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1.
Answer Key for Homework 9 (Due date: April 16, 2003 which is Wednesday)
Make sure to write your name and section number on each page.
Staple everything together. Place it on the table with your section before class starts. There is a penalty of
5 pt. if you do not staple them together.
Each question worth the same with the total 100 pt. for the homework.
Exercise 9-6 (a) Carefully compute the test statistics, (i) find the P-value and make a decision, (ii) find the
rejection region and make a decision. After this write the conclusion.
One sample size is small and population standard deviations are known. Notice that this is a upper-tailed test
and =0.01.
_ _
 x y    0
(18.12  16.87)  0



Test statistics: z 
=3.53
2
2
1  2
1.6 2 1.4 2


40
32
m
n
(i) P-value=0 < =0.01 then reject H0.
(ii) Reject H0 if z > z =2.33. The test statistics, z =3.53 > 2.33 and reject H 0.
Conclude that the true average tension bond strengths for the modified is larger than the unmodified mortars.
2.
Exercise 9-26. State your hypothesis and carefully compute the test statistics, (i) find the P-value and make a
decision, (ii) find the rejection region and make a decision. After this write the conclusion and identify the type
of error you may have made.
(i)
In the output, P-value is given as 0.0008 for the two-tailed test and we have upper tailed test. It
simply means our P-value is 0.0004. Since the P-value is less than =0.01, reject H0
(ii)
On the upper tailed test, we reject H0 if t > t 0.01;df. The df is 37 with unequal variance assumption
and 38 with equal variance assumption. The table has df either 36 or 38 and I will use 38 which
gives t 0.01;df. =2.429. The test statistics t=3.63 > 2.429 then , reject H 0.
Conclude that the true average potential drop for alloy connections (type 1) is higher than that for EC
connections.
3.
Exercise 9-27 (assume equal population variances to answer the question)
Because of equal variances, v=m+n-2=6+8-2=12
90% C.I. for
(9.5996,28.2004)
1 1
1 1
_ _
  (40.3  21.4)  t 0.05;12 9.6639
 =
 x  y   t / 2 ;v s p
m n
6 8


(m  1) s12  (n  1) s 22 5(11.32 )  7(8.32 )
t 0.05;12 =1.782 and s 2p 

then
mn2
682
1   2 :
where
is
s p  9.6639 . Notice that 0 is not in the confidence interval and 0 implies that their true means are the same.
Interval provide compelling evidence for concluding that the two 's are different. If we have computed 90% C.I. for
 2  1 , our conclusion would be the same but interval boundaries would be both negative.
4.
Exercise 9-33 (assume equal population variances to answer the question)
H 0 :  2  1  5
H a :  2  1  5
or
H 0 :  2  1  5
H a :  2  1  5
_ _
 y x    0
(40.5  32.8)  5


Test statistics: t 
= 2.2373 where

1 1
1 1
sp

2.5442 
m n
8 10
2
2
2
(m  1) s1  (n  1) s 2 7(2.6 )  9(2.5 2 )
2
then s p  2.5442
sp 

mn2
8  10  2
Since they have equal population variances, the degrees of freedom, v=8+10-2=16
0.01 < P-value = P(t>2.2373) < 0.025. Since P-value < =0.10, reject H0. Conclude that the data does suggest
enough evidence to conclude that the true average weight gain in the control situation exceeds that for the steroid
treatment by more than 5 gr.
5.
Using the data for exercise 9.40 (your textbook), I computed the following descriptive statistics for the
individual cases and for their difference.
Difference (P-B)
Mean
Standard Error
Median
Standard Deviation
Sample Variance
Skewness
Range
Minimum
Maximum
Sum
Count
-0.544375
0.17850763
-0.425
0.71403052
0.509839583
-3.342856959
3.16
-3.09
0.07
-8.71
16
Pipe
Brush
1.711875
0.619164
0.575
2.476657
6.13383
1.802189
7.89
0
7.89
27.39
16
2.25625
0.727614
0.81
2.910454
8.470745
1.906975
9.41
0.32
9.73
36.1
16
Answer the following questions using the given information in the textbook and above.
(a) Answer part (a) of the same exercise. Carefully compute the test statistics, (i) find the P-value and make a
decision, (ii) find the rejection region and make a decision. After this write the conclusion.
_
Test statistics: t 
d  0
sd / n

 0.5444  0
0.7140 / 16
= -3.0499 where the v=df=15
(i) 0.002 < P-value < 0.01. Since P-value < =0.01, reject H0.
(ii) Reject H0 if t < -t/2,v =-2.947 or t > t/2,v =2.947. The test statistics, t =-3.0499< -2.947 and reject
H0.
Conclude that one attractor appear to be more effective than the other one.
(b) Test to see if the population variances for bush and pipe are the same. Carefully compute the test statistics,
find the rejection region and make a decision. After this write the conclusion. Use the significance level of
0.1 to make a decision.
H 0 :  12   22 versus H a :  12   22
s12 6.1338
Test statistics: F  2 
 0.7241 where numerator and denominator df is 15.
s 2 8.4708
Reject H0 if F  F1-/2;m-1,n-1 = F0..95;15,15=1/2.40=0.4167 or F  F/2;m-1,n-1= F0..05;15,15=2.40
Since 0.4167 < 0.7241 <2.40, fail to reject H0 and conclude that the variances are the same.
(c) Use your finding about the population variances in (b) and then assuming the data for bush and pipe are
independent, test the hypothesis of their true means being the same versus different. Carefully compute the
test statistics, (i) find the P-value and make a decision, (ii) find the rejection region and make a decision.
After this write the conclusion.
_ _
 x y    0
(0.5444)  0


Test statistics: t 
= -0.5691 where v=df=30 and

1 1
1
1
sp

2.7036

m n
16 16
(m  1) s12  (n  1) s 22 15(2.48 2 )  15(2.912 )
s 2p 

then s p  2.7036
mn2
16  16  2
(i) P-value > 0.2. Since P-value > =0.01, fail to reject H0.
(ii) Reject H0 if t < -t/2,v =-2.75 or t > t/2,v =2.75. The test statistics, t =-0.5691 > -2.75 and < 2.75 then
fail reject H0.
Conclude that one attractor does not appear to be more effective than the other one.
6.
Exercise 9-47: When you are answering the question, carefully compute the test statistics, (i) find the P-value
and make a decision, (ii) find the rejection region and make a decision. After this write the conclusion.
Sample sizes are large enough to satisfy the assumptions and it is a lower-tailed test.
^
^
X Y
30  180
30
180

=0.2625 where p 1 
=0.15 and p 2 
=0.3
m  n 200  600
200
600
^
^

 p1  p 2   p0
(0.15  0.3)  0



Test statistics: z 
=-4.1753
^
^
0.2625(0.7375)(1 / 200  1 / 600)

 1 1 
p1  p   

 m n 
^
p
(i)
(ii)
P-value = P(Z<-4.1753)=0. Since the P-value  =0.05, reject H0.
Reject H0 if z  -z/2=-2.33 z  -2.33 and reject H0.
Conclude that someone switches brands because of financial inducement less likely to remain loyal than someone
switches without inducement.
7.
Exercise 9-61
H 0 :  L2   C2 versus H 0 :  L2   C2
s L2 54 2
 2.8477
Test statistics: F  2 
sC 32 2
Reject H 0 if F  F;m-1,n-1= F0.05;19,22  F0.05;20,22 =2.07
Since F=2.8477 > 2.07, reject H0. Conclude that there is more variability in low-dose weight gains