==⎜⎟
⎝⎠
b.
14
5
(1) (0) (0) 0.
1
f
⎛⎞
==⎜⎟
⎝⎠
c. 1 – f (0) = 1 – .9510 =.
d. No, the probability of finding one or more items in the sample defective when only 1% of the items
in the population are defective is small (only .0490). I would consider it likely that more than 1% of
the items are defective.
Expected number of defects = 1000(.0026) = 2.
c. Reducing the process standard deviation causes a substantial reduction in the number of defects.
7. Sampling and Sampling Distributions
3. 459, 147, 385, 113, 340, 401, 215, 2, 33, 348
- a. The sampling distribution is normal with
E () x = μ = 200 and / 50 / 100 5
x
σσ== = n
For ± 5, 195 ≤≤ x 205. Using Standard Normal Probability Table:
At x = 205,
5
1
5
x
x
z
μ
σ
−
=== Pz (1≤ )=.
At x = 195,
5
1
5
x
x
z
μ
σ
−−
===− Pz (<−1)=.
Px (195≤≤205)= .8413 – .1587 =.
b. For ± 10, 190 ≤≤ x 210. Using Standard Normal Probability Table:
At x = 210, z
x
x
=
−
==
μ
σ
10
5
2 Pz (2≤ )=.
At x = 190,
10
2
5
x
x
z
μ
σ
−−
===− Pz (<−2)=.
Px (190≤≤210)= .9772 – .0228 =.
- a. Normal distribution: Ep () .12= ,
(1 ) (. 1 2 ) (1. 1 2 )
.
540
p
pp
n
σ
−−
== =
b.
.
2.
p.
pp
z
σ
−
=== P ( z ≤ 2) = .9838 P ( z < -2) =.
P (.09 ≤ p ≤ .15) = .9838 – .0162 =.
c.
.
1.
p.
pp
z
σ
−
=== P ( z ≤ 1) = .8577 P ( z < -1) =.
P (.105 ≤ p ≤ .135) = .8577 – .1423 =.
- a. Normal distribution because of central limit theorem ( n > 30)
E () x = 115 ,
35
5.
40
x
n
σ
σ == =
b.
10
1.
/35/ 40
x
z
n
μ
σ
−
== = P ( z ≤ 1) = .9649, P ( z < -1) =.
P (105 ≤ x ≤ 125) = P (-1 ≤ z ≤ 1) = .9649 – .0351 =.
c. At x = 100,
100 115.
2.
35 / 40
z
−
==− P ( x ≤ 100) = P ( z ≤ -2) =.
Yes, this is an unusually low spending group of 40 alums. The probability of spending this much or
less is only .0026.
- a. Normal distribution with E ( p ) = .15 and σ p
pp
n
=
−
==
() (.)(.)
.
1015085
150
0 0292
b. P (.12 ≤ p ≤ .18) =?
8. Interval Estimation
- Margin of error =
.
z (/ )σ n = 1(600/ 50 ) = 166.
A larger sample size would be needed to reduce the margin of error to $150 or less. Section 8 can
be used to show that the sample size would need to be increased to n = 62.
1(600 / n ) 150= Solving for n yields n = 62
- x ± tsn α/2(/ ) df = 53
a. 22 ± 1(4 / 54)
22 ± 1 or 21 to 23.
b. 22 ± 2(4 / 54)
22 ± 1 or 21 to 23.
c. 22 ± 2(4 / 54)
22 ± 1 or 20 to 24.
d. As the confidence level increases, there is a larger margin of error and a wider confidence interval.
- For the JobSearch data set, x = 22
and s = 11.
a. x = 22 weeks
b. margin of error =
.
ts n / =2(11) / 40=3.
c. The 95% confidence interval is x ±margin of error = 22 ± 3 or 18 to 25.
d. Skewness = 1, data are skewed to the right.
This modest positive skewness in the data set can be expected to exist in the population.
Regardless of skewness, this is a pretty small data set. Consider using a larger sample next time.
- a. nn ==
(. ) (. )
.
196 625
2
37 52 38
22
2
Use =
b. nn ==
(. ) (. )
.
196 625
1
150 06 151
22
2
Use =
- Use planning value p* =.
nn ==
(. ) (. )(. )
(. )
.
196 050 050
003
1067 11 1068
2
2
Use =
- a. p = 46/200 =.
b.
(1 ). 2 3(1. 2 3)
.
200
pp
n
−−
==,
.
pp (1 )
pz
n
−
± = .23 ± 1(.0298)
= .23 ± .0584 or .1716 to.
- a.
22
.
22
(1 ) (1) (.156)(1 .156)
562
(.03)
zp p
n
E
∗∗
− −
== =
b.
22
.
22
(1 ) (2) (.156)(1 .156)
970.
(.03)
zp p
n
E
∗∗
− −
== = Use 971
9. Hypothesis Testing……………………………………………………………………………………………………..
- a. H 0 : μ ≤ 600 Ha: μ > 600 assuming that you give benefit of doubt to the manager.
b. We are not able to conclude that the manager’s claim is wrong.
c. The manager’s claim can be rejected. We can conclude that μ > 600.
- a. H 0 : μ ≤ 14 Ha: μ > 14 Research hypothesis
b. There is no statistical evidence that the new bonus plan increases sales volume.
c. The research hypothesis that μ > 14 is supported. We can conclude that the new bonus plan
increases the mean sales volume.
- a. H 0 : μ ≤ 8000
Ha: μ > 8000 Research hypothesis to see if the plan increases average sales.
b. Claiming μ > 8000 when the plan does not increase sales. A mistake could be implementing the
plan when it does not help.
c. Concluding μ ≤ 8000 when the plan really would increase sales. This could lead to not
implementing a plan that would increase sales.
- a.
0 26 25
1.
/6/ 40
x
z
n
μ
σ
− −
== =
b. Upper tail p -value is the area to the right of the test statistic
Using normal table with z = 1: p -value = 1 – .9306 =.
c. p -value > .01, do not reject H 0
d. Reject H 0 if z ≥ 2.
1 < 2, do not reject H 0
- a.
0
17 18
1.
/4/ 48
x
t
sn
−μ −
== =−
b. Degrees of freedom = n – 1 = 47
Because t < 0, p -value is two times the lower tail area
Using t table: area in lower tail is between .05 and .10; therefore, p -value is between .10 and .20.
Exact p -value corresponding to t = -1 is.
c. p -value > .05, do not reject H 0.
d. With df = 47, t .025 = 2.
Reject H 0 if t ≤ -2 or t ≥ 2.
t = -1; do not reject H 0
- a. H 0 : μ = 600, H a: μ ≠ 600
b.
0 612 600
1.
/65/ 40
x
t
sn
−μ −
== = df = n – 1 = 39
Because t > 0, p -value is two times the upper tail area
Using t table: area in upper tail is between .10 and .20; therefore, p -value is between .20 and .40.
Exact p -value corresponding to t = 1 is.
c. Withα= .10 or less, we cannot reject H 0. We are unable to conclude there has been a change in the
mean CNN viewing audience.
d. The sample mean of 612 thousand viewers is encouraging but not conclusive for the sample of 40
days. Recommend additional viewer audience data. A larger sample should help clarify the situation
for CNN.
- a. H 0 : μ = 2 H a: μ ≠ 2
b.
22
2.
10
i
x
x
n
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