1. No category

OPIM 5103 Stats
Prof. J. Stallaert
Solutions to Homework Set 2
CHAPTER 6
6.8
Computing Z-scores and using the table in the book yields:
(a)
P(34 < X < 50) = P(– 1.33 < Z < 0) = 0.5-0.0918=0.4082
(b)
P(X < 30) + P(X > 60) = P(Z < – 1.67) + P(Z > 0.83)
= 0.0475 + (1.0 – 0.7967) = 0.2508
(c)
(d)
6.9
P(Z < – 0.84)
P(X > A) = 0.80
 0.20
Z  –0.84 
A – 50
12
A = 50 – 0.84(12) = 39.92 thousand miles or 39,920 miles
The smaller standard deviation makes the Z-values larger.
(a)
P(34 < X < 50) = P(– 1.60 < Z < 0) = 0.5-0.0548=0.4452
(b)
P(X < 30) + P(X > 60) = P(Z < – 2.00) + P(Z > 1.00)
= 0.0228 + (1.0 – 0.8413) = 0.1815
(c)
A = 50 – 0.84(10) = 41.6 thousand miles or 41,600 miles
Computing Z-scores and using the table in the book yields:
(a)
P(12 < X < 15) = P(-0.15 < Z < 1.35) = 0.4711
(b)
P(X < 10) = P(Z < -1.15) = 0.1251
(c)
P(Xlower < X < Xupper) = 0.95
P(Z < – 1.96) = 0.0250 and P(Z < 1.96) = 0.9750
Xlow er – 5
1.5
Z  –1.96 
Z  1.96 
Xupper – 5
1.5
Xlower = 8.3801 and Xupper = 16.2199
6.10
Computing Z-scores and using the table in the book yields:
(a)
P(X < 91) = P(Z < 2.25) = 0.9878
(b)
P(65 < X < 89) = P(– 1.00 < Z < 2.00) = 0.9772 – 0.1587 = 0.8185
(c)
P(X > A) = 0.05
P(Z < 1.645) = 0.9500
Z  1.645 
(d)
A – 73
8
Option 1:
A = 73 + 1.645(8) = 86.16%
P(X > A) = 0.10
81 – 73
Z
 1.00
8
P(Z < 1.28)
 0.9000
Since your score of 81% on this exam represents a Z-score of 1.00, which is below
the minimum Z-score of 1.28, you will not earn an “A” grade on the exam under this
grading option.
Option 2: Z 
68 – 62
 2.00
3
Since your score of 68% on this exam represents a Z-score of 2.00, which is well
above the minimum Z-score of 1.28, you will earn an “A” grade on the exam under
this grading option. You should prefer Option 2.
OPIM 5103 Stats
Prof. J. Stallaert
Solutions to Homework Set 2
6.11
Computing Z-scores and using the table in the book yields:
(a)
P(X < 180) = P(Z < – 1.50) = 0.0668
(b)
P(180 < X < 300) = P(– 1.50 < Z < 1.50) = 0.9332 – 0.0668 = 0.8664
(c)
P(110 < X < 180) = P(– 3.25 < Z < – 1.50) = 0.0668 – 0.00058 = 0.06622
(d)
P(X < A) = 0.01
P(Z < – 2.33) = 0.01
A = 240 – 2.33(40) = 146.80 seconds
6.13
(a) Computing Z-scores and using the table in the book yields:
P(21.99 < X < 22.00) = P(– 2.4 < Z < – 0.4) = 0.3364
(b)
P(21.99 < X < 22.01) = P(– 2.4 < Z < 1.6) = 0.9370
(c)
P(X > A) = 0.02
Z = 2.05
A = 22.0123
(d)
(a)
P(21.99 < X < 22.00) = P(– 3.0 < Z < – 0.5) = 0.3072
(b)
P(21.99 < X < 22.01) = P(– 3.0 < Z < 2) = 0.9759
(c)
P(X > A) = 0.02
Z = 2.05
A = 22.0102
6.45
(a)
(b)
(c)
(d)
(e)
P(0.75 < X < 0.753) = P(– 0.75 < Z < 0) = 0.2734
P(0.74 < X < 0.75) = P(– 3.25 < Z < – 0.75) = 0.2266 – 0.00058 = 0.2260
P(X > 0.76) = P(Z > 1.75) = 1.0 – 0.9599 = 0.0401
P(X < 0.74) = P(Z < – 3.25) = 0.00058
P(X < A) = P(Z < – 1.48) = 0.07
A = 0.753 – 1.48(0.004) = 0.7471
6.46
(a)
(b)
(c)
(d)
(e)
P(1.90 < X < 2.00) = P(– 2.00 < Z < 0) = 0.4772
P(1.90 < X < 2.10) = P(– 2.00 < Z < 2.00) = 0.9772 – 0.0228 = 0.9544
P(X < 1.90) + P(X > 2.10) = 1 – P(1.90 < X < 2.10) = 0.0456
P(X > A) = P( Z > – 2.33) = 0.99A = 2.00 – 2.33(0.05) = 1.8835
P(A < X < B) = P(– 2.58 < Z < 2.58) = 0.99
A = 2.00 – 2.58(0.05) = 1.8710
B = 2.00 + 2.58(0.05) = 2.1290
OPIM 5103 Stats
Prof. J. Stallaert
Solutions to Homework Set 2
CHAPTER 7
7.21
7.24
7.27

2
 0.4
25
(a)
X =
(b)
P(7.8 < X < 8.2) = P(– 0.50 < Z < 0.50) = 0.6915 – 0.3085 = 0.3830
P(7.5 < X < 8.0) = P(– 1.25 < Z < 0) = 0.5 – 0.1056 = 0.3944
(c)
X =
 0.2
n
100
P(7.8 < X < 8.2) = P(– 1.00 < Z < 1.00) = 0.8413 – 0.1587 = 0.6826
(d)
With the sample size increasing from n = 25 to n = 100, more sample means will be
closer to the distribution mean. The standard error of the sampling distribution of size
100 is much smaller than that of size 25, so the likelihood that the sample mean will
fall within 0.2 minutes of the mean is much higher for samples of size 100
(probability = 0.6826) than for samples of size 25 (probability = 0. 3830).
(a)
p = 15/50 = 0.30
(b)
p =
(a)
(b)
P(0.50 < p < 0.60) = P(0 < Z < 2.83) = 0.4977
P(– 1.645 < Z < 1.645) = 0.90
p = .50 – 1.645(0.0354) = 0.4418
p = .50 + 1.645(0.0354) = 0.5582
P(p > 0.65) = P (Z > 4.24) = virtually zero
If n = 200, P(p > 0.60) = P (Z > 2.83) = 1.0 – 0.9977 = 0.0023
(c)
(d)
n



2
0.40(0.60)
= 0.0693
50
If n = 1000, P(p > 0.55) = P (Z > 3.16) = 1.0 – 0.99921 = 0.00079
More than 60% correct in a sample of 200 is more likely than more than 55% correct
in a sample of 1000.
7.32
(a)  p    0.56,  p 
 1   
n
=
0.561  0.56
= 0.04964
100
P(p < 0.5) = P (Z < -1.2087) = 0.1134
(b)
(c)
0.561  0.56
= 0.02220
n
500
P(p < 0.5) = P (Z < -2.7028) = 0.0034
 p    0.56,  p 
 1   
=
Increasing the sample size by a factor of 5 decreases the standard error by a factor of
√ . The sampling distribution of the proportion becomes more concentrated around
the true proportion of 0.56 and, hence, the probability in (b) becomes smaller than
that in (a).
OPIM 5103 Stats
7.47
 1   
 P   = 0.98,  p 
(a)
(b)
(c)
7.48
Prof. J. Stallaert
n

n

20
= 6.3246
10
PHStat output:
Probability for X <=
X Value
Z Value
P(X<=-10)
(b)
0.980.02
= 0.00626
500
P(p > 0.99) = P(Z > 1.5972) = 0.0551
P(0.97 < p < 0.99) = P(–1.5972 < Z < 1.5972 ) = 0.8898
P(p < 0.97) = P(Z < –1.5972) = 0.0551
 X   = -43.95,  X =
(a)

-10
5.3679663
1
P( X < -10) = P(Z < 5.3680) = 0.9999
PHStat output:
Probability for a Range
From X Value
To X Value
Z Value for -20
Z Value for 0
P(X<=-20)
P(X<=0)
P(-20<=X<=0)
(c)
Solutions to Homework Set 2
-20
0
3.786827
6.949105
0.9999
1.0000
0.0001
P(-20 < X < 0) = P(3.7868< Z < 6.9491) = 0.0001
PHStat output:
Probability for X >
X Value
Z Value
P(X>-30)
-30
2.2056887
0.0137
P( X > -30) = P(Z > 2.2057) = 0.0137
7.49
 = 10
(a)
  24.03
(b)
(c)
P(X < 0) = P(Z < -2.403) = 0.0081
P(10 < X < 20) = P(-1.403 < Z < -0.403) = 0.2632
P(X> 10) = P(Z > -1.403) = 0.9197
(d)
 X   = 24.03,  X =
(e)
(f)
(g)

n

10
=5
4
P( X < 0) = P(Z < -4.806) = 0.0000
P(10 < X < 20) = P(-2.806 < Z < -0.806) = 0.2076
P( X > 10) = P(Z < -2.806) = 0.9975
Since the sample mean of return of a sample of treasury bond is distributed closer than
the return of a single treasury bond to the population mean, the probabilities in (a) and
(b) are greater than those in (d) and (e) while the probability in (c) is smaller than that in
(f).
OPIM 5103 Stats
Prof. J. Stallaert
Solutions to Homework Set 2
OPIM 5103 Stats
Prof. J. Stallaert
Solutions to Homework Set 2
CHAPTER 8
8.9
0.9877    1.0023
(b)
Since the value of 1.0 is included in the interval, there is no reason to believe that the
mean is different from 1.0 gallon.
No. Since  is known and n = 50, from the Central Limit Theorem, we may assume
that the sampling distribution of X is approximately normal.
The reduced confidence level narrows the width of the confidence interval.
(d)
X  Z
(b)
(a)
n

n
 0.995  2.58 
0.02
50
X  Z
(c)
8.10

(a)
 0.995  1.96 
0.02
50
0.9895    1.0005
Since the value of 1.0 is still included in the interval, there is no reason to
believe that the mean is different from 1.0 gallon.
Using the Excel spreadsheet (“std dev known”):
Sample size
Sample Mean
Known Std Dev
Confidence
level
0.95
Point Estimate
Alpha
Margin of error
Interval
350
0.05
-24.4995
325.5005
64
350
100
374.4995
Hence: 325.5    374.50
(b)
(c)
(d)
No. The manufacturer cannot support a claim that the bulbs last an average 400
hours. Based on the data from the sample, a mean of 400 hours would represent a
distance of 4 standard deviations above the sample mean of 350 hours.
No. Since  is known and n = 64, from the Central Limit Theorem, we may assume
that the sampling distribution of X is approximately normal.
The confidence interval is narrower based on a process standard deviation of 80
hours rather than the original assumption of 100 hours.

330.4    369.6
X  Z
(b)
Based on the smaller standard deviation, a mean of 400 hours would
represent a distance of 5 standard deviations above the sample mean of 350
hours. No, the manufacturer cannot support a claim that the bulbs have a
mean life of 400 hours.
n
 350  1.96 
80
64
(a)
OPIM 5103 Stats
8.18
Solutions to Homework Set 2
One needs to find the sample mean and sample standard deviation of the data set first by
entering the data in Excel an using the functions AVERAGE and STDDEV. Using the Excel
spreadsheet (“std dev unknown”) then yields:
(a)
Sample size
Sample Mean
Sample Std Dev
Confidence level
(b)
8.19
Prof. J. Stallaert
6
36.53
4.3896
0.95
Point Estimate
Alpha
Margin of error
Interval
36.53
0.05
4.606603
31.9234
Hence: 31.9267
41.1399
41.1366
You can be 95% confident that the population mean price for two tickets with online
service charges, large popcorn, and two medium soft drinks is somewhere between
$31.93 and $41.14.
One needs to find the sample mean and sample standard deviation of the data set first by
entering the data in Excel an using the functions AVERAGE and STDDEV. Using the Excel
spreadsheet (“std dev unknown”) then yields:
Sample size
Sample Mean
Sample Std Dev
Confidence level
Point Estimate
Alpha
Margin of error
Interval
(a)
(b)
(c)
16
27.4375
2.6575
0.95
27.4375
0.05
1.416082
26.02142
28.85358
26.0214
28.8536
You can be 95% confident that the population mean miles per gallon of 2009 family
sedans priced under $20,000 is somewhere between 26.0214 and 28.8536.
Since the 95% confident interval for population mean miles per gallon of 2007 small
SUVs does not overlap with that for the population mean miles per gallon of 2007
family sedans, we are 95% confident that the population mean miles per gallon of
2007 small SUVs is lower than that of 2007 family sedans.
OPIM 5103 Stats
8.30
(a)
Prof. J. Stallaert
Solutions to Homework Set 2
Using the Excel spreadsheet (“proportions”) we get:
Sample size
Sample proportion
Confidence level
500
0.52
0.95
Point estimate
Alpha
Margin of error
Interval
0.52
0.05
-0.04379
0.476209
0.563791
Or:
X 260
= 0.52 p  Z 

n 500
Hence: 0.4762    0.5638
p
p(1– p)
0.52(1  0.52)
 0.52  1.96
n
500
(b)
Since the 95% confidence interval contains 0.50, you cannot claim that more than
half of all U.S. workers have negotiated a pay raise.
(c)
(a)
X 2600
= 0.52 p  Z 

n 5000
Hence: 0.5062    0.5338
p
p(1– p)
0.52(1  0.52)
 0.52  1.96
n
5000
(b)
(d)
8.31
Since the lower limit of the 95% confidence interval is greater than 0.50, you
can claim that more than half of all U.S. workers have negotiated a pay raise.
The larger the sample size, the narrow is the confidence interval holding everything
else constant.
Using the Excel spreadsheet (“proportions”) we get:
Sample size
Sample proportion
Confidence level
Point estimate
Alpha
Margin of error
Interval
1000
0.76
0.95
0.76
0.05
0.02647
0.73353
0.78647
Or:
p
p(1– p)
0.76(1  0.76)
 0.76  1.96
n
1000
0.7335    0.7865
X
760
= 0.76 p  Z 

n 1000
OPIM 5103 Stats
Prof. J. Stallaert
Solutions to Homework Set 2
CHAPTER 9
9.46
H0:   2.8 feet.
The mean length of steel bars produced is at least 2.8 feet and the production equipment does
not need immediate adjustment.
H1:  < 2.8 feet.
The mean length of steel bars produced is less than 2.8 feet and the production equipment
does need immediate adjustment.
(b)
From the hypo.xls spreadsheet (one sample means):
INPUT
Sample mean
Sample Std. Dev.
Hypothesized mean
Sample size
OUTPUT
t-value
P-value (two-tailed
test)
P-value (one-tailed
test)
(c)
9.57
(a)
2.73
0.2
2.8
25
-1.75
0.092895089
0.046447544
Decision: Since p value = 0.0464 is less than  = 0.05, reject H0. There is enough
evidence to conclude the production equipment needs adjustment.
The probability of obtaining a sample whose mean is 2.73 feet or less when the null
hypothesis is true is 0.0464.
H0:   0.5 H1:  < 0.5
From the hypo.xls spreadsheet (one sample proportion):
INPUT
Sample proportion
Hypothesized proportion
Sample size
OUTPUT
Std. Dev. (normal
approx.)
Z-value
P-value (two-tailed test)
P-value (one-tailed test)
ASSUMPTIONS
np
n(1-p)
(b)
0.42
0.5
500
0.022361
-3.57771
0.000347
0.000173
OK
OK
Conclusion: p-value of 0.000173 forces us to reject the null hypothesis.
H0:   0.5 H1:  < 0.5
INPUT
Sample proportion
0.42
OPIM 5103 Stats
Prof. J. Stallaert
Hypothesized proportion
Sample size
OUTPUT
Std. Dev. (normal
approx.)
Z-value
P-value (two-tailed test)
P-value (one-tailed test)
ASSUMPTIONS
np
n(1-p)
(c)
9.69
(a)
0.5
100
0.05
-1.6
0.109599
0.054799
OK
OK
Conclusion: p-value of 0.054799 means that there is not enough evidence to reject
the null hypothesis.
The larger the sample size, the smaller the sampling error. Even though the sample
proportion is the same value at 0.42 in (a) and (b), the test statistic is more negative
while the p-value is smaller in (a) compared to (b) because of the larger sample size
in (a).
H0:  < 0.5
H1:  > 0.5
From the hypo.xls spreadsheet 9one sample proportion):
INPUT
Sample proportion
Hypothesized proportion
Sample size
OUTPUT
Std. Dev. (normal
approx.)
Z-value
P-value (two-tailed test)
P-value (one-tailed test)
ASSUMPTIONS
np
n(1-p)
(b)
(c)
Solutions to Homework Set 2
0.52
0.5
1132
0.014861
1.345808
0.178364
0.089182
OK
OK
Decision: Since p-value = 0.0892 > 0.05, do not reject H0. There is not enough
evidence to show that more than half of all mobile phone users are now using the
mobile Internet.
The claim by the author is invalid.
INPUT
Sample proportion
Hypothesized proportion
Sample size
OUTPUT
Std. Dev. (normal
approx.)
Z-value
0.53
0.5
1132
0.014861
2.018712
OPIM 5103 Stats
Prof. J. Stallaert
P-value (two-tailed test)
P-value (one-tailed test)
ASSUMPTIONS
Np
n(1-p)
(d)
Solutions to Homework Set 2
0.043517
0.021759
OK
OK
H0:   0.5 H1:  > 0.5
Decision rule: If p-value < 0.05, reject H0.
p-value = 0.0218
Decision: Since p-value = 0.0218 < 0.05, reject H0. There is enough evidence
to show that more than half of all mobile phone users are now using the
mobile Internet.
(b)
The claim by the author is valid.
With the same level of significance, 53% is considered as far enough above 50% to
reject the null hypothesis of no more than half of all mobile phone users are now
using the mobile Internet while 52% is not considered as far enough above 0.5.