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Chapter 9
Exercises 9.1 – 9.10
9.1
Statistic II would be preferred because it is unbiased and has smaller variance than the other two.
9.2
An unbiased statistic is generally preferred over a biased statistic, because there is no long run
tendency for the unbiased statistic to overestimate or underestimate the true population value. That
is, there will be no systematic estimation error.
Unbiasedness by itself does not guarantee that the estimate will be close to the true value. An
unbiased statistic may have a sampling distribution that has a large variance. Thus, it would be
possible to obtain a value of the statistic that is quite inaccurate. One might choose a biased statistic
over an unbiased one if the bias of the first is small and if its sampling distribution has a small
variance.
1720
= 0.2769
6212
9.3
p=
9.4
a.
x=
∑ x = 27678 = 1845.2
b.
x=
∑ x = 26626 = 1775.1
c.
n
15
n
15
No fast food is consumed: s = 386.346, fast food is consumed: s = 620.660
9.5
The point estimate of π would be
9.6
245
p = 935 = 0.262
9.7
a.
9.8
9.9
p=
number in sample registered 14
=
= 0.70
n
20
b.
c.
19.57
= 1.957
10
s2 = 0.15945
s = 0.3993; No, this estimate is not unbiased. It underestimates the true value of σ.
a.
x=
b.
The number of cyclists in the sample whose gross efficiency is at most 20 is 4. The estimate
of all such cyclists whose gross efficiency is at most 20 is the sample proportion p = 4/19 =
0.2105.
a.
The value of σ will be estimated by using the statistic s. For this sample,
392.4
= 20.6526
19
∑ x 2 = 1757.54, ∑ x = 143.6, n = 12
( ∑ x )2
(143.6)2
1757.54 −
1757.54 − 1718.4133
n =
12
=
n− 1
12 − 1
11
∑ x2 −
2
s =
=
39.1267
= 3.557 and s = 3.557 = 1.886
11
203
9.10
b.
The population median will be estimated by the sample median. Since n = 12 is even, the
sample median equals the average of the middle two values (6th and 7th values), i.e.,
(11.3 + 11.4) = 11.35.
2
c.
In this instance, a trimmed mean will be used. First arrange the data in increasing order.
Then, trimming one observation from each end will yield an 8.3% trimmed mean. The
trimmed mean equals 117.3/10 = 11.73.
d.
The point estimate of μ would be x = 11.967. From part a, s = 1.886. Therefore the
estimate of the 90th percentile is 11.967 + 1.28(1.886) = 14.381.
a.
x J = 120.6
b.
An estimate of the total amount of gas used by all these houses in January would be
10000(120.6) = 1,206,000 therms. More generally, an estimate of the population total is
obtained by multiplying the sample mean by the size of the population.
c.
p = 8/10 = 0.8
d.
Using the sample median, an estimate of the population median usage is
(118 + 122)/2 = 120 therms.
Exercises 9.11 – 9.29
9.11
9.12
9.13
a.
As the confidence level increases, the width of the large sample confidence interval also
increases.
b.
As the sample size increases, the width of the large sample confidence interval decreases.
The values for parts a - d are found in the table for Standard Normal Probabilities (Appendix Table 2).
a.
1.96
b.
1.645
c.
2.58
d.
1.28
e.
1.44 (approximately)
For the interval to be appropriate, np ≥ 10, n(1 − p) ≥ 10 must be satisfied.
a.
np = 50(0.3) = 15, n(1 − p) = 50(0.7) = 35, yes
b.
np = 50(0.05) = 2.5, no
c.
np = 15(0.45) = 6.75, no
d.
np = 100(0.01) = 1 , no
e.
np = 100(0.70) = 70, n(1 − p) = 100(0.3) = 30, yes
f.
np = 40(0.25) = 10, n(1 − p) = 40(0.75) = 30, yes
g.
np = 60(0.25) = 15, n(1 − p) = 60(0.75) = 45, yes
h.
np = 80(0.10) = 8, no
204
9.14
9.15
9.16
a.
As the confidence level increases, the width of the confidence interval for π increases.
b.
As the sample size increases, the width of the confidence interval for π decreases.
c.
As the value of p gets farther from 0.5, either larger or smaller, the width of the confidence
interval for π decreases.
a.
Because np = 420.42 and n(1 − p ) = 580.58 , which are both greater than 10, and the
Americans in the sample were randomly selected from a large population, the large-sample
interval can be used. The sample proportion p is 0.42. The 99% confidence interval for π ,
the proportion of all Americans who made plans in May 2005 based on an incorrect weather
0.42(1 − 0.42)
⇒ 0.42 ± 0.0402 ⇒ (0.3798, 0.4602). We can
report would be 0.42 ± 2.58
1001
be 99% confident that the true proportion of adult Americans who made plans in May 2005
based on an incorrect weather report is between .38 and .46.
b.
No, weather reports may be more or less reliable during other months.
a.
Because np = 370 and n(1 − p ) = 630 , which are both greater than 10, and the students in
the sample were randomly selected from a large population, the large-sample interval can be
used. The sample proportion p is 0.37. The 90% confidence interval for π , the proportion of
all college freshman who carried a credit card balance from month to month would be
0.37(1 − 0.37)
0.37 ± 1.645
⇒ 0.37 ± 0.02511 ⇒ (0.3449, 0.3951).
1000
Because np = 480 and n(1 − p ) = 520 , which are both greater than 10, and the students in
the sample were randomly selected from a large population, the large-sample interval can be
used. The sample proportion p is 0.48. The 90% confidence interval for π , the proportion of
all college seniors who carried a credit card balance from month to month would be
0.48(1 − 0.48)
0.48 ± 1.645
⇒ 0.48 ± 0.02599 ⇒ (0.45401, 0.50599).
1000
The estimated standard deviation of the sampling distribution of p is larger when p is 0.48
than when p = 0.37 so the width of the confidence interval will be wider.
b.
c.
9.17
a.
Because np and n(1 − p ) are both greater than 10, and the potential jurors in the sample
were randomly selected from a large population, the large-sample interval can be used. The
sample proportion p is 350/500 = 0.7. The 95% confidence interval for π , the population
0.7(1 − 0.7)
⇒ 0.7 ± 0.0402 ⇒ (0.6598, 0.7402). With 95%
500
confidence we can estimate that between 66% and 74% of all potential jurors regularly watch
at least one crime-scene investigation series.
A 99% confidence interval would be wider than the 95% confidence interval in Part (a).
proportion would be 0.7 ± 1.96
b.
9.18
a.
Because np and n(1 − p ) are both greater than 10, and the adults in the sample were
randomly selected from a large population, the large-sample interval can be used. The
sample proportion p is 230/1000 = 0.23. The 95% confidence interval for π , the proportion
of all U.S. adults for whom math was the favorite subject would be
0.23(1 − 0.23)
0.23 ± 1.96
⇒ 0.23 ± 0.0261 ⇒ (0.2039, 0.2561).
1000
205
9.19
b.
Because np and n(1 − p ) are both greater than 10, and the adults in the sample were
randomly selected from a large population, the large-sample interval can be used. The
sample proportion p is 370/1000 = 0.37. The 95% confidence interval for π , the proportion
of all U.S. adults for whom math was the least favorite subject would be
0.37(1 − 0.37)
0.37 ± 1.96
⇒ 0.37 ± 0.0299 ⇒ (0.3401, 0.3999).
1000
a.
Because np and n(1 − p ) are both greater than 10, and the businesses in the sample were
randomly selected from a large population, the large-sample interval can be used. The
sample proportion p is 137/526 = 0.26. The 95% confidence interval would be
0.26(1 − 0.26)
⇒ 0.26 ± 0.0375 ⇒ (0.2225, 0.2975). With 95% confidence we
526
can estimate that between 22.25% and 29.75% of all U.S. businesses have fired workers for
misuse of the Internet.
It would be narrower because of a lower confidence level (90% instead of 95%) and because
of a smaller estimated standard error (0.0189 instead of 0.0191).
0.26 ± 1.96
b.
9.20
a.
b.
c.
9.21
Because np and n(1 − p ) are both greater than 10, and the adults in the sample were
randomly selected from a large population, the large-sample interval can be used. The
sample proportion p is 394/1000 = 0.394. The 95% confidence interval for π , the proportion
of all U.S. adults that consider themselves to be baseball fans would be:
0.394(1 − 0.394)
0.394 ± 1.96
⇒ 0.394 ± 0.0303 ⇒ (0.3637, 0.4243).
1000
Because np and n(1 − p ) are both greater than 10, and the adults in the sample were
randomly selected from a large population, the large-sample interval can be used. The
sample proportion p is 272/394 = 0.69. The 95% confidence interval for π , the proportion of
all U.S. adults who consider themselves to be baseball fans that think the designated hitter
should either be expanded to both leagues or eliminated would be:
0.69(1 − 0.69)
0.69 ± 1.96
⇒ 0.69 ± 0.0457 ⇒ (0.6443, 0.7357).
394
The intervals are not the same width (Part b is wider) because the sample size is different
(larger in Part a) and the standard error is different (larger in Part b).
Bound of error (based on 95% confidence) = 1.96
p(1 − p )
. If p = 0.82 and n = 1002,
n
0.82(1 − 0.82)
= 0.0238 . This implies a 95% confidence interval of 0.82 ± 0.0238 or (0.7962,
1002
0.8438). With 95% confidence, we can estimate that between 79.6% and 84.4% of all adults think
that reality shows are ‘totally made up’ or ‘mostly distorted’.
1.96
9.22
a.
Because np and n(1 − p ) are both greater than 10, and the adults in the sample were
randomly selected from a large population, the large-sample interval can be used. The
sample proportion p is 0.52. The 90% confidence interval for π , the proportion of all
American adults who think lying is never justified would be: 0.52 ± 1.645
0.52 ± .026 ⇒ (0.494, 0.546).
206
0.52(1 − 0.52)
⇒
1000
b.
c.
9.23
a.
b.
c.
9.24
a.
b.
c.
Because np and n(1 − p ) are both greater than 10, and the adults in the sample were
randomly selected from a large population, the large-sample interval can be used. The
sample proportion p is 650/1000 = 0.65. The 90% confidence interval for π , the proportion
of all American adults who think lying is OK if it avoids hurting someone’s feelings would be:
0.65(1 − 0.65)
0.65 ± 1.645
⇒ 0.65 ± 0.0248 ⇒ (0.6252, 0.6748).
1000
It is estimated that about half of all adult Americans think lying is never justified, yet more
than 60% of them think that it is OK to lie if it avoids hurting someone’s feelings. Obviously
there are some adults who consider “lying to avoid hurting someone’s feelings” as not really a
lie!
The sample proportion p is 38/115 = 0.330. The 95% confidence interval would be
0.330(1 − 0.330)
0.330 ± 1.96
⇒ 0.330 ± 0.0859 ⇒ (0.2441, 0.4159).
115
The sample proportion p is 22/115 = 0.1913. The 90% confidence interval would be
0.1913(1 − 0.1913)
0.1913 ± 1.645
⇒ 0.1913 ± 0.0603 ⇒ (0.1310, 0.2516).
115
The interval is wider in part a because (i) the confidence interval is higher in part (a) and (ii)
the sample proportion is more extreme; further from 0.5 in part (a) .
The sample proportion who can identify their own country is 0.9. The 90% confidence interval
0.9(1 − 0.9)
⇒ 0.9 ± 0.00901 ⇒ (0.891, 0.909).
would be 0.9 ± 1.645
3000
The sample should be a SRS of the respondents and independent of each other.
The results would only apply to the population of respondents aged 18 to 24 in the nine
different countries chosen for the study.
9.25
The sample proportion p is 18/52 = 0.3462. The 95% confidence interval would be
0.3462(1 − 0.3462)
0.3462 ± 1.96
⇒ 0.3462 ± 0.1293 ⇒ (.2169, .4755).
52
The assumption are: p is the sample proportion from a random sample, and that the sample size is
large, (np ≥ 10, and n(1 - p) ≥ 10).
9.26
With p = .36 and n = 1004, the bound on error is 1.96
(.36)(.64)
≈ .03
1004
9.27
With p = .25 and n = 1002, the bound on error is 1.96
(.25)(.75)
≈ .03
1002
9.28
An estimate of the proportion of school children in Australia who watch TV before school is p =
1060/1710 = 0.6199. A 95% confidence interval for the true proportion is
0.6199(1 − 0.6199)
0.6199 ± 1.96
⇒ 0.6199 ± 0.023 ⇒ (0.597, 0.643) .
1710
For this confidence interval to be valid, the sample must be a random sample from the population of
interest.
2
9.29
2
⎡1.96 ⎤
⎡ 1.96 ⎤
n = 0.25 ⎢
⎥ = 0.25 ⎢ 0.05 ⎥ = 384.16; take n = 385.
⎣ B ⎦
⎣
⎦
207
Exercises 9.30 – 9.50
9.30
a.
b.
c.
d.
e.
f.
g.
90%
95%
95%
99%
1%
0.5%
5%
9.31
a.
b.
c.
d.
e.
f.
2.12
1.80
2.81
1.71
1.78
2.26
9.32
a.
b.
114.4 + 115.6
= 115.0 .
2
As the confidence level increases the width of the interval increases.
Hence (114.4, 115.6) is the 90% interval and (114.1, 115.9) is the 99% interval.
x is the midpoint of the interval. So, x =
9.33
As the sample size increases, the width of the interval decreases. The interval (51.3, 52.7) has a
width of 52.7 − 51.3 = 1.4 and the interval (49.4, 50.6) has a width of 50.6 − 49.4 = 1.2. Hence, the
interval (49.4, 50.6) is based on the larger sample size.
9.34
a.
The 90% confidence interval would have been narrower, since its z critical value would have
been smaller.
b.
The statement is incorrect. The 95% refers to the percentage of all possible samples that
result in an interval that includes μ, not to the chance (probability) that a specific interval
contains μ.
c.
Again this statement is incorrect. While we would expect approximately 95 of the 100
intervals constructed to contain μ, we cannot be certain that exactly 95 out of 100 of them
will. The 95% refers to the percentage of all possible intervals that include μ.
9.35
Because the specimens were randomly selected and the distribution of the breaking force is
approximately normal, the t confidence interval formula for the mean can be used. The 95%
confidence interval for μ , the population mean breaking force is:
⎛ 41.97 ⎞
306.09 ± 2.57 ⎜
⎟ ⇒ 306.09 ± 44.035 ⇒ ( 262.06, 350.125 ) . With 95% confidence we can
6 ⎠
⎝
estimate the average breaking force for acrylic bone cement to be between 272.5 and 229.7
Newtons.
9.36
a.
b.
c.
The two groups ‘12 to 23 months’ and ‘24 to 35 months’ would both have the same variability
as each other but greater variability than the ‘less than 12 months group’ because the interval
width for both groups is 0.4 which is wider than the ‘less than 12 month’ group with an interval
width of 0.2.
The group ‘less than 12 months’ would have the largest sample size because the interval
width for that group is narrower that for the other groups.
It would be a 99% confidence level because if everything else remains constant, an increase
in the confidence level results in an increase in the interval width. The new confidence
interval is wider.
208
9.37
Because the adults were randomly selected and the sample size is large, the t confidence interval
formula for the mean can be used. The 90% confidence interval for μ , the population mean
24.2
⇒ 28.5 ± 1.780 ⇒ ( 26.72, 30.28 ) . With 90% confidence, we
500
estimate that the mean daily commute time for all working residents of Calgary, Canada is between
26.7 minutes and 30.3 minutes.
commute time is: 28.5 ± 1.645
9.38
9.39
a.
If the distribution of the anticipated Halloween expense is heavily skewed to the right, the
standard deviation could be greater than the mean.
b.
The distribution of anticipated Halloween expense is not approximately normal. As expense
has to be a nonnegative value, and the standard deviation is larger than the mean, the
distribution of Halloween expense is heavily skewed to the right.
c.
Even though the distribution of expense is not approximately normal the distribution for the
sampling distribution when n = 1000 will be approximately normal and we can use the t
confidence interval to estimate the mean anticipated Halloween expense for Canadian
residents.
d.
Because the adults were randomly selected and the sample size is large, the t confidence
interval formula for the mean can be used. The 99% confidence interval for μ , the
population mean anticipated expense is:
83.7
46.65 ± 2.58
⇒ 46.65 ± 6.829 ⇒ ( 39.82, 53.48 )
1000
With 99% confidence, we estimate the mean anticipated Halloween expense for all Canadian
residents to be between $39.82 and $53.48.
a.
The t critical value for a 95% confidence interval when df = 99 is 1.99. The confidence
interval based on this sample data is
⎛ 20 ⎞
s
x ± (t critical)
⇒ 183 ± (1.99) ⎜
⎟ ⇒ (179.03, 186.97) .
n
⎝ 100 ⎠
⎛ 23 ⎞
s
x ± (t critical)
⇒ 190 ± (1.99) ⎜
⎟ ⇒ (185.44, 194.56)
n
⎝ 100 ⎠
b.
c.
9.40
The new FAA recommendations are above the upper level of both confidence levels so it
appears that Frontier airlines have nothing to worry about.
range 20.3 − 19.9
=
= 0.1 . The required sample size
4
4
A reasonable estimate of σ is
2
⎡ (1.96 )( 0.1) ⎤
⎡ 1.96σ ⎤
n= ⎢
⎥ = 3.84 . A sample size of 4 or large would be required.
⎥ =⎢
B
0.1
⎣
⎦
⎣
⎦
2
9.41
a.
b.
The t critical value for a 90% confidence interval when df = 9 is 1.83. The confidence interval
based on this sample data is
⎛ 3.6757 ⎞
s
x ± (t critical)
⇒ 54.2 ± (1.83) ⎜
⎟ ⇒ 54.2 ± 2.1271 ⇒ (52.073, 56.327) .
n
⎝ 10 ⎠
If the same sampling method was used to obtain other samples of the same size and
confidence intervals were calculated from these samples, 90% of them would contain the true
population mean.
209
c.
9.42
As airlines are often rated by how often their flights are late, I would recommend the
published arrival time to be close to the upper bound of the confidence interval of the journey
time: 10: 57 a.m.
The standard error is
s
n
, and so is dependent on the sample size. Consider the following table:
Standard
Error
s
Hispanic
Native American
n
s
n
s
= 3011
654
s
= 29577
13
Standard
deviation
$77001.58
$106641.39
Now the variability doesn’t seem quite as different.
9.43
A boxplot (below) shows the distribution to be slightly skewed with no outliers. It seems plausible
that the population distribution is approximately normal. Calculation of a confidence interval for the
population mean cadence requires sample mean and sample standard deviation:
x = 0.926
s = 0.0809 t critical value with 19 df is 2.58.
⎛ 0.0809 ⎞
The interval is 0.926 ± (2.86) ⎜
⎟ = 0.926 ± 0.052 = (0.874,0.978) . With 99% confidence, we
⎝ 20 ⎠
estimate the mean cadence of all healthy men to be between 0.874 and 0.978 strides per second.
0.80
9.44
0.85
0.90
0.95
Cadence
1.00
1.05
The t critical value for a 90% confidence interval when df = 10 − 1 = 9 is 1.83. From the given data, n
219
= 10, ∑ x = 219, and ∑ x 2 = 4949.92 . From the summary statistics, x =
= 21.9
10
(219)2
10 = 4949.92 − 4796.1 = 153.82 = 17.09
9
9
9
4949.92 −
2
s =
s = 17.09 = 4.134 .
The 90% confidence interval based on this sample data is
x ± (t critical )
s
n
⇒ 21.9 ± (1.83)
4.134
10
⇒ 21.9 ± 2.39 ⇒ (19.51, 24.29 ) .
210
9.45
Summary statistics for the sample are: n = 5, x = 17, s = 9.03
The 95% confidence interval is given by
x ± (t critical)
9.46
a.
s
n
⇒ 17 ± (2.78)
9.03
5
⇒ 17 ± 11.23 ⇒ (5.77, 28.23) .
Summary statistics for the sample are: n = 900, x = 3.73, s = 0.45
The 95% confidence interval is given by
x ± (t critical)
b.
9.47
s
n
⇒ 3.73 ± (1.96)
0.45
900
⇒ 3.73 ± 0.0294 ⇒ (3.70, 3.76) .
No. With 95% confidence we can state that the mean GPA for the students at this university
is between 3.70 and 3.76. The individual GPAs of the students will vary quite a bit more.
Since approximately 95% of the population values will lie between μ − 2σ and μ + 2σ by the
empirical rule, we estimate that approximately 95% of the students will have their GPAs in the
interval x − 2s and x + 2s , i.e., in the interval 3.73 ± 2(0.45) ⇒ (2.83, 4.63) ; since the maximum
possible GPA is 4.0, we can say that roughly 95% of the Caucasian students will have their
GPAs in the interval from 2.83 to 4.0.
Since the sample size is small (n = 17), it would be reasonable to use the t confidence interval only if
the population distribution is normal (at least approximately). A histogram of the sample data (see
figure below) suggests that the normality assumption is not reasonable for these data. In particular,
the values 270 and 290 are much larger than the rest of the data and the distribution is skewed to the
right. Under the circumstances the use of the t confidence interval for this problem is not reasonable.
6
Frequency
5
4
3
2
1
0
120
140
160
180
200
220
240
260
280
Calories per half cup
9.48
An estimate σ is (700 − 50)/4 = 650/4 = 162.5. The required sample size is
⎡1.96 (162.5 ) ⎤
2
n=⎢
⎥ = ( 31.85 ) = 1014.42. So take n = 1015.
10
⎣
⎦
2
211
9.49
⎡ ( z critical ) σ ⎤
⎡ (1.96 )(1) ⎤
2
n=⎢
⎥ =⎢
⎥ = (19.6 ) = 384.16 . Hence, n should be 385.
B
0.1
⎣
⎦
⎣
⎦
9.50
⎡ (1.645 ) σ ⎤
For 90% confidence level: n= ⎢
⎥
B
⎣
⎦
2
2
⎡ ( 2.33 ) σ ⎤
For 98% confidence level: n = ⎢
⎥
B
⎣
⎦
2
2
Exercises 9.51 – 9.73
466
= 0.4596
1014
9.51
p=
9.52
⎡ 1.96 ⎤
The required sample size is n = (.27)(.73) ⎢
⎥ = 302.87 ; at least 303
⎣ 0.05 ⎦
2
2
⎡ 1.96 ⎤
The required sample size is n = (.5)(.5) ⎢
⎥ = 384.16 ; at least 385
⎣ 0.05 ⎦
I would recommend the larger sample size; if I was going to the trouble of surveying 303 adult
residents in my city, it wouldn’t take much more effort to find 82 more!
9.53
9.54
9.55
0.65(1 − 0.65)
⇒ 0.65 ± 0.064 ⇒ (0.589, 0.714)
150
Thus, we can be 90% confident that between 58.9% and 71.4% of Utah residents favor fluoridation.
This is consistent with the statement that a clear majority of Utah residents favor fluoridation.
A 90% confidence interval is 0.65 ± 1.645
0.4
⇒ 0.5 ± 0.089 ⇒ ( 0.411, 0.589 )
a.
0.5 ± 1.96
b.
The fact that 0 is not contained in the confidence interval does not imply that all students lie
to their mothers. There may be students in the population and even in this sample of 77 who
did not lie to their mothers. Even though the mean may not be zero, some of the individual
data values may be zero. However, if the mean is nonzero, it does imply that some students
tell lies to their mothers.
a.
b.
77
⎛ 14.41 ⎞
25.62 ± 2.33 ⎜
⎟ ⇒ 25.62 ± 5.062 ⇒ ( 20.558, 30.682 )
⎝ 44 ⎠
⎛ 15.31 ⎞
18.10 ± 2.33 ⎜
⎟ ⇒ 18.10 ± 2.225 ⇒ (15.875, 20.325 )
⎝ 257 ⎠
c.
It is based on a larger sample.
d.
Since the interval in a gives the plausible values for μ and the lower endpoint is greater than
20, this suggests that the mean number of hours worked per week for non-persistors is
greater than 20.
212
9.56
Let π = the true proportion of full-time workers that felt like hitting a co-worker during the past year.
125
For this sample p =
= 0.167 . Because np = 125 and n(1 − p ) = 625 are both greater than or
750
equal to 10, the sample size is large enough to use the formula for a large-sample confidence
interval. The 90% confidence interval is:
0.167 ± 1.645
0.167 (1 − 0.167 )
750
⇒ 0.167 ± 0.0224 ⇒ ( 0.1446, 0.1894 ) .
Based on these sample data, we can be 90% confident that the true proportion of full time workers
who have been angry enough in the last year to consider hitting a coworker is between .144 and .190.
9.57
Using a conservative value of π = .5 in the formula for required sample size gives:
2
2
⎛ 1.96 ⎞
⎛ 1.96 ⎞
n = π (1 − π ) ⎜
⎟ = .25 ⎜
⎟ = 384.16 . A sample size of at least 385 should be used.
⎝ B ⎠
⎝ .05 ⎠
0.77(0.23)
⇒ 0.77 ± 0.029 ⇒ (0.741, 0.799) .
800
9.58
The 95% confidence interval would be 0.77 ±1.96
9.59
⎡ 2.576 ⎤
⎡ 2.576 ⎤
n = 0.25 ⎢
⎥ = 0.25 ⎢ 0.10 ⎥ = 165.9; take n = 166.
B
⎣
⎦
⎣
⎦
2
9.60
2
Based on the information given, a 95% confidence interval for π , the proportion of the population
who felt that their financial situation had improved during the last year, is calculated as follows.
0.43(1 − 0.43)
0.43 ± 1.96
⇒ 0.43 ± 0.0318 ⇒ (0.398, 0.462) .
930
Hence, with 95% confidence, the percentage of people who felt their financial situation had improved
during the last year is between 39.8% and 46.2%. These end points of the confidence interval are
3.2% away on either side of the estimated value of 43%. In the report, the value 3.2% has been
rounded down to 3%. Thus, the claim of a 3% “margin of error” in the report is statistically justified.
An alternative way to justify the statement in the article is as follows. The mean of the sampling
π (1 − π )
. The largest possible standard
distribution of p is π and the standard deviation is equal to
930
deviation occurs when π is equal to 0.5, in which case the standard deviation is equal to 0.0164.
Hence the probability that p will be within 0.03 (within 3 percent) of π is greater than or equal to P( (0.03/0.0164) < z < (0.03/0.0164) ) = P(-1.8297 < z < 1.8297) = 0.933 which is approximately equal to
0.95. Hence the probability is 93.3% or greater that the value of p is within 3 percent of the true value
of π .
9.61
a.
The 95% confidence interval for μ ABC is
⎛ 5 ⎞
15.6 ± (1.96) ⎜
⎟ ⇒ 15.6 ± 1.39 ⇒ (14.21, 16.99) .
⎝ 50 ⎠
b.
For μ CBS : 11.9 ± 1.39 ⇒ (10.51, 13.29)
For μ FOX : 11.7 ± 1.39 ⇒ (10.31, 13.09)
For μ NBC : 11.0 ± 1.39 ⇒ (9.61, 12.39)
213
9.62
c.
Yes, because the plausible values for μ ABC are larger than the plausible values for the other
means. That is, μ ABC is plausibly at least 14.21, while the other means are plausibly no
greater than 13.29, 13.09, and 12.39.
a.
0.721 ± 2.58
b.
0.279 ± 1.96
0.721( 0.279 )
500
0.279 ( 0.721)
500
⇒ 0.721 ± 0.052 ⇒ ( 0.669, 0.773 )
⇒ 0.279 ± 0.039 ⇒ ( 0.24, 0.318 )
Based on this interval, we conclude that between 24% and 31.8% of college freshman are
not attending their first choice of college.
c.
9.63
It would have been narrower.
Since n = 18, the degrees of freedom is n − 1 = 18 − 1 = 17 .
From the t-table, the t critical value is 2.11.
⎛ 4.3 ⎞
The confidence interval is 9.7 ± 2.11⎜
⎟ ⇒ 9.7 ± 2.14 ⇒ ( 7.56, 11.84 ) .
⎝ 18 ⎠
Based on this interval we can conclude with 95% confidence that the true average rating of
acceptable load is between 7.56 kg and 11.84 kg.
9.64
p=
101
= 0.0996 . The 90% confidence interval is
1014
0.0996 ± 1.645
0.0996 ( 0.9004 )
1014
⇒ 0.0996 ± 0.0155 ⇒ ( 0.0841, 0.1151) .
Based on this interval, we conclude that between 8.41% and 11.51% of the population agree with the
statement “Landlords should have the right to evict a tenant from an apartment because that person
has AIDS.”
9.65
B = 0.1, σ = 0.8
2
⎡ (1.96 )( 0.8 ) ⎤
2
⎡ 1.96σ ⎤
=⎢
n=⎢
⎥ = (15.68 ) = 245.86
⎥
B
0.1
⎣
⎦
⎣
⎦
2
Since a partial observation cannot be taken, n should be rounded up to n = 246.
9.66
The 99% confidence interval computed from this sample data is
0.71 ± 2.58
0.71( 0.29 )
900
⇒ 0.71 ± 0.039 ⇒ ( 0.671, 0.749 ) .
Based on this interval, we conclude that between 67.1% and 74.9% of the population of Californians
support allowing 10-2 verdicts in criminal cases not involving the death penalty.
214
9.67
The 99% confidence interval for the mean commuting distance based on this sample is
⎛ 6.2 ⎞
s
x ± (t critical)
⇒ 10.9 ± (2.58) ⎜
⎟ ⇒ 10.9 ± 0.924 ⇒ ( 9.976, 11.824 ) .
n
⎝ 300 ⎠
9.68
p = 445/602 = 0.739. The 95% confidence interval for the true proportion of California registered
voters who prefer the cigarette tax increase is
p(1 − p)
0.739(0.261)
p± (z critical)
⇒ 0.739 ± 1.96
⇒ 0.739 ± 0.035 ⇒ ( 0.704, 0.774 ) .
n
602
the distribution of systolic blood pressure of anabolic-steroid-using athletes be approximately like a
normal distribution.
9.69
Example 9.8 has shown that the sample data meets the conditions for the t confidence interval.
99% upper confidence bound for the true mean wait time for bypass patients in Ontario:
s
⎛ 10 ⎞
x + 2.33
= 19 + 2.33 ⎜
⎟ = 19 + 1.004 = 20.004
n
⎝ 539 ⎠
9.70
p=
142
= 0.2801
507
The 95% confidence interval for the proportion of the entire population that could correctly describe
the Bill of Rights as the first ten amendments to the U.S. Constitution is
0.2801(0.7199)
0.2801 ± 1.96
⇒ 0.2801 ± 1.96(0.0199) ⇒ 0.2801 ± 0.0391 ⇒ (0.2410, 0.3192).
507
9.71
The 95% confidence interval for the population standard deviation of commuting distance is
⎛ 6.2 ⎞
6.2 ± 1.96 ⎜
⇒ 6.2 ± 0.496 ⇒ (5.704, 6.696).
⎜ 2(300) ⎟⎟
⎝
⎠
9.72
The width of the interval discussed in the text is
s ⎞ ⎛
s ⎞
s
s
⎛
.
= 3.92
⎜ x + (1.96)
⎟ − ⎜ x − (1.96)
⎟ = 2(1.96)
n⎠ ⎝
n⎠
n
n
⎝
The width of the interval suggested in this problem is
s ⎞ ⎛
s ⎞
s
s
⎛
.
= 4.08
⎜ x + (1.75)
⎟ − ⎜ x − (2.33)
⎟ = (1.75 + 2.33)
n⎠ ⎝
n⎠
n
n
⎝
Since this latter interval is wider (less precise) than the one discussed in the text, its use is not
recommended.
9.73
The t critical value for a 90% confidence interval when df = 12 − 1 = 11 is 1.80. The confidence
interval based on this sample data is
s
⎛ 7.7 ⎞
⇒ 21.9 ± (1.80) ⎜
x ± (t critical)
⎟ ⇒ 21.9 ± 4.00 ⇒ (17.9, 25.9) .
n
⎝ 12 ⎠
With 90% confidence, the mean time to consume a frog by Indian False Vampire bats is between
17.9 and 25.9 minutes.
215