Arrhythmia Care Co-ordinator Services - NHS Alliance 2012

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

5.2 a. z = 1.96, using Table IV, Appendix B, P(0 ≤ z ≤ 1.96) = .4750. Thus, α/2 =
α/2
.5000 − .4750 = .025, α = 2(.025) = .05, and 1 - α = 1 - .05 = .95. The confidence level is
100% × .95 = 95%.

b. z = 1.645, using Table IV, Appendix B, P(0 ≤ z ≤ 1.645) = .45. Thus, α/2 = .50 − .45 =
α/2
.05, α = 2(.05) = .1, and 1 − α = 1 − .1 = .90. The confidence level is 100% × .90 = 90%.

c. z = 2.575, using Table IV, Appendix B, P(0 ≤ z ≤ 2.575) = .495. Thus, α/2 = .500 −
α/2
.495 = .005, α = 2(.005) = .01, and 1 − α = 1 − .01 = .99. The confidence level is
100% × .99 = 99%.

d. z = 1.282, using Table IV, Appendix B, P(0 ≤ z ≤ 1.282) = .4. Thus, α/2 = .5 − .4 = .1,
α/2
α = 2(.1) = .2, and 1 − α = 1 − .2 = .80. The confidence level is 100% × .80 = 80%.

e. z = .99, using Table IV, Appendix B, P(0 ≤ z ≤ .99) = .3389. Thus, α/2 = .5000 − .3389
α/2
= .1611, α = 2(.1611) = .3222, and 1 − α = 1 − .3222 = .6778. The confidence level is
100% × .6778 = 67.78%.

5.4 a. For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix
B, z = 1.96. The confidence interval is:
.025

s 2.7
x ± z ⇒ 25.9 ± 1.96 ⇒ 25.9 ± .56 ⇒ (25.34, 26.46)
.025
n 90

b. For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table IV, Appendix
B, z = 1.645. The confidence interval is:
.05

s 2.7
x ± z ⇒ 25.9 ± 1.645 ⇒ 25.9 ± .47 ⇒ (25.43, 26.37)
.05
n 90

c. For confidence coefficient .99, α = .01 and α/2 = .01/2 = .005. From Table IV, Appendix
B, z = 2.58. The confidence interval is:
.005

s 2.7
x ± z ⇒ 25.9 ± 2.58 ⇒ 25.9 ± .73 ⇒ (25.17, 26.63)
.005
n 90

5.6 If we were to repeatedly draw samples from the population and form the interval x ± 1.96σ
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.

Inferences Based on a Single Sample: Estimation with Confidence Intervals 137
5.8 a. For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix
B, z = 1.96. The confidence interval is:
.025

s 3.3
x ± z ⇒ 33.9 ± 1.96 ⇒ 33.9 ± .647 ⇒ (33.253, 34.547)
.025
n 100

s 3.3
b. x ± z ⇒ 33.9 ± 1.96 ⇒ 33.9 ± .323 ⇒ (33.577, 34.223)
.025
n 400

c. 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.

5.10 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. For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix
B, z = 1.96. The confidence interval is:
.025

s 11.9
x ±z ⇒19.3±1.96 ⇒19.3±3.44⇒(15.86, 22.74)
α/2
n 46

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. The sample selected was randomly selected from the target population.
b. The sample size is sufficiently large, i.e. n > 30.

5.12 a. The point estimate for the mean charitable commitment of tax-exempt organizations is
x = 74.9667.

b. From the printout, the 95% confidence interval is (68.2371, 81.6962).

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.

138 Chapter 5
5.14 Using MINITAB, the descriptive statistics are:

Descriptive Statistics: r

Variable N Mean Median TrMean StDev SE Mean
r 34 0.4224 0.4300 0.4310 0.1998 0.0343

Variable Minimum Maximum Q1 Q3
r -0.0800 0.7400 0.2925 0.6000

For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B,
z = 1.96. The confidence interval is:
.025

s .1998
x ± z ⇒.4224±1.96 ⇒.4224±.0672
α/2 n 34
⇒(.3552, .4895)

We are 95% confident that the mean value of r is between .3552 and .4895.

5.16 a. Using MINITAB, the descriptive statistics are:

Descriptive Statistics: Rate

Variable N Mean Median TrMean StDev SE Mean
Rate 30 79.73 80.00 80.15 5.96 1.09

Variable Minimum Maximum Q1 Q3
Rate 60.00 90.00 76.75 84.00

For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B,
z = 1.645. The confidence interval is:
.05

s 5.96
x ±z ⇒79.73±1.645 ⇒79.73±1.79
α/2 n 30
⇒(77.94, 81.52)

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.

Inferences Based on a Single Sample: Estimation with Confidence Intervals 139
5.18 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 N Mean Median TrMean StDev SE Mean
carats-s 30 0.6910 0.7000 0.6965 0.2620 0.0478

Variable Minimum Maximum Q1 Q3
carats-s 0.2100 1.1000 0.5150 1.0000

From above, x =.6910 and s = .2620.

c. For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix
B, z = 1.96. The confidence interval is:
.025

s .262
x ±z ⇒.691±1.96 ⇒.691±.094⇒(.597, .785)
α/2
n 30

d. We are 95% confident that the mean number of carats is between .597 and .785.

e. From Exercise 2.47, we computed the “population” mean to be .631. This mean does fall
in the 95% confidence interval we computed in part d.

11,298
5.20 x = = 2.26
5,000

For confidence coefficient, .95, α = .05 and α/2 = .025. From Table IV, Appendix B,
z = 1.96. The confidence interval is:
.025

s 1.5
x ± z ⇒ 2.26 ± 1.96 ⇒ 2.26 ± .04 ⇒ (2.22, 2.30)
α/2
n 5000

We are 95% confident the mean number of roaches produced per roach per week is between
2.22 and 2.30.

140 Chapter 5
5.22 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.

5.24 a. P(t ≥ t ) = .025 where df = 11
0
t = 2.201
0

b. P(t ≥ t ) = .01 where df = 9
0
t = 2.821
0

c. P(t ≤ t ) = .005 where df = 6
0

Because of symmetry, the statement can be rewritten

P(t ≥ −t ) = .005 where df = 6
0
t = −3.707
0

d. P(t ≤ t ) = .05 where df = 18
0
t = −1.734
0

5.26 For this sample,
∑x 1567
x = = = 97.9375
n 16
(∑x)2 15672
∑x2 − 155,867−
s2 = n = 16 = 159.9292
n−1 16−1
s = s2 = 12.6463

a. For confidence coefficient, .80, α = 1 − .80 = .20 and α/2 = .20/2 = .10. From Table VI,
Appendix B, with df = n − 1 = 16 − 1 = 15, t = 1.341. The 80% confidence interval for
.10
μ is:
s 12.6463
x ± t ⇒ 97.94 ± 1.341 ⇒ 97.94 ± 4.240 ⇒ (93.700, 102.180)
.10
n 16

b. For confidence coefficient, .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table VI,
Appendix B, with df = n − 1 = 24 − 1 = 23, t = 2.131. The 95% confidence interval for
.025
μ is:

s 12.6463
x ± t ⇒ 97.94 ± 2.131 ⇒ 97.94 ± 6.737 ⇒ (91.203, 104.677)
.025
n 16

The 95% confidence interval for μ is wider than the 80% confidence interval for μ found
in part a.

Inferences Based on a Single Sample: Estimation with Confidence Intervals 141
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.

5.28 a. Using MINITAB, the descriptive statistics are:

Descriptive Statistics: MTBE

Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximum
MTBE 12 0 97.2 32.8 113.8 8.00 12.0 50.5 146.0 367.0

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. For confidence coefficient .99, α = .01 and α/2 = .01/2 = .005. From Table VI, Appendix
B, with df = n – 1 = 12 – 1 = 11, t = 3.106. The 99% confidence interval is:
.005

s 113.8
x ±t ⇒97.2±3.106 ⇒97.2±102.04⇒(−4.84, 199.24)
.005
n 12

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.

c. 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
y
nc 3
e
u
q
e
r
F 2
1
0
0 50 100 150 200 250 300 350
MTBE

142 Chapter 5
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 ±2s⇒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.

Next, we look at the ratio of the IQR to s. IQR = Q – Q = 146.0 – 12.0 = 134.0.
U L

IQR 134.0
= =1.18 This is somewhat smaller than the 1.3 we would expect if the data
s 113.8
were normal. This method indicates the data may not be normal.

Finally, using MINITAB, the normal probability plot is:

Probability Plot of MTBE
Normal - 95% CI
99
Mean 97.17
StDev 113.8
95 N 12
AD 0.929
90 P-Value 0.012
80
70
nt 60
e
c 50
r
e 40
P
30
20
10
5
1
-300 -200 -100 0 100 200 300 400 500
MTBE

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.

Inferences Based on a Single Sample: Estimation with Confidence Intervals 143
5.30 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 VI,
Appendix B, with df = n − 1 = 20 − 1 = 19, t = 1.729. The 90% confidence interval is:
.05

s 1.2
x ±t ⇒ 3.8 ± 1.729 ⇒ 3.8 ± .464 ⇒ (3.336, 4.264)
.05
n 20

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.

5.32 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.

5.34 a. The population is the set of all DOT permanent count stations in the state of Florida.

b. Yes. There are several types of routes included in the sample. There are 3 recreational
areas, 7 rural areas, 5 small cities, and 5 urban areas.

c. Using MINITAB, the descriptive statistics are:

Descriptive Statistics: 30th hour, 100th hour

Variable N Mean Median TrMean StDev SE Mean
30th hou 20 2206 2064 2165 1224 274
100th ho 20 2096 1999 2048 1203 269

Variable Minimum Maximum Q1 Q3
30th hou 252 4905 1429 3068
100th ho 229 4815 1318 2877

For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table VI,
Appendix B, with df = n – 1 = 20 – 1 = 19, t = 2.093. The 95% confidence interval is:
.025

s 1,224
x ±t ⇒2,206±2.093 ⇒2,206±572.84⇒(1,633.16, 2,778.84)
.025
n 20

We are 95% confident that the mean traffic count at the 30th highest hour is between
1,633.16 and 2,778.84.

d. We must assume that the distribution of the traffic counts at the 30th highest hour is
normal. From the stem-and-leaf display, the data look fairly mound-shaped. Thus, the
assumption of normality is probably met.

144 Chapter 5
e. For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table VI, Appendix
B, with df = n – 1 = 20 – 1 = 19, t.025 = 2.093. The 95% confidence interval is:

s 1,203
x ±t ⇒2,096±2.093 ⇒2,096±563.01⇒(1,532.99, 2,659.01)
.025
n 20

We are 95% confident that the mean traffic count at the 100th highest hour is between
1,532.99 and 2,659.01.

We must assume that the distribution of the traffic counts at the 100th highest hour is
normal. From the stem-and-leaf display, the data look fairly mound-shaped. Thus, the
assumption of normality is probably met.

f. If μ = 2,700, it is very possible that it is the mean count for the 30th highest hour. It falls
in the 95% confidence interval for the mean count for the 30th highest hour. It is not very
likely that the mean count for the 100th highest hour is 2,700. It does not fall in the 95%
confidence interval for the mean count for the 100th highest hour. (See parts c and e
above.)

5.36 By the Central Limit Theorem, the sampling distribution of is approximately normal with
pq
mean μ = p and standard deviation σ = .
pˆ pˆ n

5.38 a. The sample size is large enough if the interval pˆ ± 3σ does not include 0 or 1.
pˆ

pq pˆqˆ .88(1−.88)
pˆ ± 3σ ⇒ pˆ ± 3 ⇒ pˆ ± 3 ⇒ .88 ± ⇒ .88 ± .089
pˆ n n 121
⇒ (.791, .969)

Since the interval lies within the interval (0, 1), the normal approximation will be
adequate.

b. For confidence coefficient .90, α = .10 and α/2 = .05. From Table IV, Appendix B,
z = 1.645. The 90% confidence interval is:
.05

pq pˆqˆ .88(.12)
pˆ ± z ⇒ pˆ ± 1.645 ⇒ .88 ± 1.6451.645 ⇒ .88 ± .049
.05
n n 121
⇒ (.831, .929)

c. We must assume that the sample is a random sample from the population of interest.

Inferences Based on a Single Sample: Estimation with Confidence Intervals 145
15
5.40 a. Of the 50 observations, 15 like the product ⇒ pˆ = = .30.
30

To see if the sample size is sufficiently large:

pˆqˆ .3(.7)
pˆ ± 3σ ≈ pˆ ± 3 ⇒ .3 ± 3 ⇒ .3 ± .194 ⇒ (.106, .494)
pˆ n 50

Since this interval is wholly contained in the interval (0, 1), we may conclude that the
normal approximation is reasonable.

For the confidence coefficient .80, α = .20 and α/2 = .10. From Table IV, Appendix B,
z = 1.28. The confidence interval is:
.10

pˆqˆ .3(.7)
pˆ ± z ⇒ .3 ± 1.28 ⇒ .3 ± .083 ⇒ (.217, .383)
.10
n 50

b. We are 80% confident the proportion of all consumers who like the new snack food is
between .217 and .383.

5.42 a. The point estimate of p is pˆ =.11.

b. To see if the sample size is sufficiently large:

pˆqˆ .11(.89)
pˆ ±3σ ≈ pˆ ±3 ⇒.11±3 ⇒.11±.077⇒(.033, .187)
pˆ n 150
Since the interval is wholly contained in the interval (0, 1), we may conclude that the
normal approximation is reasonable.

For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix
B, z = 1.96. The confidence interval is:
.025

pˆqˆ .11(.89)
pˆ ±z ⇒.11±1.645 ⇒.11±.05⇒(.06, .16)
.025 n 150

c. We are 95% confident that the true proportion of MSDS that are satisfactorily completed
is between .06 and .16.

x 16
5.44 a. The point estimate of p is pˆ = = =.052.
n 308

To see if the sample size is sufficiently large:
pˆqˆ .052(.948)
pˆ ±3σ ≈ pˆ ±3 ⇒.052±3 ⇒.052±.038⇒(.014, .090)
pˆ n 308
Since the interval is wholly contained in the interval (0, 1), we may conclude that the
normal approximation is reasonable.

146 Chapter 5