STAT 511 - 001
Homework 13 Solution
9.1
a.
b.
E X  Y   E X   E Y   4.1  4.5  .4 , irrespective of sample sizes.
V  X  Y   V  X   V Y  
of
c.
 12
m

 22
n
2
2


1.8
2.0 


100
100
 .0724 , and the s.d.
X  Y  .0724  .2691 .
A normal curve with mean and s.d. as given in a and b (because m = n = 100, the CLT
implies that both X and Y have approximately normal distributions, so X  Y does
also). The shape is not necessarily that of a normal curve when m = n = 10, because the
CLT cannot be invoked. So if the two lifetime population distributions are not normal,
the distribution of
X  Y will typically be quite complicated.
9.3
The test statistic value is z 
z  2.33 . We compute z 
x  y   5000 , and H
0
will be rejected at level .01 if
s12 s 22

m n
42,500  36,800  5000
2
2200 2 1500

45
45

700
 1.76 , which is not
396.93
> 2.33, so we don’t reject H0 and conclude that the true average life for radials does not
exceed that for economy brand by significantly more than 500.
9.4
a.
From Exercise 2, the C.I. is
s12 s 22

 2100  1.96433.33  2100  849.33
m n
 1250.67,2949.33 . In the context of this problem situation, the interval is
x  y   1.96
moderately wide (a consequence of the standard deviations being large), so the
information about  1 and  2 is not as precise as might be desirable.
b.
From Exercise 3, the upper bound is
5700  1.645396.93  5700  652.95  6352.95 .
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STAT 511 - 001
9.6
a.
H0 should be rejected if
z  2.33 . Since z 
18.12  16.87  3.53  2.33 , H
0
2.56 1.96

40
32
should be rejected at level .01.


1 0 
   .50  .3085
.3539 
b.
 1   2.33 
c.
2.56 1.96
1
1.96


 .1169 
 .0529  n  37.06 , so use
2
40
n
n
1.645  1.28
n = 38.
d.
Since n = 32 is not a large sample, it would no longer be appropriate to use the large
sample z test of Section 9.1. A small sample t procedure should be used (Section 9.2),
and the appropriate conclusion would follow. Note, however, that the test statistic of 3.53
would not change, and thus it shouldn’t come as a surprise that we would still reject H 0 at
the .01 significance level.
b.
As
9.15
 decreases, z  increases, and since z  is the numerator of n , n increases also.
9.17
a.


52
10
2 2

2
6
 10
2
   
2
5
10
6
10
9
b.


52
10
2 2
2
9
37.21
 17.43  17
.694  1.44

24.01
 21.7  21
.694  .411
9
6
 15
2

2
   
5
10

2
6
15
2
14
9.19
For the given hypotheses, the test statistic
the d.f. is  
t  t.01,9
4.2168  4.82412
4.21682  4.82412
t
115.7  129.3  10
5.032
6
 5.38
6
2

 3.6
 1.20 , and
3.007
 9.96 , so use d.f. = 9. We will reject H0 if
5
5
 2.764; since –1.20 > -2.764, we don’t reject H0.
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STAT 511 - 001
9.20
We want a 95% confidence interval for
1   2 . t.025,9  2.262 , so the interval is
 13.6  2.2623.007   20.40,6.80. Because the interval is so wide, it does not
appear that precise information is available.
9.25
We calculate the degrees of freedom 


5.5 2
28
2 2
5.5
28
.8
 731
2

2
   
2
7.8 2
31
27
 53.95 , or about 54 (normally
30
we would round down to 53, but this number is very close to 54 – of course for this large
number of df, using either 53 or 54 won’t make much difference in the critical t value) so the
91.5  88.3  1.68 528.5  731.8
 3.2  2.931  .269,6.131 . Because 0 does not lie inside this interval, we can be
2
desired confidence interval is
2
reasonably certain that the true difference 1   2 is not 0 and, therefore, that the two
population means are not equal. For a 95% interval, the t value increases to about 2.01 or so,
which results in the interval 3.2  3.506 . Since this interval does contain 0, we can no
longer conclude that the means are different if we use a 95% confidence interval.
9.31
a.
Comparative Box Plot for High Range and Mid Range
470
mid range
460
450
440
430
420
mid range
high range
The most notable feature of these boxplots is the larger amount of variation present in the
mid-range data compared to the high-range data. Otherwise, both look reasonably
symmetric with no outliers present.
b.
Using df = 23, a 95% confidence interval for
438.3  437.45  2.069
plausible values for
15.12
17
 mid  range   high range is
83
 6.11
 .85  8.69   7.84,9.54 . Since
2
 mid  range   high range are both positive and negative (i.e., the
interval spans zero) we would conclude that there is not sufficient evidence to suggest
that the average value for mid-range and the average value for high-range differ.
3