Chapter 5 Inferences Based on a Single Sample: Estimation with Confidence Intervals α

Transcription

Chapter 5
Inferences Based on a Single Sample:
Estimation with Confidence Intervals
5.1
5.2
5.3
a.
For α = .10, α/2 = .10/2 = .05. zα/2 = z.05 is the z score with .05 of the area to the right of it. The area
between 0 and z.05 is .5 − .05 = .4500. Using Table IV, Appendix B, z.05 = 1.645.
b.
For α = .01, /2 = .01/2 = .005. z /2 = z.005 is the z score with .005 of the area to the right of it. The
area between 0 and z.005 is .5 − .005 = .4950. Using Table IV, Appendix B, z.005 = 2.575.
c.
For α = .05, α/2 = .05/2 = .025. zα/2 = z.025 is the z score with .025 of the area to the right of it. The
area between 0 and z.025 is .5 − .025 = .4750. Using Table IV, Appendix B, z.025 = 1.96.
d.
For α = .20, α/2 = .20/2 = .10. zα/2 = z.10 is the z score with .10 of the area to the right of it. The area
between 0 and z.10 is .5 − .10 = .4000. Using Table IV, Appendix B, z.10 = 1.28.
a.
zα/2 = 1.96, using Table IV, Appendix B, P(0 ≤ z ≤ 1.96) = .4750. Thus, α/2 = .5000 − .4750 = .025,
α = 2(.025) = .05, and 1 − α = 1 - .05 = .95. The confidence level is 100% × .95 = 95%.
b.
α = 2(.05) = .1, and 1 − α = 1 − .1 = .90. The confidence level is 100% × .90 = 90%.
c.
d.
e.
zα/2 = .99, using Table IV, Appendix B, P(0 ≤ z ≤ .99) = .3389. Thus, α/2 = .5000 − .3389 = .1611,
α = 2(.1611) = .3222, and 1 − α = 1 − .3222 = .6778. The confidence level is 100% × .6778
= 67.78%.
a.
For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B,
z.025 = 1.96. The confidence interval is:
σ  28 ± 1.96 12  28 ± .784  (27.216, 28.784)
x ± z.025
75
n
b.
x ± z.025
σ  102 ± 1.96
n
c.
x ± z.025
x ± z.025
e.
100
σ  4.05 ± 1.96 .83
n
 102 ± .65  (101.35, 102.65)
200
σ  15 ± 1.96 .3
n
d.
22
 15 ± .0588  (14.9412, 15.0588)
100
 4.05 ± .163  (3.887, 4.213)
No. Since the sample size in each part was large (n ranged from 75 to 200), the Central Limit
Theorem indicates that the sampling distribution of is approximately normal.
267
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
268
5.4
Chapter 5
a.
x ± z.025
b.
s
 25.9 ± 1.645
2.7
90
n
s
 25.9 ± 2.58
n
a.
90
n
 25.9 ± .56  (25.34, 26.46)
 25.9 ± .47  (25.43, 26.37)
x ± z.005
5.5
2.7
 25.9 ± 1.96
x ± z.05
c.
s
2.7
90
 25.9 ± .73  (25.17, 26.63)
For confidence coefficient .95, = .05 and /2 = .05/2 = .025. From Table IV, Appendix B,
x  zα / 2
s
n
 26.2  1.96
4.1
70
 26.2  .96  (25.24, 27.16)
b.
The confidence coefficient of .95 means that in repeated sampling, 95% of all confidence intervals
constructed will includeμ.
c.
x  zα / 2
s
n
 26.2  2.58
4.1
70
 26.2  1.26  (24.94, 27.46)
d.
As the confidence coefficient increases, the width of the confidence interval also increases.
e.
Yes. Since the sample size is 70, the Central Limit Theorem applies. This ensures the distribution of
is normal, regardless of the original distribution.
5.6
If we were to repeatedly draw samples from the population and form the interval x ± 1.96 σ x each time,
approximately 95% of the intervals would contain μ. We have no way of knowing whether our interval
estimate is one of the 95% that contain μ or one of the 5% that do not.
5.7
A point estimator is a single value used to estimate the parameter, μ. An interval estimator is two values,
an upper and lower bound, which define an interval with which we attempt to enclose the parameter, . An
interval estimate also has a measure of confidence associated with it.
Inferences Based on a Single Sample: Estimation with Confidence Intervals
5.8
a.
For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 =
1.96. The confidence interval is:
x ± z.025
s
 33.9 ± 1.96
3.3
 33.9 ± 1.96
3.3
100
n
b.
x ± z.025
s
400
n
c.
269
 33.9 ± .647  (33.253, 34.547)
 33.9 ± .323  (33.577, 34.223)
For part a, the width of the interval is 2(.647) = 1.294. For part b, the width of the interval is 2(.323)
= .646. When the sample size is quadrupled, the width of the confidence interval is halved.
5.9
Yes. As long as the sample size is sufficiently large, the Central Limit Theorem says the distribution of is
approximately normal regardless of the original distribution.
5.10
a.
A point estimate for the average number of latex gloves used per week by all healthcare workers with
latex allergy is x = 19.3 .
b.
x  zα / 2
n
 19.3  1.96
11.9
46
 19.3  3.44  (15.86, 22.74)
c.
We are 95% confident that the true average number of latex gloves used per week by all healthcare
workers with a latex allergy is between 15.86 and 22.74.
d.
The conditions required for the interval to be valid are:
a.
b.
5.11
s
The sample selected was randomly selected from the target population.
The sample size is sufficiently large, i.e. n > 30.
For confidence coefficient .90, α = 1 - .90 = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B,
z.05 = 1.645. The 90% confidence interval is:
x  zα / 2
s
2, 484
 6,563  97.65  (6, 465.35, 6,660.65)
1, 751
We are 90% confident that the true mean expenses per full-time equivalent employees of all U.S. Army
hospitals is between $6,465.35 and $6,660.65.
5.12
n
 6,563  1.645
a.
The point estimate for the mean charitable commitment of tax-exempt organizations is
x = 79.67.
b.
From the printout, the 95% confidence interval is (75.84, 83.49).
c.
The probability of estimating the true mean charitable commitment with a single number is 0. By
estimating the true mean charitable commitment with an interval, we can be pretty confident that the
true mean is in the interval.
270
5.13
Chapter 5
a.
s
x  zα / 2
5.14
5.15
n
 4.25  2.58
12.02
56
 4.25  4.14  (0.11, 8.39)
b.
We are 99% confident that the true mean number of blogs/forums per site of all Fortune 500 firms
that provide blogs and forums for marketing tools is between 0.11 and 8.39.
c.
No. Since our sample size is 56, the sampling distribution of x is approximately normal by the
Central Limit Theorem.
a.
From the printout, the 95% confidence interval is (1.6711, 2.1989).
b.
We are 95% confident that the true mean failure time of used colored display panels is between
1.6711 and 2.1989.
c.
If 95% confidence intervals are formed, then approximately .95 of the intervals will contain the true
mean failure time.
a.
The target parameter is the population mean 2008 salary of these 500 CEOs who participated in the
Forbes’ survey.
b.
Using MINITAB, a sample of 50 CEOs was selected. The ranks of the 50 selected are: 9, 10, 14, 18,
19, 22, 25, 32, 38, 39, 45, 49, 50, 55, 60, 66, 69, 77, 96, 104, 106,
115, 147, 152, 192, 197, 209, 213, 229, 241, 245, 261, 268, 278, 283, 292, 305, 309,
325, 337, 342, 358, 364, 370, 376, 384, 405, 417, 433, 470.
c.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: PAY2005 ($mil)
Variable
PAY ($mil)
N
50
Mean
19.56
StDev
19.95
Minimum
1.23
Q1
4.70
Median
8.99
Q3
32.78
Maximum
73.17
The sample mean is x = 19.56 and the sample standard deviation is s = 19.95.
d.
Using MINITAB, the descriptive statistics for the entire data set is:
Descriptive Statistics: Pay ($mil)
Variable
Pay ($mil)
N
497
Mean
12.874
StDev
18.502
Minimum
0.000000000
Q1
3.395
Median
6.470
Q3
14.250
Maximum
192.920
From the above, the standard deviation of the population is $18.5 million.
e.
x  zα / 2
s
n
 19.56  2.58
19.95
50
 19.56  7.28  (12.28, 26.84)
5.16
271
f.
We are 99% confident that the true mean salary of all 500 CEOs in the Forbes’ survey is between
$12.28 million and $26.84 million.
g.
From part d, the true mean salary of all 500 CEOs is $12.87 million. This value does fall within the
99% confidence interval that we found in part e.
a.
Descriptive Statistics: Rate
Variable
Rate
N
30
Mean
79.73
Median
80.00
TrMean
80.15
Variable
Rate
Minimum
60.00
Maximum
90.00
Q1
76.75
Q3
84.00
StDev
5.96
SE Mean
1.09
x  zα / 2
s
n
 79.73  1.645
5.96
30
 79.73  1.79
 (77.94, 81.52)
5.17
b.
We are 90% confident that the mean participation rate for all companies that have 401(k) plans is
between 77.94% and 81.52%.
c.
We must assume that the sample size (n = 30) is sufficiently large so that the Central Limit Theorem
applies.
d.
Yes. Since 71% is not included in the 90% confidence interval, it can be concluded that this
company's participation rate is lower than the population mean.
e.
The center of the confidence interval is . If 60% is changed to 80%, the value of will increase, thus
indicating that the center point will be larger. The value of s2 will decrease if 60% is replaced by
80%, thus causing the width of the interval to decrease.
a.
An estimate of the true mean Mach rating score of all purchasing managers is x =99.6.
b.
x  zα / 2
s
n
 99.6  1.96
12.6
122
 99.6  2.24  (97.36, 101.84)
c.
We are 95% confident that the true Mach rating score of all purchasing managers is between 97.36
and 101.84.
d.
Yes, there is evidence to dispute this claim. We are 95% confident that the true mean Mach rating
score is between 97.36 and 101.84. It would be very unlikely that the true means Mach scores is as
low as 85.
272
5.18
Chapter 5
a.
Using MINITAB, I generated 30 random numbers using the uniform distribution from 1 to 308. The
random numbers were:
9, 15, 19, 36, 46, 47, 63, 73, 90, 92, 108, 112, 117, 127, 144, 145, 150, 151, 172, 178, 218, 229, 230,
241, 242, 246, 252, 267, 274, 282
I numbered the 308 observations in the order that they appear in the file. Using the random numbers
generated above, I selected the 9th, 15th, 19th, etc. observations for the sample. The selected sample is:
.31, .34, .34, .50, .52, .53, .64, .72, .70, .70, .75, .78, 1.00, 1.00, 1.03, 1.04, 1.07, 1.10, .21, .24, .58,
1.01, .50, .57, .58, .61, .70, .81, .85, 1.00
b.
Using MINITAB, the descriptive statistics for the sample of 30 observations are:
Descriptive Statistics: carats-samp
Variable
carats-s
N
30
Mean
0.6910
Median
0.7000
TrMean
0.6965
Variable
carats-s
Minimum
0.2100
Maximum
1.1000
Q1
0.5150
Q3
1.0000
StDev
0.2620
SE Mean
0.0478
From above, x =.6910 and s = .2620.
c.
x  zα / 2
d.
e.
5.19
s
n
 .691  1.96
.262
30
 .691  .094  (.597, .785)
We are 95% confident that the mean number of carats is between .597 and .785.
From Exercise 2.49, we computed the “population” mean to be .631. This mean does fall
95% confidence interval we computed in part d.
in the
To answer the question, we will first form 90% confidence intervals for each of the 2 SAT scores.
For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B, z.05 = 1.645.
The confidence interval is:
x  zα / 2
s
n
 19  1.645
65
265
 19  6.57  (12.43, 25.57)
We are 90% confident that the mean change in SAT-Mathematics score is between 12.43 and 25.57 points.
x  zα / 2
s
n
 7  1.645
49
265
 7  4.95  (2.05, 11.95)
We are 90% confident that the mean change in SAT-Verbal score is between 2.05 and 11.95 points.
The SAT-Mathematics test would be the most likely of the two to have 15 as the mean change in score.
This value of 15 is in the 90% confidence interval for the mean change in SAT-Mathematics score.
However, 15 does not fall in the 90% confidence interval for the mean SAT-Verbal test.
5.20
x
273
11, 298
= 2.26
5, 000
For confidence coefficient, .95, α = .05 and α/2 = .025. From Table IV, Appendix B,
x ± zα/2
s
n
 2.26 ± 1.96
1.5
5000
 2.26 ± .04  (2.22, 2.30)
We are 95% confident the mean number of roaches produced per roach per week is between 2.22 and 2.30.
5.21
5.22
5.23
a.
For confidence coefficient .80, α = 1  .80 = .20 and α/2 = .20/2 = .10. From Table IV, Appendix B,
z.10 = 1.28. From Table V, with df = n  1 = 5  1 = 4, t.10 = 1.533.
b.
c.
For confidence coefficient .95, α = 1  .95 = .05 and α/2 = .05/2 = .025. From Table IV, Appendix
B, z.025 = 1.96. From Table V, with df = n  1 = 5  1 = 4, t.025 = 2.776.
d.
e.
For confidence coefficient .99, α = 1  .99 = .02 and α/2 = .02/2 = .005. From Table IV, Appendix
B, z.005 = 2.575. From Table V, with df = n  1 = 5  1 = 4, t.005 = 4.604.
f.
Both the t- and z-distributions are symmetric around 0
and mound-shaped. The t-distribution is more spread
out than the z-distribution.
a.
If x is normally distributed, the sampling distribution
of x is normal, regardless of the sample size.
b.
If nothing is known about the distribution of x, the sampling distribution of x is approximately
normal if n is sufficiently large. If n is not large, the distribution of x is unknown if the distribution
of x is not known.
a.
P(−t0 < t < t0) = .95 where df = 10
Because of symmetry, the statement can be written
P(0 < t < t0) = .475 where df = 10
 P(t ≥ t0) = .025
t0 = 2.228
P(t ≤ (−t0 or t ≥ t0) = .05 where df = 10
 2P(t ≥ t0) = .05
 P(t ≥ t0) = .025 where df = 10
t0 = 2.228
b.
c.
P(t ≤ t0) = .05 where df = 10
Because of symmetry, the statement can be written
 P(t ≥ −t0) = .05 where df = 10
t0 = −1.812
274
5.24
Chapter 5
d.
P(t < −t0 or t > t0) = .10 where df = 20
 2P(t > t0) = .10
 P(t > t0) = .05 where df = 20
t0 = 1.725
e.
P(t ≤ −t0 or t ≥ t0) = .01 where df = 5
 2P(t ≥ t0) = .01
 P(t ≥ t0) = .005 where df = 5
t0 = 4.032
a.
P(t ≥ t0) = .025 where df = 11
t0 = 2.201
b.
P(t ≥ t0) = .01 where df = 9
t0 = 2.821
c.
P(t ≤ t0) = .005 where df = 6
Because of symmetry, the statement can be rewritten
P(t ≥ −t0) = .005 where df = 6
t0 = −3.707
d.
5.25
P(t ≤ t0) = .05 where df = 18
t0 = −1.734
First, we must compute x and s.
x
s2 

x
2
 x  30  5
n
 x 

n
n 1
6
2

(30) 2
6  26  5.2
6 1
5
176 
s  5.2  2.2804
a.
For confidence coefficient .90, α = 1  .90 = .10 and α/2 = .10/2 = .05. From Table V, Appendix B,
with df = n  1 = 6  1 = 5, t.05 = 2.015. The 90% confidence interval is:
s
x  t0.5
b.
n
 5 ± 2.015
2.2804
6
 5 ± 1.88  (3.12, 6.88)
For confidence coefficient .95, α = 1  .95 = .05 and α/2 = .05/2 = .025. From Table V, Appendix
B, with df = n − 1 = 6 − 1 = 5, t.025 = 2.571. The 95% confidence interval is:
x  t.025
s
n
 5 ± 2.571
2.2804
6
 5 ± 2.39  (2.61, 7.39)
c.
For confidence coefficient .99, α = 1  .99 = .01 and α/2 = .01/2 = .005. From Table V, Appendix
B, with df = n  1 = 6  1 = 5, t.005 = 4.032. The 99% confidence interval is:
x  t.005
d.
275
s
 5 ± 4.032
n
2.2804
6
 5 ± 3.75  (1.25, 8.75)
For confidence coefficient .90, α = 1  .90 = .10 and α/2 = .10/2 = .05. From Table V,
Appendix B, with df = n  1 = 25  1 = 24, t.05 = 1.711. The 90% confidence interval is:
a)
s
x  t.05
2.2804
 5 ± .78  (4.22, 5.78)
n
25
b)
 5 ± 1.711
s
x  t.025
n
 5 ± 2.064
2.2804
25
 5 ± .94  (4.06, 5.94)
c)
s
x  t.005
n
 5 ± 2.797
2.2804
25
 5 ± 1.28  (3.72, 6.28)
Increasing the sample size decreases the width of the confidence interval.
5.26
For this sample,
x
s2 =
 x  1567
n

16
x2
= 97.9375
 x 

n 1
n
2

1567 2
16 = 159.9292
16  1
155,867 
s=
s 2 = 12.6463
a.
For confidence coefficient, .80, α = 1 − .80 = .20 and α/2 = .20/2 = .10. From Table V, Appendix B,
with df = n − 1 = 16 − 1 = 15, t.10 = 1.341. The 80% confidence interval for μ is:
12.6463
s
x ± t.10
 97.94 ± 1.341
 97.94 ± 4.240  (93.700, 102.180)
16
n
b.
For confidence coefficient, .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table V, Appendix
B, with df = n − 1 = 24 − 1 = 23, t.025 = 2.131. The 95% confidence interval for μ is:
x ± t.025
s
n
 97.94 ± 2.131
12.6463
16
 97.94 ± 6.737  (91.203, 104.677)
The 95% confidence interval for μ is wider than the 80% confidence interval for μ found in part a.
276
Chapter 5
c.
For part a:
We are 80% confident that the true population mean lies in the interval 93.700 to 102.180.
For part b:
We are 95% confident that the true population mean lies in the interval 91.203 to 104.677.
The 95% confidence interval is wider than the 80% confidence interval because the more confident
you want to be that μ lies in an interval, the wider the range of possible values.
5.27
For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table V, Appendix B, with
df = n – 1 = 25 – 1 = 24, t.05 = 1.711. The 90% confidence interval is:
x  t.05
s
n
 75.4  1.711
10.9
25
 75.4  3.73  (71.67, 79.13)
We are 90% confident that the mean breaking strength of the white wood is between 71.67 and 79.13.
5.28
We must assume that the distribution of the LOS's for all patients is normal.
a.
For confidence coefficient .90, α = 1 − .90 = .10 and α/2 = .10/2 = .05. From Table V, Appendix B,
with df = n − 1 = 20 − 1 = 19, t.05 = 1.729. The 90% confidence interval is:
x  t.05
5.29
5.30
s
n
 3.8 ± 1.729
1.2
20
 3.8 ± .464  (3.336, 4.264)
b.
We are 90% confident that the mean LOS is between 3.336 and 4.264 days.
c.
“90% confidence” means that if repeated samples of size n are selected from a population and 90%
confidence intervals are constructed, 90% of all intervals thus constructed will contain the population
mean.
a.
From the printout, the 95% confidence interval is (−320%, 5,922%). We are 95% confident that the
true mean 5-year revenue growth rate for the 12 companies is between −320% and 5,922%.
b.
The population being sampled from must be normally distributed.
c.
From the stem-and-leaf display in the printout, the data appear to be skewed to the right. The data do
not appear to have come from a normal distribution. Therefore, the 95% confidence interval in part a
may not be valid.
a.
The 95% confidence interval for the mean surface roughness of coated interior pipe is
(1.63580, 2.12620).
b.
No. Since 2.5 does not fall in the 95% confidence interval, it would be very unlikely that the average
surface roughness would be as high as 2.5 micrometers.
a.
Descriptive Statistics: Skid
Variable
Skid
N
20
N*
0
Mean
358.5
SE Mean
26.3
StDev
117.8
Minimum
141.0
Q1
276.0
Median
367.5
Q3
438.0
Maximum
574.0
x  t.05
s
n
 358.5  2.093
117.8
20
 358.5  55.13  (303.37, 413.63)
b.
We are 95% confident that the mean skidding distance is between 303.37 and 413.63 meters.
c.
In order for the inference to be valid, the skidding distances must be from a normal distribution. We
will use the four methods to check for normality. First, we will look at a histogram of the data.
Using MINITAB, the histogram of the data is:
Histogram of Skid
4
3
Fr equency
5.31
277
2
1
0
200
300
400
500
Skid
From the histogram, the data appear to be fairly mound-shaped. This indicates that the data may be
normal.
Next, we look at the intervals x  s, x  2s, x  3s . If the proportions of observations falling in each
interval are approximately .68, .95, and 1.00, then the data are approximately normal. Using
MINITAB, the summary statistics are:
x  s  358.5  117.8  (240.7, 476.3) 14 of the 20 values fall in this interval. The proportion is
.70. This is very close to the .68 we would expect if the data were normal.
x  2s  358.5  2(117.8)  358.5  235.6  (122.9, 594.1) 20 of the 20 values fall in this
interval. The proportion is 1.00. This is a larger than the .95 we would expect if the data were
normal.
x  3s  358.5  3(117.8)  358.5  353.4  (5.1, 711.9) 20 of the 20 values fall in this interval.
278
Chapter 5
The proportion is 1.00. This is exactly the 1.00 we would expect if the data were normal.
From this method, it appears that the data may be normal.
Next, we look at the ratio of the IQR to s. IQR = QU – QL = 438 – 276 = 162.
IQR
162

 1.37 This is fairly close to the 1.3 we would expect if the data were normal. This
s
117.8
method indicates the data may be normal.
Finally, using MINITAB, the normal probability plot is:
Probability Plot of Skid
N ormal - 95% C I
99
95
90
Mean
StDev
358.5
117.8
N
AD
P-Value
20
0.170
0.921
P er cent
80
70
60
50
40
30
20
10
5
1
0
100
200
300
400
Skid
500
600
700
800
Since the data form a fairly straight line, the data may be normal.
From above, all the methods indicate the data may be normal. It appears that the assumption that the
data come from a normal distribution is probably valid.
5.32
d.
No. A distance of 425 meters falls above the 95% confidence interval that was computed in part a.
It would be very unlikely to observe a mean skidding distance of at least 425 meters.
a.
Descriptive Statistics: MTBE
Variable
MTBE
N
12
N*
0
Mean
97.2
SE Mean
32.8
StDev
113.8
Minimum
8.00
Q1
12.0
Median
50.5
Q3
146.0
Maximum
367.0
A point estimate for the true mean MTBE level for all well sites located near the New Jersey gasoline
service station is x  97.2 .
b.
279
x  t.005
s
n
 97.2  3.106
113.8
12
 97.2  102.04  (4.84, 199.24)
We are 99% confident that the true mean MTBE level for all well sites located near the New Jersey
gasoline service station is between −4.84 and 199.24.
We must assume that the data were sampled from a normal distribution. We will use the four
methods to check for normality. First, we will look at a histogram of the data. Using MINITAB, the
histogram of the data is:
Histogram of MTBE
5
4
Fr equency
c.
3
2
1
0
0
50
100
150
200
M T BE
250
300
350
From the histogram, the data do not appear to be mound-shaped. This indicates that the data may not
be normal.
Next, we look at the intervals x  s, x  2s, x  3s . If the proportions of observations falling in each
interval are approximately .68, .95, and 1.00, then the data are approximately normal. Using
MINITAB, the summary statistics are:
x  s  97.2  113.8  (16.6, 211.0) 10 of the 12 values fall in this interval. The proportion is .83.
This is not very close to the .68 we would expect if the data were normal.
x  2s  97.2  2(113.8)  97.2  227.6  (130.4, 324.8) 11 of the 12 values fall in this interval.
The proportion is .92. This is a somewhat smaller than the .95 we would expect if the data were
normal.
x  3s  97.2  3(113.8)  97.2  341.4  (244.2, 438.6) 12 of the 12 values fall in this interval.
The proportion is 1.00. This is exactly the 1.00 we would expect if the data were normal.
From this method, it appears that the data may not be normal.
280
Chapter 5
Next, we look at the ratio of the IQR to s. IQR = QU – QL = 146.0 – 12.0 = 134.0.
IQR 134.0

 1.18 This is somewhat smaller than the 1.3 we would expect if the data were normal.
s
113.8
This method indicates the data may not be normal.
Finally, using MINITAB, the normal probability plot is:
Probability Plot of MTBE
N ormal - 95% C I
99
95
90
Mean
StDev
97.17
113.8
N
AD
P-Value
12
0.929
0.012
P er cent
80
70
60
50
40
30
20
10
5
1
-300
-200
-100
0
100
200
M T BE
300
400
500
Since the data do not form a fairly straight line, the data may not be normal.
From above, the all methods indicate the data may not be normal. It appears that the data probably
are not normal.
5.33
a.
Descriptive Statistics: Alqaeda
Variable
Alqaeda
N
21
Mean
1.857
StDev
1.195
Minimum
1.000
Q1
1.000
Median
1.000
Q3
2.000
The sample mean is x = 1.857 and the sample standard deviation is s = 1.195.
Maximum
5.000
b.
281
Using MINITAB, a histogram of the data is:
Histogram of Alqaeda
12
10
Frequency
8
6
4
2
0
1
2
3
Alqaeda
4
5
The data are skewed to the right.
c.
Since the sample size is small (n = 21), the Central Limit Theorem does not apply. In order to use a
small sample confidence interval, the population being sampled from must be normal. In this case, it
does not look like the population being sampled from is normal. Thus, the confidence interval may
not be valid.
For confidence coefficient .90, α = 1 - .90 = .10 and α/2 = .10/2 = .05. From Table V, Appendix B,
with df = n – 1 = 21 – 1 = 20, t.05 = 1.725. The 90% confidence interval is:
x  tα / 2
5.34
s
n
 1.857  1.725
1.195
21
 1.857  .450  (1.407, 2.307)
d.
We are 90% confident that the mean number of individual suicide bombings and/or attacks per
incident is between 1.407 and 2.307.
e.
If the distribution being sampled from is normal, then .90 of all intervals formed will contain the true
mean.
Descriptive Statistics: Comp
Variable
Comp
N
10
Mean
1368
StDev
463
Minimum
720
Q1
1016
Median
1352
Q3
1652
Maximum
2112
For confidence coefficient .90, α = 1 - .90 = .10 and α/2 = .10/2 = .05. From Table V, Appendix B, with df
= n – 1 = 10 – 1 = 9, t.05 = 1.833. The 90% confidence interval is:
x  tα / 2
s
n
 1368  1.833
463
10
 1368  268.38  (1, 099.62, 1, 636.38)
We are 90% confident that the true mean threshold compensation level for all major airlines is between
$1,099.62 and $1,636.38.
282
5.35
Chapter 5
a.
The population from which the sample was drawn is the Forbes 441 Biggest Private companies.
b.
Descriptive Statistics: Revenue
Variable
Revenue
N
15
Mean
4.61
StDev
4.51
Minimum
1.10
Q1
1.75
Median
2.20
Q3
5.05
Maximum
14.32
x  t.025
s
n
 4.61  2.624
4.51
15
 4.61  3.06  (1.55, 7.67)
c.
We are 98% confident that the mean revenue is between $1.554 and $7.666 billion .
d.
The population must be normally distributed in order for the procedure used in part b to be valid.
e.
Yes. The value of $5.0 billion dollars falls in the 98% confidence interval computed in part b.
Therefore, we should believe the claim.
5.36
By the Central Limit Theorem, the sampling distribution of is approximately normal with mean
5.37
pq
.
n
The sample size is large enough if both npˆ  15 and nqˆ  15 .
μ pˆ = p and standard deviation σ pˆ 
a.
When n = 400, pˆ = .10:
npˆ = 400(.10) = 40 and nqˆ = 400(.90) = 360
Since both numbers are greater than or equal to 15, the sample size is sufficiently large to conclude
the normal approximation is reasonable.
b.
When n = 50, pˆ = .10:
npˆ = 50(.10) = 5 and nqˆ = 50(.90) = 45
Since npˆ is less than 15, the sample size is not large enough to conclude the normal approximation is
reasonable.
c.
When n = 20, pˆ = .5:
npˆ = 20(.5) = 10 and nqˆ = 20(.5) = 10
Since both numbers are less than 15, the sample size is not large enough to conclude the normal
approximation is reasonable.
d.
When n = 20, pˆ = .3:
npˆ = 20(.3) = 6 and nqˆ = 20(.7) = 14
Since both numbers are less than 15, the sample size is not large enough to conclude the normal
approximation is reasonable.
5.38
a.
npˆ = 121(.88) = 106.48 and nqˆ = 121(.12) = 14.52
Since nqˆ is less than 15, the sample size is not large enough to conclude the normal approximation is
reasonable. However, 14.52 is very close to 15, so the normal approximation may work fairly well.
b.
For confidence coefficient .90, α = .10 and α/2 = .05. From Table IV, Appendix B, z.05 = 1.645. The
90% confidence interval is:
pˆ  z .05
5.39
283
pq
 pˆ ± 1.645
n
ˆˆ
pq
.88(.12)
 .88 ± 1.645 1.645
 .88 ± .049
n
121
 (.831, .929)
c.
We must assume that the sample is a random sample from the population of interest.
a.
npˆ = 225(.46) = 103.5 and nqˆ = 225(.54) = 121.5
b.
For confidence coefficient .95, α = .05 and α/2 = .025. From Table IV, Appendix B, z.025 = 1.96.
The 95% confidence interval is:
pˆ ± z.025
5.40
pq
 pˆ ± 1.96
n
ˆˆ
pq
.46(1  .46)
 .46 ± 1.96
 .46 ± .065  (.395, .525)
n
225
c.
We are 95% confident the true value of p will fall between .395 and .525.
d.
"95% confidence interval" means that if repeated samples of size 225 were selected from the
population and 95% confidence intervals formed, 95% of all confidence intervals will contain the
true value of p.
a.
Of the 50 observations, 15 like the product  pˆ 
15
= .30.
50
npˆ = 50(.3) = 15 and nqˆ = 50(.7) = 35
For the confidence coefficient .80, α = .20 and α/2 = .10. From Table IV, Appendix B, z.10 = 1.28.
pˆ ± z.10
5.41
.3(.7)
ˆˆ
pq
 .3 ± 1.28
 .3 ± .083  (.217, .383)
50
n
b.
We are 80% confident the proportion of all consumers who like the new snack food is between .217
and .383.
a.
The population of interest is all American adults.
b.
The sample is the 1,000 adults surveyed.
c.
The parameter of interest is the proportion of all American adults who think Starbucks coffee is
overpriced.
284
Chapter 5
d.
npˆ = 1,000(.73) = 730 and nqˆ = 1,000(.27) = 270
For confidence coefficient .95, α = 1 - .95 = .05 and α/2 = .05/2 = .025. From Table IV, Appendix
B, z.025 = 1.96. The 95% confidence interval is:
pˆ  zα / 2
pq
 pˆ  zα / 2
n
ˆˆ
.73(.27)
pq
 .73  1.96
 .73  .028  (.702, .758)
n
1000
We are 95% confident that the true proportion of all American adults who say Starbucks coffee is
overpriced is between .702 and .758.
5.42
a.
An estimate of the true proportion of satellite radio subscribers who have a satellite radio receiver
x 396
in their car is pˆ  
 .79 .
n 501
b.
npˆ = 501(.79) = 395.79 and nqˆ = 501(.21) = 105.21
pˆ  zα / 2
5.43
pq
 pˆ  zα / 2
n
ˆˆ
.79(.21)
pq
 .79  1.645
 .79  .036  (.754, .826)
n
501
c.
We are 90% confident that the true proportion of satellite radio subscribers who have a satellite
receiver in their car is between .754 and .826.
a.
The point estimate of p is pˆ 
b.
x 414

 .46 .
n 900
npˆ = 900(.46) = 414 and nqˆ = 900(.54) = 486
pˆ  z.05
ˆˆ
.46(.54)
pq
 .46  1.645
 .46  .027  (.433, .487)
n
900
5.44
c.
We are 90% confident that the true proportion of contractors in the U.S. who have a company
website or will have one by the end of the year is between .433 and .487.
d.
The meaning of “90% confident” is that in repeated sampling, 90% of all confidence intervals
constructed will contain the true proportion and 10% will not.
a.
The point estimate of p is pˆ  .11 .
b.
285
npˆ = 150(.11) = 16.5 and nqˆ = 150(.89) = 133.5
pˆ  z.025
5.45
ˆˆ
.11(.89)
pq
 .11  1.645
 .11  .05  (.06, .16)
n
150
c.
We are 95% confident that the true proportion of MSDS that are satisfactorily completed is between
.06 and .16.
a.
b.
x
35

 .03 .
n 1,165
npˆ = 1,165(.03) = 34.95 and nqˆ = 1,165(.97) = 1130.05
pˆ  z.025
5.46
ˆˆ
pq
.03(.97)
 .03  1.96
 .03  .01  (.02, .04)
n
1,165
c.
We are 95% confident that the true proportion of drivers that use their cell phone while driving is
between .02 and .04.
a.
Since all the people surveyed were from Muncie, Indiana, the population of interest is all
consumers in Muncie, Indiana.
b.
The characteristic of interest in the population is the proportion of shoppers who believe that “Made
in the USA” means that 100% of labor and materials are from the USA.
286
Chapter 5
c.
x 64

 .604 .
n 106
npˆ = 106(.604) = 64.024 and nqˆ = 106 (.396) = 41.976
pˆ  z.05
5.47
ˆˆ
pq
.604(.396)
 .604  1.645
 .604  .078  (.526, .682)
n
106
d.
We are 90% confident that the true proportion of shoppers who believe that “Made in the USA” means
that 100% of labor and materials are from the USA is between .526 and .682.
e.
90% confidence means that if we took repeated samples of size 106 and computed 90% confidence
intervals for the true proportion shoppers who believe that “Made in the USA” means that 100% of
labor and materials are from the USA, 90% of the intervals computed will contain the true
proportion.
a.
A point estimate for the true percentage of all American adults who have access to a high-speed
internet connection is 42%. Changing this to a proportion, the point estimate is pˆ  .42 .
b.
npˆ = 4,000(.42) = 1,680 and nqˆ = 4,000(.58) = 2,320
Since both numbers are greater than or equal to 15, the sample size is sufficiently large to conclude the
normal approximation is reasonable.
pˆ  zα / 2
pq
 pˆ  zα / 2
n
ˆˆ
pq
.42(.58)
 .42  1.96
 .42  .015  (.405, .435)
n
4, 000
We are 95% confident that the true proportion of all American adults who have access to a high-speed
internet connection at home is between .405 and .435.
c.
Yes, since the interval constructed in part b contains values greater than .30, we could conclude that
the percentage has increased since 2005.
5.48
a.
The population is all senior human resource executives at U.S. companies.
b.
The population parameter of interest is p, the proportion of all senior human resource executives at
U.S. companies who believe that their hiring managers are interviewing too many people to find
qualified candidates for the job.
287
x 211

 .42 .
n 502
npˆ = 502(.42) = 241.84 and nqˆ = 502(.58) = 291.16
d.
pˆ  z.01
ˆˆ
pq
.42(.58)
 .42  2.33
 .42  .051  (.369, .471)
n
502
We are 98% confident that the true proportion of all senior human resource executives at U.S.
companies who believe that their hiring managers are interviewing too many people to find qualified
candidates for the job is between .369 and .471.
e.
5.49
A 90% confidence interval would be narrower. If the interval was narrower, it would contain fewer
values, thus, we would be less confident.
Of the 2,778 sampled firms, 748 announced one or more acquisitions during the year 2000. Thus,
pˆ 
x
748

 .269
n 2, 778
npˆ = 2,778(.269) = 747 and nqˆ = 2,778(.731) = 2031
For confidence coefficient .90, α = 1 - .90 = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B, z.05 =
1.645. The 90% confidence interval is:
pˆ  zα / 2
pq
 pˆ  zα / 2
n
ˆˆ
pq
.269(.731)
 .269  1.645
 .269  .014  (.255, .283)
n
2, 778
We are 90% confident that the true proportion of all firms that announced one or more acquisitions during
the year 2000 is between .255 and .283. Changing these to percentages, the results would be 25.5% and
28.3%.
288
5.50
Chapter 5
a.
x 16

 .052 .
n 308
npˆ = 308(.052) = 16 and nqˆ = 502(.948) = 476
pˆ  z.005
b.
ˆˆ
.052(.948)
pq
 .052  2.58
 .052  .033  (.019, .085)
308
n
We are 99% confident that the true proportion of diamonds for sale that are classified as “D” color is
x
81
 .263 .
The point estimate of p is pˆ  
n 308
npˆ = 308(.263) = 81 and nqˆ = 308(.737) = 227
pˆ  z.005
ˆˆ
.263(.737)
pq
 .263  2.58
 .263  .065  (.198, .328)
308
n
We are 99% confident that the true proportion of diamonds for sale that are classified as “VS1”
clarity, is between .198 and .328.
5.51
x 52

 .867 .
n 60
a.
b.
pˆ  z.025
c.
ˆˆ
pq
.867(.133)
 .867  1.96
 .867  .085  (.781, .953)
n
60
We are 95% confident that the true proportion of Wal-Mart stores in California that have more than 2
inaccurately priced items per 100 scanned is between .781 and .953.
d.
e.
289
If 99% of the California Wal-Mart stores are in compliance, then only 1% or .01 would not be.
However, we found the 95% confidence interval for the proportion that are not in compliance is
between .781 and .953. The value of .01 is not in this interval. Thus, it is not a likely value. This
claim is not believable.
npˆ = 60(.867) = 52 and nqˆ = 60(.133) = 8
Since nqˆ is less than 15, the sample size is not large enough to conclude the normal approximation is
reasonable. Thus, the confidence interval constructed in part b may not be valid. Any inference based
on this interval is questionable.
5.52
Using Wilson’s adjustment, the point estimate of the true proportion of all working nannies who are
men is
x2
24  2
26
p 


 .0062
n  4 4,176  4 4,180
z.025 = 1.96. Wilson’s adjusted 95% confidence interval is:
p  zα / 2

pq
.0062(.9938)
 .0062  1.96
 .0062  .0024  (.0038, .0086)
n
4,176  4
We are 95% confident that the true proportion of all working nannies who are men is between .0038 and
.0086.
5.53
a.
The parameter of interest is p, the proportion of all fillets that are red snapper.
b.
The estimate of p is pˆ 
x 22  17

 .23
n
22
npˆ = 22(.23) = 5 and nqˆ = 22(.77) = 17
Since npˆ is less than 15, the sample size is not large enough to conclude the normal approximation is
reasonable.
c.
We will use Wilson’s adjustment to form the confidence interval.
Using Wilson’s adjustment, the point estimate of the true proportion of all fillets that are not red
snapper is
p 
x2 5 2
7


 .27
n  4 22  4 26
For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table IV, Appendix
B, z.025 = 1.96. Wilson’s adjusted 95% confidence interval is:
p  zα / 2

.27(.73)
pq
 .27  1.96
 .27  .170  (.10, .44)
22  4
n
290
Chapter 5
d.
5.54
We are 95% confident that the true proportion of all fillets that are red snapper is between .10 and
.44.
x 36

 .434 .
n 83
npˆ = 83(.434) = 36 and nqˆ = 83(.566) = 47
pˆ  z.025
ˆˆ
pq
.434(.566)
 .434  1.96
 .434  .107  (.327, .541)
n
83
We are 95% confident that the true proportion of healthcare workers with latex allergies actually suspects
the he or she actually has the allergy is between .327 and .541.
5.55
We will use a 99% confidence interval to estimate the true proportion of mailed items that are delivered on
time.
First, we must compute pˆ : pˆ =
x
282, 200
=
= .85
n
332, 000
npˆ = 332,000(.85) = 282,200 and nqˆ = 332,000(.15) = 49,800
pˆ ± z.005
pq
≈ pˆ ± 2.58
n
ˆˆ
.85(.15)
pq
 .85 ± 2.58
 .85 ± .002  (.848, .852)
332,000
n
We are 99% confident that the true percentage of items delivered on time by the U.S. Postal Service is
between 84.8% and 85.2%.
5.56
To compute the necessary sample size, use
n=
 zα / 2 2 σ 2
SE 2
where α = 1 − .95 = .05 and α/2 = .05/2 = .025.
From Table IV, Appendix B, z.025 = 1.96. Thus,
n=
(1.96) 2 (7.2)
= 307.328 ≈ 308
.32
You would need to take 308 samples.
5.57
a.
An estimate of σ is obtained from:
range ≈ 4s
range 34  30

=1
s≈
4
4
n=
 zα / 2 2σ 2
( SE ) 2
where α = 1 − .90 = .10 and α/2 = .05.
n=
b.
(1.645) 2(1) 2
= 67.65 ≈ 68
.2 2
A less conservative estimate of σ is obtained from:
range ≈ 6s
range 34  30

= .6667
s≈
6
6
Thus, n =
5.58
a.
 zα / 2 2σ 2
( SE )
2
=
(1.645) 2(.6667) 2
= 30.07 ≈ 31
.2 2
To compute the needed sample size, use:
n=
 z α / 22 pq
SE 2
Thus, n =
where z.025 = 1.96 from Table IV, Appendix B.
(1.96) 2 (.2)(.8)
= 96.04 ≈ 97
.08 2
You would need to take a sample of size 97.
b.
To compute the needed sample size, use:
n=
 z α / 22 pq
SE 2
=
(1.96) 2(.5)(.5)
= 150.0625 ≈ 151
.08 2
You would need to take a sample of size 151.
291
292
5.59
Chapter 5
We know pˆ is in the middle of the interval, so pˆ 
The confidence interval is pˆ  z.05
We know .4  1.645
 .4 −
.8059
.4(.6)
= .26
n
.8059

n
 n = 5.756² = 33.1 ≈ 34
5.60
a.
ˆˆ
pq
.4(.6)
 .4  1.645
n
n
= .26
n
 .4 − .26 =
.54  .26
= .4
2
n
.8059
= 5.756
.14
For a width of 5 units, SE = 5/2 = 2.5.
To compute the needed sample size, use
n=
 z α / 22σ 2
SE 2
where α = 1 − .95 = .05 and α/2 = .025.
n=
(1.96)2 (14)2
2.52
= 120.47 ≈ 121
You would need to take 121 samples at a cost of 121($10) = $1210.
Yes, you do have sufficient funds.
b.
For confidence coefficient .90, α = 1 − .90 = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B,
z.05 = 1.645.
n=
(1.645) 2 (14)2
2.52
= 84.86 ≈ 85
You would need to take 85 samples at a cost of 85($10) = $850.
You still have sufficient funds but have an increased risk of error.
5.61
a.
The width of a confidence interval is 2(SE) = 2zα/2
293
σ
n
B, z.025 = 1.96.
For n = 16,
W = 2zα/2
σ = 2(1.96) 1
16
n
= 0.98
For n = 25,
W = 2zα/2
σ = 2(1.96) 1
25
n
= 0.784
For n = 49,
W = 2zα/2
σ = 2(1.96) 1
49
n
= 0.56
For n = 100,
W = 2zα/2
σ = 2(1.96)
n
1
100
= 0.392
For n = 400,
W = 2zα/2
σ = 2(1.96)
n
1
400
= 0.196
b.
5.62
The sample size will be larger than necessary for any p other than .5.
5.63
Thus,
n
( zα / 2 )2 σ 2
( SE ) 2

1.6452 (10.9) 2
42
 20.09  21
Thus, we would need a sample of size 21.
294
5.64
Chapter 5
To compute the necessary sample size, use:
n
 zα / 2 2 σ 2
SE 2
where α = 1 - .95 = .05 and α/2 = .05/2 = .025.
n
5.65
1.52
 245.86  246
z.05 = 1.645. Since we have no estimate given for the value of p, we will use .5. The sample size is:
n=
5.66
1.962 122
zα2 / 2 pq
( SE ) 2

1.6452 (.5)(.5)
.022
= 1,691.3 ≈ 1,692
a.
The confidence level desired by the researchers is 90%.
b.
The sampling error desired by the researchers is SE = .05.
c.
x 64
z.05 = 1.645. From the problem, we will use pˆ  
 .604 to estimate p.
n 106
Thus,
n
( zα / 2 )2 pq
( SE )2

1.6452.604(.396)
.052
 258.9  259
Thus, we would need a sample of size 259.
5.67
From the previous estimate, we will use pˆ = 1/3 to estimate p.
n
5.68
zα2 / 2 pq
( SE )2

2.5752 (1 / 3)(2 / 3)
.012
 14, 734.7  14, 735
To compute the necessary sample size, use:
n
 zα / 2 2 pq
SE 2
where α = 1 - .90 = .10 and α/2 = .10/2 = .05.
From Table IV, Appendix B, z.05 = 1.645. We will use pˆ = .867 from Exercise 5.51 to estimate p.
Thus, n 
1.962 .867(.133)
.052
 177.2  178
5.69
295
n=
 zα / 2 2σ 2
( SE ) 2
where α = 1 − .95 = .05 and α/2 = .05/2 = .025.
From Table IV, Appendix B, z.025 = 1.96.
Thus, for s = 10, n =
For s = 20, n =
For s = 30, n =
5.70
(1.96) 2(10) 2
32
(1.96) 2(20) 2
32
(1.96) 2(30) 2
32
= 42.68 ≈ 43
= 170.74 ≈ 171
= 384.16 ≈ 385
n=
( zα / 2 ) 2 σ 2
SE 2
where α = 1 − .90 = .10 and α/2 = .05.
n=
5.71
(1.645) 2 (10)2
12
= 270.6 ≈ 271
The bound is SE = .05. For confidence coefficient .99, α = 1 − .99 = .01 and α/2 = .01/2 = .005. From
Table IV, Appendix B, z.005 = 2.575.
We estimate p with = 11/27 = .407. Thus,
n=
 zα / 2 2 pq
( SE )2

2.5752 (.407)(.593)
.052
≈ 640.1  641
The necessary sample size would be 641. The sample was not large enough.
5.72
a.
n=
( zα / 2 ) 2 σ 2
SE 2
where α = 1 − .90 = .10 and α/2 = .05.
n=
b.
(1.645)2 (2) 2
.12
= 1,082.41 ≈ 1,083
As the sample size decreases, the width of the confidence interval increases. Therefore, if we sample
100 parts instead of 1,083, the confidence interval would be wider.
296
Chapter 5
c.
To compute the maximum confidence level that could be attained meeting the management's
specifications,
n=
( zα / 2 ) 2 σ 2
SE
2
 100 =
( zα / 2 )(2)2
2
.1
 ( zα / 2 )2 
100(.01)
= .25  zα/2 = .5
4
Using Table IV, Appendix B, P(0 ≤ z ≤ .5) = .1915. Thus, α/2 = .5000 − .1915 = .3085, α = 2(.3085)
= .617, and 1 − α = 1 − .617 = .383.
The maximum confidence level would be 38.3%.
5.73
a.
Percentage sampled =
n
1000
(100%) 
(100%) = 40%
N
2500
Finite population correction factor:
N n
2500  1000

 .6 = .7746
N
2500
b.
n
1000
(100%) 
(100%) = 20%
N
5000
N n
5000  1000

 .8 = .8944
N
5000
c.
n
1000
(100%) 
(100%) = 10%
N
10, 000
N n
10, 000  1000

 .9 = .9487
N
10, 000
d.
n
1000
(100%) 
(100%) = 1%
N
100, 000
N n
100, 000  1000

 .99 = .995
N
100, 000
5.74
σx =
σ
n
a.
σx
b.
σx 
N n
N
200
1000
200
1000
2500  1000
= 4.90
2500
5000  1000
= 5.66
5000
5.75
5.76
200
10, 000  1000
= 6.00
10, 000
c.
σx =
d.
σx 
a.
σˆ x 
b.
σˆ x 
c.
σˆ x 
d.
As n increases, σ x decreases.
e.
We are computing the standard error of x . If the entire population is sampled, then
x = μ. There is no sampling error, so σ x = 0.
a.
For n = 36, with the finite population correction factor:
 N  n  24  5000  64 
σˆ x  s / n 

 3 .9872 = 2.9807

N 
5000 
64 

1000
200
1000
100, 000  1000
= 6.293
100, 000
N n

N
s
n
50
4000
50
10,000
50
2000
10, 000  2000
= 1.00
10, 000
10, 000  4000
= .6124
10, 000
10, 000  10, 000
=0
10, 000
without the finite population correction factor:
24
σˆ x  s / n 
=3
64
σˆ x without the finite population correction factor is slightly larger.
b.
For n = 400, with the finite population correction factor:
 N n
24  5000  400 
σˆ x  s / n 

 1.2 .92 = 1.1510


5000 
N 
400 

without the finite population correction factor:
24
σˆ x  s / n 
= 1.2
400
c.
In part a, n is smaller relative to N than in part b. Therefore, the finite population correction factor
did not make as much difference in the answer in part a as in part b.
297
298
5.77
Chapter 5
The approximate 95% confidence interval for p is
pˆ ± 2σˆ pˆ  pˆ ± 2
 .42 ± 2
5.78
5.79
pˆ (1  pˆ )
n
N  n
N
.42(.58) 6000  1600
 .42 ± .011  (.409, .431)
1600
6000
An approximate 95% confidence interval for μ is:
s
N n
14
x ± 2σˆ x  x ± 2
 422 ± 2
N
40
n
a.
x=
2
s =
375  40
 422 ± 4.184  (417.816, 426.184)
375
 x  1081 = 36.03
30
n
x
2
( x)
−
2
n
= 41, 747 −
n −1
μˆ = x = 36.03
x ± 2σˆ x
s
 x ±2
n
 36.03 ± 2
2
1081
= 96.3782
30
N n
N
96.3782
30
300  30
 36.03 ± 3.40
300
 (32.63, 39.43)
b.
pˆ =
x 21
= .7

n 30
pˆ (1  pˆ )
n
x ± 2σˆ pˆ  pˆ ± 2
.7(.3) 300  30
 .7 ± .159  (.541, .859)
30
300
 .7 ± 2
5.80
a.
N n
N
For N = 2,193, n = 223, x =116,754, and s = 39,185, the 95% confidence interval is:
N n
39,185 2,193  223
 116, 754  2
N
2,193
n
223
 116, 754  4,974.06  (111, 779.94, 121, 728.06)
x  2σˆ x  x  2
b.
s
We are 95% confident that the mean salary of all vice presidents who subscribe to
Quality Progress is between $111,777.94 and $121,728.06.
5.81
a.
First, we must estimate p: pˆ 
299
x
759

 .560
n 1,355
For confidence coefficient .95, α = 1 - .95 = .05 and α/2 = .05/2 = .025. From Table IV,
Appendix B, z.025 = 1.96. Since n/N = 1,355/1,696 = .799 > .05, we must use the finite population
correction factor. The 95% confidence interval is:
pˆ  zα / 2
ˆˆ  N n
pq
.560(.440)  1, 696  1,355 

  .560  1.96

n
N
1,355 
1, 696
 .560  .012  (.548, .572)
5.82
b.
We used the finite correction factor because the sample size was very large compared to the
population size.
c.
We are 95% confident that the true proportion of active NFL players who select a professional coach
as the most influential in their career is between .548 and .572.
a.
The population of interest is the set of all households headed by women that have incomes of
$25,000 or more in the database.
b.
Yes. Since n/N = 1,333/25,000 = .053 exceeds .05, we need to apply the finite population correction.
c.
The standard error for pˆ should be:
pˆ (1  pˆ )  N  n 
.708(1  .708)  25, 000  1,333 


 = .012
 N 
n
1333
25, 000
σˆ pˆ 
d.
For confidence coefficient .90, α = 1 − .90 = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B,
z.05 = 1.645. The approximate 90% confidence interval is:
pˆ ± 1.645σˆ pˆ  .708 ± 1.645(.012)  (.688, .728)
5.83
a.
First, we must calculate the sample mean:
15
fx
i i
x
i 1
n

3(108)  2(55)  1(500)    19(100) 15,646

 156.46
100
100
The point estimate of the mean value of the parts inventory is x = 156.46.
300
Chapter 5
b.
The sample variance and standard deviation are:
s2 
 f x 

2
15

f i xi2
i 1
i i
n
3(108)2  2(55) 2    19(100)2 

n 1
15, 6462
6, 776,336 
100  43, 720.83677

99
100  1
15, 6462
100
s  s 2  43,720.83677  209.10
The estimated standard error is:
σˆ x =
c.
s
n
N n
209.10 500  100
=
= 18.7025
N
500
100
The approximate 95% confidence interval is:
 s  N n
x ± 2σˆ x  x ± 2 
 156.46 ± 2(18.7025)  156.46 ± 37.405
N
 n 
 (119.055, 193.865)
We are 95% confident that the mean value of the parts inventory is between $119.06 and $193.87.
d.
5.84
Since the interval in part c does not include $300, the value of $300 is not a reasonable value for the
mean value of the parts inventory.
For N = 1,500, n = 35, x = 1, and s = 124, the 95% confidence interval is:
 s  N n
 124  1,500  35
x ± 2σˆ x  x ± 2 
 1 ± 2
 1 ± 41.43

N
1,500
 n
 35 
 (−40.43, 42.43)
We are 95% confident that the mean error of the new system is between -$40.43 and $42.43.
5.85
pˆ 
x 15

 .086
n 175
The standard error of pˆ is:
σˆ pˆ =
pˆ (1  pˆ )  N  n 

 =
n
N 
.086(1  .086)  3000  175 

 = .0206
175
3000 
An approximate 95% confidence interval for p is:
pˆ ± 2 σˆ pˆ  .086 ± 2(.0206)  .086 ± .041  (.045, .127)
Since .07 falls in the 95% confidence interval, it is not an uncommon value. Thus, there is no evidence that
more than 7% of the corn-related products in this state have to be removed from shelves and warehouses.
5.86
5.87
5.88
301
a.
For a small sample from a normal distribution with unknown standard deviation, we use the t
statistic. For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table V,
Appendix B, with df = n − 1 = 23 − 1 = 22, t.025 = 2.074.
b.
For a large sample from a distribution with an unknown standard deviation, we can estimate the
population standard deviation with s and use the z statistic. For confidence coefficient .95, α = 1 −
.95 = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96.
c.
For a small sample from a normal distribution with known standard deviation, we use the z statistic.
B, z.025 = 1.96.
d.
For a large sample from a distribution about which nothing is known, we can estimate the population
standard deviation with s and use the z statistic. For confidence coefficient .95, α = 1 − .95 = .05 and
α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96.
e.
For a small sample from a distribution about which nothing is known, we can use neither z nor t.
a.
P(t ≤ t0) = .05 where df = 20
t0 = −1.725
b.
P(t ≥ t0) = .005 where df = 9
t0 = 3.250
c.
P(t ≤ −t0 or t ≥ t0) = .10 where df = 8 is equivalent to
P(t ≥ t0) = .10/2 = .05 where df = 8
t0 = 1.860
d.
P(t ≤ −t0 or t ≥ t0) = .01 where df = 17 is equivalent to
P(t ≥ t0) = .01/2 = .005 where df = 17
t0 = 2.898
a.
Of the 400 observations, 227 had the characteristic  pˆ = 227/400 = .5675.
npˆ = 400(.5675) = 227 and nqˆ = 400(.4325) = 173
pˆ ± z.025
pq
 ± 1.96
n
ˆˆ
.5675(.4325)
pq
 .5675 ± 1.96
 .5675 ± .0486
400
n
 (.5189, .6161)
302
Chapter 5
b.
For this problem, SE = .02. For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From
Table IV, Appendix B, z.025 = 1.96. Thus,
n=
( zα / 2 ) 2 pq
SE 2

(1.96)2 (.5675)(.4325)
.022
= 2,357.2 ≈ 2,358
Thus, the sample size was 2,358.
5.89
a.
x ± zα/2
5.90
s
n
30
225
 zα / 2  σ 2

 32.5 ± 5.15  (27.35, 37.65)
2.5752 (30) 2
= 27,870.25 ≈ 23,871.
.5 2
b.
The sample size is n =
c.
"99% confidence" means that if repeated samples of size 225 were selected from the population and
99% confidence intervals constructed for the population mean, then 99% of all the intervals
constructed will contain the population mean.
a.
The finite population correction factor is:
( N  n)
=
N
b.
c.
( SE ) 2
(2, 000  50)
= .9874
2, 000
( N  n)
=
N
(100  20)
= .8944
100
( N  n)
=
N
5.91
 32.5 ± 2.575
(1,500  300)
= .8944
1,500
The parameters of interest for the problems are:
(1)
(2)
(3)
(4)
The question requires a categorical response. One parameter of interest might be the proportion, p,
of all Americans over 18 years of age who think their health is generally very good or excellent.
A parameter of interest might be the mean number of days, μ, in the previous 30 days that all
Americans over 18 years of age felt that their physical health was not good because of injury or
illness.
Americans over 18 years of age felt that their mental health was not good because of stress,
depression, or problems with emotions.
Americans over 18 years of age felt that their physical or mental health prevented them from
performing their usual activities.
5.92
a.
For confidence coefficient .99, α = .01 and α/2 = .01/2 = .005. From Table IV,
Appendix B, z.005 = 2.58. The 99% confidence interval is:
x  zα / 2
5.93
303
 17.77 
 141.31  2.58 
  141.31  .145  (141.165, 141.455)
n
 100, 000 
s
b.
We are 99% confident that the mean number of semester hours taken by all first-time candidates for
the CPA exam is between 141.165 and 145.455.
c.
Since the sample size was so large, no conditions must hold for the interpretation in part b to be
valid.
a.
Of the 1000 observations, 29% said they would never give personal information to a company
 pˆ = .29
npˆ = 1000(.29) = 290 and nqˆ = 1000(.71) = 710
b.
B, z.025 = 1.96. The 95% confidence interval is:
pˆ ± z.025
ˆˆ
pq
 .29 ± 1.96
n
.29(.71)
 .29 ± .028  (.262, .318)
1000
We are 95% confident that the proportion of Internet users who would never give personal
information to a company is between .262 and .318.
5.94
x 67

 .638 .
n 105
a.
b.
npˆ = 105(.638) = 67 and nqˆ = 105(.362) = 38
pˆ  z.025
c.
ˆˆ
pq
.638(.362)
 .638  1.96
 .638  .092  (.546, .730)
n
105
We are 95% confident that the true proportion of on-the-job homicide cases that occurred at night is
304
5.95
Chapter 5
For this study,
n=
( zα / 2 ) 2 σ 2
SE 2

1.962 (5) 2
12
= 96.04 ≈ 97
The sample size needed is 97.
5.96
a.
From the printout, the 90% confidence interval for the mean lead level is (0.61, 5.16).
b.
From the printout, the 90% confidence interval for the mean copper level is (0.2637, 0.5529).
c.
We are 95% confident that the mean lead level in water specimens from Crystal Lakes Manors is
between .61 and 5.16.
We are 95% confident that the mean copper level in water specimens from Crystal Lakes Manors is
d.
5.97
99% confidence means that if repeated samples of size n are selected and 99% confidence intervals
formed, 99% of all confidence intervals will contain the true mean.
First, we must estimate p: pˆ 
pˆ  2
x 50

 .694 . The 95% confidence interval is:
n 72
ˆˆ N n
pq
.694(.306)  251  72 

  .694  2

  .694  .092  (.602, .786)
n
N
72
251 
We are 95% confident that the proportion of all New Jersey Governor’s Council business members that
have employees with substance abuse problems is between .602 and .786.
5.98
a.
For confidence coefficient .90, α = .10 and α/2 = .05. From Table IV, Appendix B,
x ± z.05
σ  x ± 1.645 s  12.2 ± 1.645 10
n
n
100
 12.2 ± 1.645  (10.555, 13.845)
We are 90% confident that the mean number of days of sick leave taken by all its employees is
between 10.555 and 13.845.
b.
 zα / 2 2 σ 2
SE
2
=
(2.58) 2 (10) 2
22
= 166.4 ≈ 167
You would need to take n = 167 samples.
5.99
305
There are a total of 96 channel catfish in the sample. The point estimate of p is
x 96
pˆ  
 .667 .
n 144
npˆ = 144(.667) = 96 and nqˆ = 144(.333) = 48
pˆ  z.05
ˆˆ
pq
.667(.333)
 .667  1.645
 .667  .065  (.602, .732)
n
144
We are 90% confident that the true proportion of channel catfish in the population is between .602 and
.732.
5.100
a.
x  zα / 2
s
n
 1.13  2.58
2.21
72
 1.13  .67
 (.46, 1.80)
We are 99% confident that the mean number of pecks at the blue string is between .46 and 1.80.
5.101
b.
Yes. The mean number of pecks at the white string is 7.5. This value does not fall in the 99%
confident interval for the blue string found in part a. Thus, the chickens are more apt to peck at white
string.
a.
For confidence coefficient .99, α = .01 and α/2 = .01/2 = .005. From Table V, Appendix B, with df
= n − 1 = 3 − 1 = 2, t.005 = 9.925. The confidence interval is:
x ± t.005
s
n
 49.3 ± 9.925
1.5
3
 49.3 ± 8.60  (40.70, 57.90)
b.
We are 99% confident that the mean percentage of B(a)p removed from all soil specimens using the
poison is between 40.70% and 57.90%.
c.
We must assume that the distribution of the percentages of B(a)p removed from all soil specimens
using the poison is normal.
d.
Since the 99% confidence interval for the mean percent removed contains 50%, this would be a very
possible value.
306
5.102
Chapter 5
For a width of .06, SE = .06/2 = .03
( zα / 2 ) 2 pq
=
SE 2
(.1645)2 (.17)(.83)
.032
= 424.2 ≈ 425
5.103
a.
Descriptive Statistics: IQ25, IQ60
Variable
IQ25
IQ60
N
36
36
Mean
66.83
45.39
Median
66.50
45.00
TrMean
66.69
45.22
Variable
IQ25
IQ60
Minimum
41.00
22.00
Maximum
94.00
73.00
Q1
54.25
36.25
Q3
80.00
58.00
StDev
14.36
12.67
SE Mean
2.39
2.11
x  zα / 2
s
n
 66.83  2.58
14.36
36
 66.83  6.17
 (60.66, 73.00)
We are 99% confident that the mean raw IQ score for all 25 year olds is between 60.66 and 73.00.
b.
We must assume that the sample is random and that the observations are independent.
c.
x  zα / 2
s
n
 45.39  1.96
12.67
36
 45.39  4.14
 (41.25, 49.53)
We are 95% confident that the mean raw IQ score for all 60 year olds is between 41.25 and 49.53.
5.104
pˆ = x/n = 35/55 = .636.
a.
The point estimate of p is
b.
npˆ = 55(.636) = 35 and nqˆ = 55(.364) = 20
For confidence coefficient, .99, α = .01 and α/2 = .01/2 = .005. From Table IV, Appendix B, z.005 =
pˆ ± z.005
.636(.364)
ˆˆ
pq
 .636 ± 2.575
 .636 ± .167  (.469, .803)
55
n
5.105
307
c.
We are 99% confident that the true proportion of fatal accidents involving children is between .469
and .803.
d.
The sample proportion of children killed by air bags who were not wearing seat belts or were
improperly restrained is 24/35 = .686. This is rather large proportion. Whether a child is killed by an
airbag could be related to whether or not he/she was properly restrained. Thus, the number of
children killed by air bags could possibly be reduced if the child were properly restrained.
The bound is SE = .1. For confidence coefficient .99, α = 1 − .99 = .01 and α/2 = .01/2 = .005.
We estimate p with from Exercise 5.104 which is pˆ = .636. Thus,
n=
 zα / 2 2 pq
( SE ) 2

2.5752 (.636)(.364)
.12
1 = 153.5  154
The necessary sample size would be 154.
5.106
a.
Of the 100 cancer patients, 7 were fired or laid off  pˆ = 7/100 = .07.
npˆ = 100(.07) = 7 and nqˆ = 100(.93) = 93
Since npˆ is less than 15, the sample size is not sufficiently large to conclude the normal
approximation is reasonable. We will iuse Wilson’s adjustment to form the confidence interval.
p 
x2
72
9


 .087
n  4 100  4 104
p ± z.05

pq
 p ± 1.645
n

pq
.087(.913)
 .087 ± 1.645
 .087 ± .045  (.042, .132)
n
100  4
Converting these to percentages, we get (4.2%, 13.2%).
5.107
b.
We are 90% confident that the percentage of all cancer patients who are fired or laid off due to their
illness is between 4.2% and 13.2%.
c.
Since the rate of being fired or laid off for all Americans is 1.3% and this value falls outside the
confidence interval in part b, there is evidence to indicate that employees with cancer are fired or laid
off at a rate that is greater than that of all Americans.
a.
Of the 24 observations, 20 were 2 weeks of vacation  pˆ = 20/24 = .833.
For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV,
Appendix B, z.025 = 1.96. The confidence interval is:
pˆ  z.025
ˆˆ
pq
.833(.167)
 .833  1.96
 .833  .149  (.683, .982)
n
24
308
Chapter 5
b.
npˆ = 24(.833) = 20 and nqˆ = 24(.167) = 4
Since nqˆ is less than 15, the sample size is not sufficiently large to conclude the normal
approximation is reasonable. The validity of the confidence interval is in question.
c.
The bound is SE = .02. For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table
IV, Appendix B, z.025 = 1.96. Thus,
n=
 zα / 2 2 pq
( SE )2
=
1.962 (.833)(.167)
.022
= 1,336.02 ≈ 1,337.
Thus, we would need a sample size of 1,337.
5.108
Descriptive Statistics: r
Variable
r
N
34
Mean
0.4224
Median
0.4300
TrMean
0.4310
Variable
r
Minimum
-0.0800
Maximum
0.7400
Q1
0.2925
Q3
0.6000
StDev
0.1998
SE Mean
0.0343
x  zα / 2
s
n
 .4224  1.96
.1998
34
 .4224  .0672
 (.3552, .4895)
We are 95% confident that the mean value of r is between .3552 and .4895.
5.109
Sampling error has to do with chance. In a population, there is variation – not all observations are the
same. The sampling error has to do with the variation within a sample. By chance, one might get a sample
that overestimates the mean just because all the observations in the sample happen to be high.
Nonsampling error has to do with errors that have nothing to do with the sampling. These errors could be
due to misunderstanding the question being asked, asking a question that the respondent does not know
how to answer, etc.
5.110
a.
The point estimate for the fraction of the entire market who refuse to purchase bars is:
pˆ 
b.
309
x 23

 .094
n 244
npˆ = 244(.094) = 22.9 and nqˆ = 244(.906) = 221.1
c.
B, z.025 = 1.96. The confidence interval is:
ˆˆ
pq
(.094)(.906)
 .094  1.96
 .094  .037  (.057, .131)
n
244
pˆ  z.025
d.
5.111
The best estimate of the true fraction of the entire market who refuse to purchase bars six months
after the poisoning is .094. We are 95% confident the true fraction of the entire market who refuse to
purchase bars six months after the poisoning is between .057 and .131.
From Exercise 5.110, a good approximation for p is .094. Also, SE = .02.
 zα / 2 2 pq
( SE )
2
=
(1.96) 2(.094)(.906)
.022
= 817.9 ≈ 818
5.112
The bound is SE = .1. For confidence coefficient .99, α = 1 − .99 = .01 and α/2 = .01/2 = .005.
We estimate p with from Exercise 7.48 which is = .636. Thus,
n=
( zα / 2 ) 2 pq
SE 2

2.5752 (.636)(.364)
= 153.5  154
.12
The necessary sample size would be 154.
5.113
Solution will vary. See page 1037 for Guided Solutions.
310
5.114
Chapter 5
Since the manufacturer wants to be reasonably certain the process is really out of control before shutting
down the process, we would want to use a high level of confidence for our inference. We will form a 99%
confidence interval for the mean breaking strength.
x  t.005
s
n
 985.6  3.355
22.9
9
 985.6  25.61  (959.99, 1,011.21)
We are 99% confident that the true mean breaking strength is between 959.99 and 1,011.21. Since 1,000 is
contained in this interval, it is not an unusual value for the true mean breaking strength. Thus, we would
recommend that the process is not out of control.
5.115
a.
As long as the sample is random (and thus representative), a reliable estimate of the mean weight of
all the scallops can be obtained.
b.
The government is using only the sample mean to make a decision. Rather than using a point
estimate, they should probably use a confidence interval to estimate the true mean weight of the
scallops so they can include a measure of reliability.
c.
We will form a 95% confidence interval for the mean weight of the scallops. Using MINITAB, the
descriptive statistics are:
Descriptive Statistics: Weight
Variable
Weight
N
18
N*
0
Variable
Weight
Maximum
1.1400
Mean
0.9317
SE Mean
0.0178
StDev
0.0753
Minimum
0.8400
Q1
0.8800
Median
0.9100
Q3
9800
For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table V, Appendix A, with
df = n – 1 = 18 – 1 = 17, t.025 = 2.110.
The 95% confidence interval is:
x  t.025
s
n
 .932  2.110
.0753
18
 .932  .037  (.895, .969)
We are 95% confident that the true mean weight of the scallops is between .8943 and .9691. Recall
that the weights have been scaled so that a mean weight of 1 corresponds to 1/36 of a pound. Since
the above confidence interval does not include 1, we have sufficient evidence to indicate that the
minimum weight restriction was violated.

Chapter 5 Inferences Based on a Single Sample: Estimation with Confidence Intervals α

Transcription

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