Homework 11 Solution
Total 30
7.12
s
x  2.58
n
 .81  2.58
.34
110
 .81  .08  .73,.89
(3)
7.14
a.
89.10  1.96
3.73
 89.10  .56  88.54,89.66. Yes, this is a very narrow
169
(3)
interval. It appears quite precise.
 1.964
n
 245.86  n  246 .
 .5 
2
b.
7.22
(2)
pˆ  .072 ;
the 99% upper confidence bound is:
.072 
2.332  2.33 .072.928  2.332
2
2487 
487
4487 
2

2.33
1

.0776  .0279
 .1043
1.0111
(5)
487
7.23
a.
pˆ 
24
 .6486 ; The 99% confidence interval for p is
37
2

2.58
.6486 
237 
2

.6486 .3514  2.58
 2.58

2
37
437 
2

2.58
1

.7386  .2216
 .438,.814 
1.1799
37
22.58 .25  2.58 .01  42.58 .25.25  .01  .012.58
b. n 
.01
3.261636  3.3282

 659
.01
2
2
4
4
7.30
a.
t.025,10  2.228
(1.5)
b.
t .025,15  2.131
(1.5)
c.
t.005,15  2.947
(1.5)
d.
t.005, 4  4.604
(1.5)
e.
t.01, 24  2.492
(1.5)
f.
t.005,37  2.712
(1.5)
c.
With d.f. = n – 1 = 16, the critical value for a 95% C.I. is t .025,16  2.120 , and the
7.33c
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.
7.36a
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 
(4)
7.45
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
8.033 
 41.399
this interval requires that fracture toughness be (at least approximately) normally distributed.
STAT 511 - 001
7.46 b
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

is
(4)
3