St361HW7sol
7.14.
(a) A 95% two-sided confidence interval for the true average dye-layer density for all such trees is:
 s 

x  1.96 

 n
 .163
1.28  1.96 
 69
1.028  0.03846
.9895 ,1.0665 
1.96 s 
(b) n  

 B 
2




1.96 .16  
 

 .025 
2
 158
A sample size of 158 trees would be required.
(Note: The researchers wanted an interval width of .05. So, the bound on the error of estimation, B, is
half of the width. B = .025)
7.15.
(a) The large sample size (n = 169) allows us to use the large-sample formula x  z
95% confidence is z = 1.96, so the interval is x  z
s
n
= 89.10  (1.96)
3.73
169
s
n
. The z-value for
= 89.10  .5624 =
[88.5376, 89.6624] or, approximately, [88.54, 89.66]. The interval is very narrow (i.e., its width 89.6688.54 = 1.12 is a small fraction of the mean value, 89.10) so our knowledge about  is quite precise.
(b) Given that  = 4, the minimum sample size needed to estimate  to within .5 Mpa is: n =
2
2
 1.96 
 1.96 (4) 

 =
 = (15.68)2 = 245.86, or, n  246.
 .5 
 B 
7.38.
Given: n = 8
x  30 .2
s  3.1
A 95% confidence interval for the true average stress is:
 s 

x  t critica l value 

 n
The t critical value is obtained from Table IV with df = (n – 1) = 7 for a two-sided interval.
 3.1 

30 .2  2.365 

 8
30 .2  2.6
27 .6, 32 .8
7.49.

The approximate degrees of freedom for this estimate are: df 
 
11.32
6
11.32
6
2
2
 8.83

2
 

8.32
8
2
= 893.586/101.175
5
7
= 8.83, so we round down and use df  8. For a 95% 2-sided confidence interval with 8 df, the critical t
value is 2.306, so the desired interval is: ( x1 - x2 ) - t
s12 s 22
= (40.3-21.4)  (2.306)

n1 n2
11 .32 8.32
=

6
8
18.9  (2.306)(5.4674) = 18.9  12.607 = [6.3, 31.5]. Because 0 is not contained in this interval, there is
strong evidence that 1-2  0; i.e., we can conclude that the population means are not equal. Calculating
a confidence interval for 2-1 would change only the order of subtraction of the sample means (i.e., x2 -
x1 = -18.9), but the standard error calculation would give the same result as before. Therefore, the 95%
interval estimate of 2-1 would be [-31.5, -6.3], just the negatives of the endpoints of the interval for 12. Since 0 is not in this interval, we reach exactly the same conclusion as before: the population means
are not equal.
7.74.
(a) A 95% lower confidence bound for the true average strength of joints with a side coating is:
 s 

x  t critical value 

 n
The t critical value is obtained from Table IV with df = (n – 1) = 9 for a one-sided interval.
Thus:
 5.96 

63 .23  1.833 

 10 
63 .23  3.45  59 .78
That is, with a confidence level of 95%, the value of  lies in the interval (59 .78, ) .
Note: Our above interval is valid only if the distribution of joint strength is normal.
(b)
A 95% lower prediction bound for the strength of a single joint with a side coating is:

1
x  t critical value  s 1  

n 

The t critical value is obtained from Table IV with df = (n – 1) = 9 for a one-sided interval, just as in
part (a) of this exercise.
Thus:

1 
63 .23  1.833  5.96 1  

10 

63 .23  11 .46  51 .77
That is, with a confidence level of 95%, the strength of a single joint with a side coating would be in
the interval (59 .77 , ) .
(c)
For a confidence level of 95%, a two-sided tolerance interval for capturing at least 95% of the
strength values of joints with side coating is:
x  (tolerance critical value) s
The tolerance critical value is obtained from Table V for a two-sided interval, with 95% confidence, k
= 95%, and n = 10.
Thus,
63.23  3.379 5.96 
63.23  20.14
(43.09, 83.37 )
That is, we can be highly confident that at least 95% of all joints with side coatings have strength
values between 43.09 and 83.37.
(d)
A 95% confidence interval for the difference between the true average strengths for the two types of
joints is:
(80 .95  63 .23)  ( t critical value)
9.59 2  5.96 2
10
10
The t critical value requires the df be calculated.
2

 91 .9681 35 .5216 




10 
 10
df  
 91 .9681 10 2 35 .5216 10 2


9
9



  15 .05



Thus, we use df = 15 and the t critical value from Table IV is 2.131.
So,
17.72  (2.131)(3.57 )
17 .72  7.61
(10 .11, 25 .33)
With 95% confidence, we can say that the true average strength for joints without side coating exceeds
that of joints with side coating by between 10.11 and 25.33 lb-in./in.