CHAPTER 7
Section 7.1
1.
a.
z 2  2.81 implies that
level is

2
 1  2.81  .0025 , so   .005 and the confidence
1001   %  99.5% .
b.
z 2  1.44 for   21  1.44  .15 , and 1001   %  85% .
c.
99.7% implies that
  .003 ,  2  .0015 , and z.0015  2.96 . (Look for cumulative
area .9985 in the main body of table A.3, the Z table.)
  .25 ,  2  .125 , and z.125  1.15 .
d.
75% implies
a.
The sample mean is the center of the interval, so x 
2.
114.4  115.6
 115 .
2
b.
The interval (114.4, 115.6) has the 90% confidence level. The higher confidence level
will produce a wider interval.
a.
A 90% confidence interval will be narrower. The z critical value for a 90% confidence
level is 1.645, smaller than the z of 1.96 for the 95% confidence level, thus producing a
narrower interval.
b.
Not a correct statement. Once and interval has been created from a sample, the mean μ is
either enclosed by it, or not. We have 95% confidence in the general procedure, under
repeated and independent sampling.
c.
Not a correct statement. The interval is an estimate for the population mean, not a
boundary for population values.
d.
Not a correct statement. In theory, if the process were repeated an infinite number of
times, 95% of the intervals would contain the population mean μ. We expect 95 out of
100 intervals will contain μ, but we don’t know this to be true.
3.
223
Chapter 7: Statistical Intervals Based on a Single Sample
4.
a.
b.
c.
d.
58.3 
58.3 
1.963
25
1.963
58.3 
100
2.583
100
 58.3  1.18  57.1,59.5
 58.3  .59  57.7,58.9
 58.3  .77  57.5,59.1
82% confidence  1    .82    .18   2  .09 , so
the interval is
58.3 
1.343
100
z 2  z.09  1.34 and
 57.9,58.7  .
e.
 22.583 
n
  239.62 so n = 240.
 1
a.
4.85 
2
5.
b.
1.96.75  4.85  .33  (4.52, 5.18).
20
z 2  z.022  z.01  2.33 , so the interval is 4.56 
2.33.75 
16
 21.96 .75 
n
  54.02 , so n = 55.
.40

2
c.
 22.58.75 
d. w = 2(.2) = .4, so n  
  93.61 , so n = 94.
.4

2
6.
a.
b.
8439 
1.645100  8439  32.9  (8406.1, 8471.9).
25
1    .92    .08   2  .04 so z 2  z.04  1.75
224
(4.12, 5.00).
Chapter 7: Statistical Intervals Based on a Single Sample
7.
If
L  2 z 2

and we increase the sample size by a factor of 4, the new length is
n

  1  L
 2 z 2
   . Thus halving the length requires n to be
4n 
n  2  2
L
increased fourfold. If n  25n , then L   , so the length is decreased by a factor of 5.
5
L  2 z 2

8.
  
  z 2 . These inequalities can be
1   , z1  X     
 n
manipulated exactly as was done in the text to isolate  ; the result is


, so a 1001   % interval is
X  z 2
   X  z 1
n
n


 
 X  z 2
, X  z1

n
n

a.
With probability
b.
The usual 95% interval has length
z
1
 z 2

n
3.92

, while this interval will have length
n
. With z1  z.0125  2.24 and z 2  z .0375  1.78 , the length is
2.24  1.78 
n
 4.02

n
, which is longer.
9.
a.
b.
c.



 x  1.645
,   . From 5a, x  4.85 ,   .75 , n = 20;
n 

.75
4.85  1.645
 4.5741 , so the interval is 4.5741,  .
20



 x  z
,  
n 


 
  , x  z
 ; From 4a, x  58.3 ,   3.0 , n = 25;
n

3
58.3  2.33
  ,59.70
25
225
Chapter 7: Statistical Intervals Based on a Single Sample
10.
a.
When n = 15,
2  X i has a chi-squared distribution with 30 d.f. From the 30 d.f. row
of Table A.6, the critical values that capture lower and upper tail areas of .025 (and thus a
central area of .95) are 16.791 and 46.979. An argument parallel to that given in
 2  x i 2 x i 
 as a 95% C. I. for   1 . Since
,
 46.979 16.791 



Example 7.5 gives 
x
b.
c.
i
 63.2 the interval is (2.69, 7.53).
A 99% confidence level requires using critical values that capture area .005 in each tail of
the chi-squared curve with 30 d.f.; these are 13.787 and 53.672, which replace 16.791 and
46.979 in a.
V X  
1

2
when X has an exponential distribution, so the standard deviation is
1

,
the same as the mean. Thus the interval of a is also a 95% C.I. for the standard deviation
of the lifetime distribution.
11.
Y is a binomial r.v. with n = 1000 and p = .95, so E(Y) = np = 950, the expected number of
intervals that capture
 , and  Y  npq  6.892 . Using the normal approximation to
the binomial distribution, P(940  Y  960) = P(939.5  Ynormal  960.5) = P(-1.52  Z  1.52)
= .9357 - .0643 = .8714.
Section 7.2
12.
x  2.58
s
n
 .81  2.58
.34
110
 .81  .08  .73,.89
13.
x  z .025
s
 654 .16  1.96
164 .43
a.
= (608.58, 699.74). We are 95% confident that the
n
50
true average CO2 level in this population of homes with gas cooking appliances is
between 608.58ppm and 699.74ppm
b.
w  50 
21.96 175 
n
 n
21.96 175 
 13 .72  n = (13.72)2 = 188.24 ↑ 189
50
226
Chapter 7: Statistical Intervals Based on a Single Sample
14.
a.
89.10  1.96
3.73
 89.10  .56  88.54,89.66. Yes, this is a very narrow
169
interval. It appears quite precise.
 1.964
 245.86  n  246 .
b. n  
 .5 
2
15.
a.
z  .84 , and .84  .7995  .80 , so the confidence level is 80%.
b.
z  2.05 , and 2.05  .9798  .98 , so the confidence level is 98%.
c.
z  .67 , and .67  .7486  .75 , so the confidence level is 75%.
x  382.1 , s = 31.5; The 95% upper confidence bound =
s
31.5
x  z
 382.1  1.645
 382.1  7.64  389.74
n
46
16.
n = 46,
17.
x  z.01
s
n
 135.39  2.33
4.59
153
 135.39  .865  134.53 . We are 99%
confident that the true average ultimate tensile strength is greater than 134.53.
18.
19.
90% lower bound:
pˆ 
x  z.10
s
1.30
 4.25  1.28
 4.06
n
78
201
 .5646 ; We calculate a 95% confidence interval for the proportion of all dies
356
that pass the probe:
.5646 
1.962  1.96 .5646.4354  1.962
2
2356 
356
4356 
1.962
1
356
227

.5700  .0518
 .513,.615
1.01079
Chapter 7: Statistical Intervals Based on a Single Sample
20.
Because the sample size is so large, the simpler formula (7.11) for the confidence interval for
p is sufficient.
.15  2.58
21.
pˆ 
.15.85  .15  .013  .137,.163
4722
133
 .2468 ; the 95% lower confidence bound is:
539
2

1.645
.2468 
2539 
2

.2468.7532  1.645
 1.645

2
539
4539 
2

1.645
1

.2493  .0307
 .218
1.005
539
22.
pˆ  .072 ; the 99% upper confidence bound is:
2
2


2.33
.072.928 2.33
.072 
 2.33

2
2487 
487
4487 
2

2.33
1

.0776  .0279
 .1043
1.0111
487
23.
a.
pˆ 
24
 .6486 ; The 99% confidence interval for p is
37
.6486 
2.582
237 
.6486.3514  2.582
2
37
437 
2.582
1
 2.58

.7386  .2216
 .438,.814 
1.1799
37
22.58 .25  2.58 .01  42.58 .25.25  .01  .012.58
.01
3.261636  3.3282

 659
.01
2
2
4
4
b. n 
24.
n = 56,
x  8.17 , s = 1.42; For a 95% C.I., z 2  1.96 . The interval is
 1.42 
8.17  1.96
  7.798,8.542 . We make no assumptions about the distribution if
 56 
percentage elongation.
228
Chapter 7: Statistical Intervals Based on a Single Sample
25.
21.96 .25  1.96 .01  41.96 .25.25  .01  .011.96
n
 381
.01
2
a.
b.
26.
n
21.96

2 1
3
2
4
 23   1.96 .01  41.96
2

4 1
3
 23 13  23  .01  .011.96
4
.01
Using the quadratic equation to solve the limits
X
4
 339
X 
  Z gives the solutions
/n
Z2
Z 2  4nX
. For the data provided, n = 50 and x =203/50 = 4.06.
Z
2n
2n
Substituting these and Z = 1.96, we are 95% confident that the true value of λ is in the interval
1.96 2  4(50)(4.06)
1.96 2
= (3.54, 4.66). Note that for large n, this is
4.06 
 1.96
2(50)
2(50)
approximately equivalent to X  Z
X
ˆ
 ˆ  Z
(since μ = σ2 = λ for the Poisson
n
n
distribution).
z2
2 , which is roughly x  2 with a
Note that the midpoint of the new interval is
n4
n  z2
confidence level of 95% and approximating 1.96  2 . The variance of this quantity is
p 1  p 
x2
np1  p 
, or roughly
. Now replacing p with
, we have
2
n4
n4
n  z 2 
x
27.
 x  2

  z 2
 n  4
pˆ * 
 x  2  x  2 

1 

 n  4  n  4  ; For clarity, let x *  x  2 and n *  n  4 , then
n4
x*
ˆ *  z 2
and the formula reduces to p
*
n
further discussion, see the Agresti article.
229
pˆ * qˆ *
, the desired conclusion. For
n*
Chapter 7: Statistical Intervals Based on a Single Sample
Section 7.3
28.
a.
1.341
b.
1.753
c.
1.708
a.
d.
1.684
e.
2.704
t.025,10  2.228
d.
t.005,50  2.678
b.
t.025, 20  2.086
e.
t.01, 25  2.485
c.
t.005, 20  2.845
f.
 t.025,5  2.571
a.
t.025,10  2.228
d.
t.005, 4  4.604
b.
t .025,15  2.131
e.
t.01, 24  2.492
c.
t.005,15  2.947
f.
t.005,37  2.712
a.
t 05,10  1.812
d.
t.01, 4  3.747
b.
t.05,15  1.753
e.
 t .025, 24  2.064
c.
t .01,15  2.602
f.
t.01,37  2.429
29.
30.
31.
32.
d.f. = n – 1 = 7, so the critical value for a 95% C.I. is t .025, 7  2.365 . The interval is
 3.1 
30.2  2.365
  30.2  2.6  27.6,32.8 .
 8
230
Chapter 7: Statistical Intervals Based on a Single Sample
33.
a.
The boxplot indicates a very slight positive skew, with no outliers. The data appears to
center near 438.
420
430
440
450
460
470
polymer
b.
c.
Based on a normal probability plot, it is reasonable to assume the sample observations
came from a normal distribution.
With d.f. = n – 1 = 16, the critical value for a 95% C.I. is t .025,16  2.120 , and the
interval is
 15.14 
438.29  2.120
  438.29  7.785  430.51,446.08 .
 17 
Since 440 is within the interval, 440 is a plausible value for the true mean. 450, however,
is not, since it lies outside the interval.
34.
n = 14,
a.
x  8.48 , s = .79; t.05,13  1.771
A 95% lower confidence bound:
 .79 
8.48  1.771
  8.48  .37  8.11 . With
 14 
95% confidence, the value of the true mean proportional limit stress of all such joints is
greater than 8.11 MPa. We must assume that the sample observations were taken from a
normally distributed population.
b.
A 95% lower prediction bound:
8.48  1.771.79 1 
1
 8.48  1.45  7.03 . If
14
this bound is calculated for sample after sample, in the long run 95% of these bounds will
provide a lower bound for the corresponding future values of the proportional limit stress
of a single joint of this type.
231
Chapter 7: Statistical Intervals Based on a Single Sample
35.
n = 15, x = 25.0, s = 3.5; t .025,15  2.131
a.
A 95% C.I. for the mean: 25 .0  2.131
3.5
= (23.1, 26.9)
15
b.
36.
1
= (17.3, 32.7). The prediction
15
interval is about 4 times wider than the confidence interval.
A 95% Prediction Interval: 25 .0  2.131(3.5) 1 
n = 26,
a.
x  370.69 , s = 24.36; t .05, 25  1.708
A 95% upper confidence bound:
 24.36 
370.69  1.708
  370.69  8.16  378.85
 26 
b.
A 95% upper prediction bound:
370.69  1.70824.36 1 
c.
1
 370.69  42.45  413.14
26
Following a similar argument as that on p. 300 of the text, we need to find the variance of
X  X new : V  X  X new   V  X   V  X new   V  X   V  12  X 27  X 28 
 V  X   V  12 X 27   V  12 X 28   V  X   14 V  X 27   14 V  X 28 

2
X  X new
1
1
1 1
~t
  2   2   2    . We eventually arrive at T 
n 4
4
s 12  1n
2 n
distribution with n – 1 d.f., so the new prediction interval is
x  t / 2,n1  s
1
2
 1n . For
this situation, we have
370.69  1.70824.36
37.
1 1

 370.69  30.53  340.16,401.22
2 26
a.
A 95% C.I. :
.9255  2.093.0181  .9255  .0379  .8876,.9634
b.
A 95% P.I. :
.9255  2.093.0809 1  201  .9255  .1735  .7520,1.0990
c.
A tolerance interval is requested, with k = 99, confidence level 95%, and n = 20. The
tolerance critical value, from Table A.6, is 3.615. The interval is
.9255  3.615 .0809  .6330,1.2180 .




232
Chapter 7: Statistical Intervals Based on a Single Sample
x  .0635 , s = .0065
1
 .0635  .0137  .0498,.0772 .
a. 95% P.I. : .0635  2.064.0065 1  25
38.
n = 25,
b.
99% Tolerance Interval, with k = 95, critical value 2.972 (table A.6):
.0635  2.972 .0065  .0442,.0828 .




39.
a.
Normal Probability Plot
.999
.99
Probability
.95
.80
.50
.20
.05
.01
.001
30
50
70
volume
Average: 52.2308
StDev: 14.8557
N: 13
Anderson-Darling Normality Test
A-Squared: 0.360
P-Value: 0.392
Based on the above plot, generated by Minitab, it is plausible that the population
distribution is normal.
b.
We require a tolerance interval. (from table A6, with 95% confidence, k = 95, and n=13,
the tcv = 3.081.
x  tcvs  52.231  3.08114.856  52.231  45.771  6.460,98.002
c.
A prediction interval, with t .025,12  2.179 :
52.231  2.17914.856 1  131  52.231  33.593  18.638,85.824
233
Chapter 7: Statistical Intervals Based on a Single Sample
40.
a.
We need to assume the samples came from a normally distributed population.
b.
A Normal Probability plot, generated by Minitab:
Normal Probability Plot
.999
.99
Probability
.95
.80
.50
.20
.05
.01
.001
125
135
145
strength
Average: 134.902
StDev: 4.54186
N: 153
Anderson-Darling Normality Test
A-Squared: 1.065
P-Value: 0.008
The very small p-value indicates that the population distribution from which this data was
taken is most likely not normal.
c.
Despite the apparent lack of normality, we will proceed with a 95% lower prediction
bound: with df = 153-1 = 152, we estimate t = -1.98:
1
135.39  1.984.59 1  153
 135.39  9.12  126.27
41.
The 20 d.f. row of Table A.5 shows that 1.725 captures upper tail area .05 and 1.325 captures
upper tail area .10 The confidence level for each interval is 100(central area)%. For the first
interval, central area = 1 – sum of tail areas = 1 – (.25 + .05) = .70, and for the second and
third intervals the central areas are 1 – (.20 + .10) = .70 and 1 – (.15 + .15) = 70. Thus each
interval has confidence level 70%. The width of the first interval is
s.687  1.725 2.412s
, whereas the widths of the second and third intervals are 2.185

n
n
and 2.128 standard errors respectively. The third interval, with symmetrically placed critical
values, is the shortest, so it should be used. This will always be true for a t interval.
234
Chapter 7: Statistical Intervals Based on a Single Sample
Section 7.4
42.
a.
 .21,15  22 .307 (.1 column, 15
d.
 .2005,25  46 .925
e.
 .299,25  11.523 (from .99
d.f. row)
b.
 .21,25
c.
 .201,25  44.313
f.
 .2995,25  10.519
a.
 .205,10  18 .307
b.
 .295,10  3.940
c.
Since 10 .987   .2975, 22 and 36.78   .2025,22 , P  .2975,22   2   .2025,22  .95 .
d.
Since 14.611   .295,25 and 37 .652   .205, 25 , P(χ2 < 14.611 or χ2 > 37.652) =
 34 .381
column, 25 d.f. row)
43.


1 – P(χ2 > 14.611) + P(χ2 > 37.652) = (1 – .95) + .05 = .10.
44.
n – 1 = 8,
.2025,8  17.534 ,  .2975,8  2.180 , so the 95% interval for  2
 8(7.90 ) 8(7.90 ) 
,

  3.60,28 .98  . The 95% interval for
 17 .534 2.180 
45.

is
 3.60 ,
is

28 .98  1.90,5.38  .
n = 22 implies that d.f. = n – 1 = 21, so the .995 and .005 columns of Table A.7 give the
necessary chi-squared critical values as 8.033 and 41.399. xi  1701.3 and
xi2  132,097.35 , so s 2  25.368 . The interval for  2 is
 2125.368  2125.368  
,

  12.868 ,66.317  and that for  is 3.6,8.1 Validity of this
8.033 
 41.399
interval requires that fracture toughness be (at least approximately) normally distributed.
46.
a.
Using a normal probability plot, we ascertain that it is plausible that this sample was
taken from a normal population distribution.
b.
With s = 1.579 , n = 15, and  .295,14  6.571 the 95% upper confidence bound for
14 1.579 2
 2.305
6.571
235

is
Chapter 7: Statistical Intervals Based on a Single Sample
Supplementary Exercises
47.
a.
n = 48, x  8.079 , s2 = 23.7017, and s = 4.868.
A 95% C.I. for  = the true average strength is
x  1.96
b.
pˆ 
s
n
 8.079  1.96
48
 8.079  1.377  6.702,9.456
13
 .2708 . A 95% C.I. is
48
.2708 
.2708.7292  1.96 2
1.96 2
 1.96
2
248
48
448
1
48.
4.868
1.96
48
2

.3108  .1319
 .166,.410
1.0800
A 98% t C.I. requires t / 2, n 1  t .01,8  2.896 . The interval is
188.0  2.896
7.2
9
 188.0  7.0  181.0,195.0 .
49.
a.
There appears to be a slight positive skew in the middle half of the sample, but the lower
whisker is much longer than the upper whisker. The extent of variability is rather
substantial, although there are no outliers.
20
30
40
50
%porevolume
b.
The pattern of points in a normal probability plot is reasonably linear, so, yes, normality
is plausible.
c.
n = 18,
x  38.66 , s = 8.473, and t.01,17  2.567 . The 98% confidence interval is
38.66  2.567
8.473
 38.66  5.13  33.53,43.79 .
18
236
Chapter 7: Statistical Intervals Based on a Single Sample
50.
x  the middle of the interval =
229.764  233.504
 231.634. To find s we use
2
 s 
width  2t.025, 4 
 , and solve for s. Here, n = 5, t.025, 4  2.776 , and width = upper
 n
s
5 3.74
limit – lower limit = 3.74. 3.74  22776 
s
 1.5063 . So for a
22.776
5
99% C.I., t .005, 4  4.604 , and the interval is
231.634  4.604
1.5063
 213.634  3.101  228.533,234.735.
5
51.
a.
p̂ 
136
 .680  a 90% C.I. is
200
.680 
.680.320  1.645 2
1.645 2
 1.645
2
2200
200
4200
1
1.645
200
2

.6868  .0547
 .624,.732
1.01353
21.645 .25  1.645 .05  41.645 .25.25  .0025  .05 2 1.645
b. n 
.0025
1.3462  1.3530

 1079.7  use n = 1080
.0025
2
c.
52.
2
2
4
4
No, it gives a 95% upper bound.
n = 5, x = 24.3, s = 4.1
4.1
a.
t.025,4 = 2.776: 24 .3  2.776
 24 .3  5.09 = (19.21, 29.39). We are 95% confident that
5
the true average arsenic concentration in all such water specimens is between 19.21 μg/L
and 29.39 μg/L.
b.
A 90% upper bound for σ, with  .290,4  1.064 , is
c.
A 95% prediction interval is 24 .3  2.776 (4.1) 1 
237
4(4.1) 2
= 7.95 μg/L
1.064
1
= (11.83, 36.77).
5
Chapter 7: Statistical Intervals Based on a Single Sample
53.
With ˆ 
1
3
X
1
 X 2  X 3   X 4 ,  2ˆ  19 VarX 1  X 2  X 3   VarX 4  =
1   12  22  32   42 ; ˆ is obtained by replacing each  2 by s 2 and taking the




i
i
ˆ
9  n1
n2
n3  n4
square root. The large-sample interval for  is then
1
3
s32
s 22
1  s12



9  n1 n2 n3
 x1  x2  x3   x4  z / 2
 s 42 . For the given data, ˆ
  .50 ,

 n
4

ˆ ˆ  .1718 , so the interval is  .50  1.96.1718   .84,.16 .
54.
11
 .2  a 90% C.I. is
55
p̂ 
.2 
.2.8  1.645 2
1.645 2
 1.645
2
255
55
455
1
1.645
55
2

.2246  .0887
 .1295,.2986 .
1.0492
 21.96.8 
The specified condition is that the interval be length .2, so n  
  245.86 , so
.2

2
55.
n = 246 should be used.
56.
a.
b.
A normal probability plot lends support to the assumption that pulmonary compliance is
normally distributed. Note also that the lower and upper fourths are 192.3 and 228,1, so
the fourth spread is 35.8, and the sample contains no outliers.
t .025,15  2.131 , so the C.I. is
209.75  2.131
c.
57.
24.156
16
 209.75  12.87  196.88,222.62 .
K = 95, n = 16, and the tolerance critical value is 2.903, so the 95% tolerance interval is
209.75  2.903 24.156  209.75  70.125  139.625,279.875 .



Proceeding as in Example 7.5 with T r replacing


X i , the C.I. for

1

is
 2t
2t
r

, 2r
2
  1 , 2 r   , 2 r
2
2





where t r  y1  ...  yr  n  r yr . In Example 6.7, n = 20, r = 10, and tr = 1115. With
d.f. = 20, the necessary critical values are 9.591 and 34.170, giving the interval (65.3, 232.5).
This is obviously an extremely wide interval. The censored experiment provides less
information about 1 than would an uncensored experiment with n = 20.
238
Chapter 7: Statistical Intervals Based on a Single Sample
58.
a.
P(min( X i )  ~  max( X i ))  1  P(~  min( X i )or max( X i )  ~)
 1  P(~  min( X ))  P(max( X )  ~)
i
i
 1  P(~  X 1 ,..., ~  X n )  P( X 1  ~,..., X n  ~)
 1  .5  .5  1  .5
n
b.
c.
n 1
n
, from which the confidence interval follows.
min( xi )  1.44 and max( xi )  3.54, the C.I. is (1.44, 3.54).
Since
P( X ( 2)  ~  X ( n1) )  1  P( ~  X ( 2) )  P( X ( n1)  ~)
~ ) – P(at most one XI exceeds ~ )
= 1 – P( at most one XI is below 
 n  1 n 1
 n  n 1
n
n
1  .5   .5 .5  .5   .5 .5 .
1
 1
n
n 1
 1  2n  1.5  1  n  1.5
Thus the confidence coefficient is 1  n  1.5

100 1  n  1.5
n 1
% confidence interval.
59.
1 / 2 1 / n
a.

/ 2
1/ n
nu n 1 du  u n
statement,

1 / 2 1 / n
 / 2 1 / n
 1
 2   1  1   2 
max  X i   max  X i 
1
n
n 1
1
n

2


2
, or in another way, a
 1   . From the probability
with probability 1   , so taking the
 max  X i  max  X i  

 1    1n ,   1n 
2
2


reciprocal of each endpoint and interchanging gives the C.I. 
for
b.
.
 
1
n
max( X i )
probability
c.

 1 with probability 1   , so 1 
1   , which yields the interval

max  X i 

1

max  X i  

 max  X i ,
.



1
1
with
n
n
It is easily verified that the interval of b is shorter – draw a graph of
f U u  and verify
that the shortest interval which captures area 1   under the curve is the rightmost such
interval, which leads to the C.I. of b. With   .05, n = 5, max(xI)=4.2; this yields (4.2,
7.65).
239
Chapter 7: Statistical Intervals Based on a Single Sample
60.
The length of the interval is
minimized, i.e. when
61.
 

2

 z  
s
n
, which is minimized when z  z  is
 1 1      1 1      is minimized. Taking
d
and
d
1
1
where  is the standard normal p.d.f.,

1    1     
equating to 0 yields
whence
z
.
~
x  76.2, the lower and upper fourths are 73.5 and 79.7, respectively, and f s  6.2. The
 6.2 
76.2  1.93
  76.2  2.6  73.6,78.8 .
 22 
x  77.33 , s = 5.037, and t.025, 21  2.080 , so the t interval is
robust interval is
 5.037 
77.33  2.080
  77.33  2.23  75.1,79.6 . The t interval is centered at
 22 
x , which is pulled out to the right of ~
x by the single mild outlier 93.7; the interval widths
are comparable.
62.
a.
Since
2X i has a chi-squared distribution with 2n d.f. and the area under this chi-
squared curve to the right of
that

2
.95, 2 n
2X i
 .295, 2n


is .95, P  .295, 2n  2X i  .95 . This implies
is a lower confidence bound for
 with confidence coefficient 95%.
Table
A.7 gives the chi-squared critical value for 20 d.f. as 10.851, so the bound is
10.851
 .0098 . We can be 95% confident that  exceeds .0098.
2550.87 
b.
Arguing as in a,


P 2X i   .205, 2n  .95 . The following inequalities are equivalent
to the one in parentheses:

 .205, 2 n
2 X i
Replacing the
 t 
 t .205, 2 n
2X i
  t .205, 2 n 
 e t  exp 
.
 2X i 
X i by xi in the expression on the right hand side of the last inequality
gives a 95% lower confidence bound for e
and
 t
. Substituting t = 100,
 .205, 20  31.410
xi  550.87 gives .058 as the lower bound for the probability that time until
breakdown exceeds 100 minutes.
240