252solnD1 10/18/06
(Open this document in 'Page Layout' view!)Edited to replace  or  with D .
D. COMPARISON OF TWO SAMPLES
1. Two Means, Two Independent Samples, Large Samples.
Text 10.1-10.3, 10.7 [10.1 – 10.3, 10.5] (10.1 – 10.3, 10.5)
2. Two Means, Two Independent Samples, Populations Normally Distributed, Population Variances Assumed Equal.
Text 10.4, 10.13a, 10.20a, b, e [10.4, 10.15a, 10.13a,b,e]. For the last
problem: x1  17 .5571 , s1  1.9333 , x 2  19.8905 , s 2  4.5767 (10.4, 10.14, 10.12a,b,e)


3. Two Means, Two independent Samples, Populations Normally Distributed, Population Variances not Assumed Equal.
Optional Text 10.20[10.13c,d] (10.12c,d) See data above. D3, D4
4. Two Means, Paired Samples (If samples are small, populations should be normally distributed).
Text 10.26, 10.29[10.36, 10.37], D1, D2 (10.32*(in 252hwkadd.), [10.34] (different numbers), 10.25[10.35], D1, D2)
5. Rank Tests.
a. The Wilcoxon-Mann-Whitney Test for Two Independent Samples. Text 12.65[10.48] (10.46)
b. Wilcoxon Signed Rank Test for Paired Samples.
Text 12.74-12.76[10.57-59] (10.80-82 on CD), Downing & Clark 18-15, 18-9 (in chapter 17 in D&C 3rd edition), D5
6. Proportions.
Text 10.32, 10.38, 10.39, 12.32** [12.2, 12.7*, 12.8*] (12.2)
7. Variances.
Text 10.40, 10.43-10.48 [10.16, 10.19 - 10.24, 10.25] (10.15, 10.18 - 10.23, 10.24) D6a (below), D6, D7 (A summary problem), D8
(A summary problem)
Graded assignment 3 will be posted.
Solutions to outline points 1 through 3 are in this document.
From the formula table
Interval for
Confidence
Hypotheses
Test Ratio
Interval
Difference
H 0 : D  D0 *
  d  z  2  d
d  D0
z
between Two
H
:
D

D
,
d
1
0
Means (
 12  22



D




1
2
d
known)
n
n
1
Critical Value
d cv  D0  z  2  d
2
d  x1  x2
Difference
between Two
Means (
unknown,
variances
assumed equal)
Difference
between Two
Means(
unknown,
variances
assumed
unequal)
H 1 : D  D0 ,
1
1

n1 n 2
sd  s p
D  1   2
DF  n1  n2  2
H 0 : D  D0 *
D  d  t 2 s d
DF 
H 1 : D  D0 ,
s12 s22

n1 n2
sd 
 s12 s22 
  
n

 1 n2 
D  1   2
t
sˆ 2p 
t
d  D0
sd
d cv  D0  t  2 s d
n1  1s12  n2  1s22
n1  n2  2
d  D0
sd
d cv  D0  t  2 s d
2
   
s12
2
n1
n1  1
* Same as
H 0 : D  D0 *
D  d  t 2 s d
s 22
2
n2
n2  1
H 0 : 1   2
H 1 : 1   2
if D0  0.
1
(Open this document in 'Page Layout' view!)Edited to replace  or  with D .
252solnD1 10/18/06
Problems with 2 means and known population variances or 2 large independent
samples
1  20
 2  10


Exercise 10.1: If for sample 1 n1  40 and for sample 2 n2  50 , what is the value of the z-test
 x  72
 x  66
1
 2
statistic for testing differences between the population means.
Solution: From the outline if D   1   2 and d  x1  x 2 , the formula for a test ratio is
t
x  x 2   10   20 
d  D0
or t  1
where, for large samples we use s d 
sd
sd
with z . If the population variances are known, replace sd with  d 
1  20

the value of z when for sample 1 n1  40 and for sample 2
 x  72
1
d  x1  x2  72  66  6 , D0  10  20  0 , and  d 
 3.4641 . So z 
d  D0
d

12
n1

 22
n2
s12 s 22
and replace t

n1 n 2
. The problem asks for
 2  10

n2  50 . Here
 x  66
 2
 12
n1

 22
n2

20 2
40

10 2
50
 10  2
x  x   10  20   72  66   0  1.732
6-0
 1.732 or z  1 2
3.4641
12  22
20 2 10 2


n1 n2
40
50
Exercise 10.2: What is your decision in 10.1 if you are testing H0: 1   2 and H1: 1   2 and
  .01 .
Solution:   .01 .   .01 , so z  z.005  2.576 . Make a diagram: since this is a 2-sided test, your
2
rejection regions are below -2.576 and above 2.576. So we have H0:
Decision rule: If z  2.576 or z  2.576 , reject H0.
Test statistic: z 
d  D0
d

1   2 and H1: 1   2 .
6-0
 1.732
3.4641
Decision: Since zcalc  1.732 is between the critical bounds of z =  2.576, do not reject H0.
There is inadequate evidence to conclude the two population means are different.
Exercise 10.3: What is the p-value for Problem 10.1?
Solution: pvalue  2Pz  1.732   2.5  .4582   .0836
2
252solnD1 10/18/06
(Open this document in 'Page Layout' view!)Edited to replace  or  with D .
Exercise 10.7 [10.5 in 9th]: We wish to determine whether there is a difference between the life expectancy
of light bulbs produced by two different machines. The problem says the following:
Population Standard deviation
Sample size
Sample Mean
Machine 1
110 hrs
25
375
Machine 2
125 hrs
25
362
a) Is there a difference between the mean life of the bulbs? b) Find and interpret a p-value.
Solution: (a) H0: 1   2 Light bulbs produced by machine 1 have the same average life expectancy as
light bulbs produced by machine 2.
H1: 1   2 Light bulbs produced by machine 1 have a different average life
expectancy as light bulbs produced by machine 2.
1  110
 2  125


Given:   .05, n1  25 and n2  25 . d  x1  x2  375  362  13 ,
 x  362
 x  375
 2
1
D0  10  20  0 , and  d 
(i) Test Ratio: t 
 12
n1

 22
n2

110 2
25

125 2
25
 484  625  33 .307 .3
x  x 2   10   20 
d  D0
or t  1
. If this test ratio lies between t    z  , do
sd
sd
2
2
not reject H 0 .
z
d  D0
13  0

 0.390 . Make a diagram with zero in the middle
sd
33 .307
showing 'reject' regions below -1.960 and above z .025  1.960 . Since 0.390 does not fall in one
of the 'reject' regions, do not reject H 0 .
According to the Instructor’s Solution Manual (edited):
Decision rule: If z < – 1.960 or z > 1.960, reject H0.
( X1 – X2 ) – ( 1 – 2 )
(375 – 362) – 0
= 0.390
 12  2 2
1102 125 2


n1
n2
25
25
Decision: Since zcalc = 0.390 is between the critical bounds of  1.960, do not reject H0.
Test statistic: Z 

(ii). Critical Value: d CV  D0  t  2 s d or x1  x 2 CV  10   20   t  2 s d . If d  x1  x 2 is between the
two critical values, do not reject H 0 . d CV  D0  z  2 s d  0  1.960 33.307   0  65 .28 , or -65.28 and
+65.28. Make a diagram with 0 in the middle showing 'reject' regions below -65.28 and above +65.28.
Since d  13 does not fall in a 'reject' region, do not reject H 0 .
(iii) Confidence Interval: D  d  z  2 s d  13  1.960 33.307   13  65 .28
or -52.28 to 87.28. Since
D0  0 falls in the confidence interval, do not reject H 0 .
According to the Instructor’s Solution Manual, there is not enough evidence to conclude
that light bulbs produced by machine 1 have a different average life expectancy from
light bulbs produced by machine 2.
(b)
pval  2Pz  0.390   2.5  .1517   .6966 . (Since the p-value is above the significance
level , do not reject H 0 .) According to the Instructor’s Solution Manual, the probability
of obtaining samples whose means differ by 13 hours or more when the null hypothesis is
true is 0.6966.
3
(Open this document in 'Page Layout' view!)Edited to replace  or  with D .
252solnD1 10/18/06
Problems with 2 means and 2 independent samples with variances assumed equal.
Text 10.4, 10.15a, 10.13a,b,e
x1  17.5571, s1  1.9333 , x2  19.8905 , s 2  4.5767  (10.4, 10.14, 10.12a,b,e)
s1  4
 s2  5


Exercise 10.4: Given: n1  8 and n2  15 . a) What is the value of the pooled-variance t statistic for
 x  42
 x  34
 1
 2
testing the equality of the population means? b) If we need the critical value of the t-ratio, how many
degrees of freedom are there? c) If we use a significance level of .01 what is the critical value for the test
H :   2
ratio for a one tailed test of the hypotheses  0 1
? d) What is your statistical decision?
 H1 : 1   2
s1  4

Solution: Given: n1  8 and
 x  42
 1
 s2  5

n2  15 .
 x  34
 2
d  x1  x2  42  34  8 . We are assuming 12   22 .
n  1s12  n2  1s22  742  14 52  22 and s  s 2  1  1 

So s p2  1
p

d
n1  n2  2
21
 n1 n2 
1 1 
 22     22 0.125  0.06667   22 0.1916667   4.2166667  2.05345
 8 15 
The test ratio is t 
(a)
t
d  D0
8-0

 3.896 . The Instructor’s Solution Manual has
sd
2.05345
( X 1  X 2 )  ( 1  2 )
1 1
Sp   
 n1 n2 

(42  34)  0
1 1 
22   
 8 15 
2
 3.8959
(e)
df = (n1 – 1) + (n2 – 1) = 7 + 14 = 21
H :   2
 H 0 : 1   2  0
H 0 : D  0
  .01,  0 1
or 
or 
D  1   2 . This is a
H
:



H
:




0
2
2
 1 1
 1 1
 H1 : D  0
right-tail test. Make a diagram. Show a Normal curve with a mean at zero. Since
21
t  t.01
 2.518 , show a 1% rejection region above 2.518.
Decision rule: If t > 2.518, reject H0.
Decision: Since tcalc = 3.896 is above the critical bound of t = 2.518, reject H0. There is
enough evidence to conclude that the first population mean is larger than the second
population mean.
We are sampling from two independent normal distributions having equal variances.
(f)
  .05. A 2-sided confidence interval has the form D  d  t 2 sd
(b)
(c)
(d)
21
1  2   x1  x2   t sd and t 21  t.025
 2.080. .
 8  2.080 2.05345   8  4.271 or 3.729 to 12.271.

2

2
or
So D  d  t 2 sd
The Instructor’s Solution Manual has
1 1
1 1 
 X 2   t S p2      42  34   2.0796 22   
 8 15 
 n1 n2 
or 3.7296<1  2  12.2704 .
X
1
4
252solnD1 10/18/06
(Open this document in 'Page Layout' view!)Edited to replace  or  with D .
Exercise 10.13a [10.15a in 9th] (10.14a in 8th edition): We have the following numbers concerning the
s1  0.012

two independent samples of fill weights of cans of pet food produced by two different lines; n1  11
 x  8.005
 1
s2  0.005

and n2  16
. If we assume that the population variances are equal is there evidence of a difference in
 x  7.997
 2
the mean fill weight from the two lines?
s1  0.012
s2  0.005


Solution: Given: n1  11
and n2  16
. d  x1  x2  8.005  7.997  0.008 .
 x  7.997
 x  8.005
 2
 1
df  11  1  16  1  25. We are assuming 12   22 .
n  1s12  n2  1s22  100.012 2  15 0.005 2  .00144  .000375  0.0000726  7.26 10 5

So s p2  1
25
25
n1  n2  2
1 
1 1
 1
  0.0000726     0.0000726 0.090909  0.0625000
and sd  s p2  

11
16 
n
n

2 
 1

H :    2
 0.0000726 0.153409   0.0000111375  0.00333729 . We are testing  0 1
or
 H 1 : 1   2
H : D  0
 H 0 : 1   2  0
or  0
where D  1   2 .   .05 .

H 1 : D  0
 H 1 : 1   2  0
(i) Test Ratio: t 
x  x 2   10   20 
d  D0
or t  1
. If this test ratio lies between
sd
sd
25
 t   t .025
 2.060 , do not reject H 0 .
2
t
d  D0
0.008  0

 2.397 . Make a
sd
0.00333729
diagram with zero in the middle showing 'reject' regions below -2.060 and above 2.060 . Since
2.397 falls in one of the 'reject' regions, reject H 0 . Or you can say that, since 2.397 falls
25
25
 2.060 and t .001
 2.485 , for a one-sided test, .010  p  value  .025 . For a
between t .025
2-sided test, we double this to get .020  p  value  .05 . If we want a more exact value, try
my available Minitab macro tareaA as below.
————— 10/20/2005 6:04:19 PM ————————————————————
Welcome to Minitab, press F1 for help.
Results for: notmuch.MTW #a nonsense worksheet used so that Mintab finds tareaA
MTB > %tareaA
Executing from file: tareaA.MAC
Graphic display of t curve areas
Finds and displays areas to the left or right of a given value
or between two values. (This macro uses C100-C116 and K100-K120)
Enter the degrees of freedom.
DATA> 25
Do you want the area to the left of a value? (Y or N)
n
Do you want the area to the right of a value? (Y or N)
y
Enter the value for which you want the area to the right.
DATA> 2.397
5
252solnD1 10/18/06
(Open this document in 'Page Layout' view!)Edited to replace  or  with D .
...working...
t Curve Area
Data Display
mode
0
median
0
What this is saying is Pt  2.397   .9878 , so 2.397  t1.9878  t .0122 and
p  value  2.0122   .0244 . This is below   .05 .
According to the Instructor’s Solution Manual (edited):
(a)
H0: 1  2
where Populations: 1 = Line A, 2 = Line B
H1: 1  2
Decision rule: df = 25. If |t| > 2.0595, reject H0.
Test statistic:
x  x 2   1   2 
8.005  7.997   0  2.3972
t 1

 1
1 1
1 
7.26 10 5   

sˆ 2p  
 11 16 
 n1 n 2 
Since t = 2.3972 > 2.0595 or p-value = 0.0243 < 0.05, reject H0. There is sufficient
evidence of a difference in the average weight of cans filled on the two lines.
(ii). Critical Value: d CV  D0  t  2 s d or x1  x 2 CV  10   20   t  2 s d . If d  x1  x 2 is between the
two critical values, do not reject H 0 . d CV  D0  t sd  0  2.0600.00333729  0  0.00687 , or
2
-0.00687 and +0.00687. Make a diagram with 0 in the middle showing 'reject' regions below -0.00867 and
above +0.00687. Since d  0.008 falls in a 'reject' region, reject H 0 .
(iii) Confidence Interval: D  d  t  2 s d  0.008  2.060 0.00333729   0.008  0.00687
or 0.0011 to
0.0149. Since D0  0 does not fall in the confidence interval, reject H 0 .
Exercise 10.20a, b, e [10.13a, b, e in 9th edition] (10.12a, b, e in 8th edition): We are to compare two
randomly selected groups of 21 employees each trained by two different methods. Data from the file
training gives the following results for time necessary to assemble a part after training:
x1  17.5571, s1  1.9333 , x2  19.8905 , s 2  4.5767  n1  21 and n2  21 a) Assume that the variances in
the two populations of assembly times are equal. Is there evidence of a difference in the mean assembly
times? Use   .05 . b) What other assumption is necessary in a)? e) On the same assumptions construct and
interpret a 95% confidence interval for the difference between the population means.
Solution: (a)
H0:
1   2 (where Populations: 1 = computer-assisted, 2 = team-based) Average assembly
times in seconds are the same between employees trained in a computer-assisted,
individual-based program and those trained in a team-based program.
6
(Open this document in 'Page Layout' view!)Edited to replace  or  with D .
252solnD1 10/18/06
H1: 1   2 Average assembly times in seconds are different between
Employees trained in a computer-assisted, individual-based program and those
trained in a team-based program.
d  x1  x 2  17.557119.8905  2.3334 . df  21  1  21  1  40 . We are assuming
12   22 .
n  1s12  n2  1s22  201.9333 2  20 4.5767 2  74.7530  418 .9237  12.3419

So s p2  1
40
40
n1  n2  2
H :    2
1 
1
1
 1
  12 .3419     1.175419  1.08417 . We are testing  0 1
and sd  s p2  

 21 21 
 H 1 : 1   2
 n1 n2 
H : D  0
H :    2  0
or  0 1
or  0
where D  1   2 .   .05 .
H 1 : D  0
 H 1 : 1   2  0
(i) Test Ratio: t 
x  x 2   10   20 
d  D0
or t  1
. If this test ratio lies between
sd
sd
40
 t   t .025
 2.021 , do not reject H 0 .
2
t
d  D0
 2.3334  0

 2.152 . Make a diagram with
sd
1.08417
zero in the middle showing 'reject' regions below -2.021 and above +2.021. Since -2.152 falls in one of the
40
40
'reject' regions, reject H 0 . Or you can say that, since 2.152 falls between t .025
 2.020 and t .01
 2.421 ,
for a one-sided test, .010  p  value  .025 . For a 2-sided test, we double this to get
.020  p  value  .05 .
According to the Instructor’s Solution Manual (edited):
Decision rule: df = 40. If t < – 2.0211 or t > 2.0211, reject H0.
Test statistic:
(n1 – 1)  S12  (n2 –1) S2 2 (20) 1.93332  (20)  4.57672
= 12.3417

(n1 –1)  (n2 – 1)
20  20
(X – X2 ) – (1 – 2 ) (17.5571– 19.8905) – 0
= – 2.1522
t 1

1
1




1
1
2
12.3417  
S p   
21 21
n
n
 1
2 
Sp 
2
Decision: Since tcalc = – 2.1522 is below the lower critical bound of – 2.0211, reject H0.
There is enough evidence to conclude that the average assembly times in seconds are
different between employees trained in a computer-assisted, individual-based program
and those trained in a team-based program.
(ii). Critical Value: d CV  D0  t  2 s d or x1  x 2 CV  10   20   t  2 s d . If d  x1  x 2 is between the
two critical values, do not reject H 0 . d CV  D0  t  2 s d  0  2.021 1.08417   0  2.1911 , or -2.1911 and
2.1911. Make a diagram with 0 in the middle showing 'reject' regions below -2.1911 and above +2.19117.
Since d  2.3334 falls in a 'reject' region, reject H 0 .
(iii) Confidence Interval: D  d  t  2 s d  -2.3334  2.021 1.08417   -2.3334  2.1911
or -4.5245 to
-0.1423. Since D0  0 does not fall in the confidence interval, reject H 0 .
(b)
One must assume that each of the two independent populations is normally distributed.
The following sections (c) and (d) were not required, and were copied from the Instructor’s Solution
7
(Open this document in 'Page Layout' view!)Edited to replace  or  with D .
252solnD1 10/18/06
Manual to allow you to see what output looks like in the unequal variances case.
10.13 (c)
(a) H0: 1   2
where Populations: 1 = computer-assisted, 2 = team-based
cont.
Average assembly times in seconds are the same between employees trained in a
computer-assisted, individual-based program and those trained in a team-based
program.
H1: 1   2
Average assembly times in seconds are different between employees trained in a
computer-assisted, individual-based program and those trained in a team-based
program.
Decision rule: df = 26. If p-value < 0.05, reject H0.
Excel output:
t-Test: Two-Sample Assuming Unequal Variances
Computer-Assisted Program Team-based Program
Mean
17.55714286
19.89047619
Variance
3.737571429
20.94590476
Observations
21
21
Hypothesized Mean Difference
0
df
27
t Stat
-2.152203195
P(T<=t) one-tail
0.020240852
t Critical one-tail
1.703288035
P(T<=t) two-tail
0.040481703
t Critical two-tail
2.051829142
Since p-value = 0.041 < 0.05, reject H0. There is enough evidence to conclude that the
average assembly times in seconds are different between employees trained in a
computer-assisted, individual-based program and those trained in a team-based program.
My solution: x1  17.5571, s1  1.9333 , x 2  19.8905 , s 2  4.5767  n1  21 , n2  21 and   .05 .
d  x1  x 2  17.557119.8905  2.3334 .
s12
n1

s 22 4.5767 2

 0.9974372 ,
n2
21
1.9333 2
 0.1779832 ,
21
s12 s 22

 0.1779832  0.9974372  1.1754205 , s d 
n1 n 2
DF 
 s12 s 22 



 n1 n 2 


2
so
s12 s 22

 1.1754205  1.0842 , and
n1 n 2
2
2

1.1754205 2
0.1779832 2  0.9974372 2

1.3816134 (21)
29 .01388

 28 .263
0.031678  0.9948809 1.0265679
 
 
 
 
21
21
 n1 
 n2 
 
 

n1  1
n2 1
Round this down to 28 degrees of freedom.
Remember d  2.3334 , df  28, D 0  0 and s d  1.0842 .
s12
(i) Test Ratio: t 
s 22
x  x 2   10   20 
d  D0
or t  1
. If this test ratio lies between
sd
sd
28
 t   t .025
 2.048 , do not reject H 0 .
2
t
d  D0
 2.3334  0

 2.152 . Make a diagram with
sd
1.08417
zero in the middle showing 'reject' regions below -2.048 and above +2.048. Since -2.152 falls in one of the
8
252solnD1 10/18/06
(Open this document in 'Page Layout' view!)Edited to replace  or  with D .
28
28
'reject' regions, reject H 0 . Or you can say that, since 2.152 falls between t .025
 2.048 and t .01
 2.467 ,
for a one-sided test, .010  p  value  .025 . For a 2-sided test, we double this to get
.020  p  value  .05 . Since these are both below the significance level, reject H 0 .
(ii). Critical Value: d CV  D0  t  2 s d or x1  x 2 CV  10   20   t  2 s d . If d  x1  x 2 is between the
two critical values, do not reject H 0 . d CV  D0  t  2 s d  0  2.048 1.08417   0  2.2203 , or -2.2203 and
2.2203. Make a diagram with 0 in the middle showing 'reject' regions below -2.2203 and above +2.2203.
Since d  2.3334 falls in a 'reject' region, reject H 0 .
(iii) Confidence Interval: D  d  t  2 s d  -2.3334  2.048 1.08417   -2.3334  2.2203
or -4.5537 to
-0.11131. Since D0  0 does not fall in the confidence interval, reject H 0 .
(d)
The results in (a) and (c) are the same.
(e)
Since my answer is above, this is copied from the Instructor’s Solution Manual.
Assuming equal variances:
1 1
1 1
 X 2   t S p2     17.557  19.891  2.0211 12.3417   
 21 21 
 n1 n2 
4.52<1  2  0.14
X
1
We are 95% confidence that the difference between the population means of the two
training methods is between –4.52 and –0.14.
9
(Open this document in 'Page Layout' view!)Edited to replace  or  with D .
252solnD1 10/18/06
Problems with 2 means and 2 independent samples with variances not assumed
equal.
Exercise 10.20 [10.13c,d in 9th] (10.12c, d in 8th edition): ): We are to compare two randomly selected
groups of 21 employees each trained by two different methods. Data from the file training gives the
following results for time necessary to assemble a part after training:
x1  17.5571, s1  1.9333 , x2  19.8905 , s 2  4.5767  n1  21 and n2  21 c) Assume that the variances in
the two populations of assembly times are not equal. Is there evidence of a difference in the mean assembly
times? Use   .05 . d) Compare the results of a) and c).
Solution: See above.
Problem D3 (Optional): A secretary types 16 pages on word processor 1 and 16 pages on word processor
2. Her times are:
x1  8.20
s12  4.10
x2  7.10 s22  4.20
If   1  2 =1-2 test D  0 at the 90 per cent confidence level. Assume that these are independent
samples and that  12   22 . (   .10 )
Solution: We have n1  16 and n2  16 .
class.
From the Syllabus supplement:
Difference
D  d  t 2 s d
Between Two
s2 s2
Means(
sd  1  2
Unknown,
n1 n2
Variances
2
 s12 s22 
  
Assumed
n

n2 
1
DF   2
Unequal)
2
s
s
The two-sided confidence interval for this problem was done in
H 0 : D  D0
t
H 1 : D  D0
D  1   2
d  D0
sd
d cv  D0  t  2 s d
   
2
1
n1
n1  1
2
2
n2
n2  1
We found the following in class:
s12 4.1
s 2 4.2
s2 s2

 0.25625 , 2 
 0.26250 , so 1  2  0.25625  0.26250  0.51875 ,
n1 16
n2 16
n1 n2
sd 
DF 
s12 s 22

 0.51875  0.720 , d  x1  x2  8.20  7.10  1.10 and
n1 n 2
 s12 s 22 



 n1 n 2 


2
2
2

0.51875 2
0.25625 2  0.26250 2

0.26910
 29 .9 .
0.00438  0.00459
 
 
 
 
15
15
 n1 
 n2 
 
 

n1  1
n2 1
I will follow my own advice this time and round the degrees of freedom down to 29. (If we had followed
29
this advice in class, we would have used t.025
 2.045 and the two-sided confidence interval would have
s12
s 22
been D  d  t  2 s d  1.10  2.045 0.720   1.10  1.47 .)
H 0 : 1   2
H 0 : D  0
We are now testing 
or 
. Since our hypotheses are one-sided we use
H1 : 1   2
H 1 : D  0
t 29  1.311
.10
10
252solnD1 10/18/06
(Open this document in 'Page Layout' view!)Edited to replace  or  with D .
d  D0 x1  x 2   1   2  1.10  0


 1.527 . Make a diagram with zero in the
sd
sd
0.720
middle showing a 'reject' region above t 29  1.311 . Since 1.527 falls in the 'reject' region, reject H .
(i) Test Ratio: t 
0
.10
(ii) Critical Value: d CV  D0  t s d  0  1.311 0.720   0.943 . Make a diagram with zero in the
middle showing a 'reject' above 0.943. Since d  1.10 falls in the 'reject' region, reject H 0 .
(iii) Confidence interval: D  d  t 2 s d becomes D  d  t s d  1.10  1.311 0.720   0.157 . D  0.157
contradicts the null hypothesis D  0 so reject
H0 .
Problem D4: (Old Minitab Manual - modified) In a study of tool life , two independent samples of wear
are taken. The first of these represents volume loss in millionths of cubic inches from 10 untreated tools.
The second represents loss in the same units from 10 tools that were treated by a new wear retardant
process.
Untreated
.56 .50 .69 .59 .47 .42 .45 .47 .50 .50
Treated
.13 .13 .18 .23 .18 .31 .35 .23 .31 .33
On the assumption that the parent populations are Normal, test the hypothesis that the means are equal
and do a confidence interval for the difference between the means ) (a) assuming that the variances are
equal and (b) assuming that the variances are not equal.
Solution: a)
x1
x

s12 
1
n`1
x
2
1
5.15

 0.515
10
 nx12

2.709  10 0.515 2
9

0.628  10 0.238 2
9
n1  1
 0.00625 .
x2 2.38
x2 

 0.238
n2`
10

s22 
x
2
2
 nx2 2
n2  1
 0.00684 .
x1
0.56
0.50
0.69
0.59
0.47
0.42
0.45
0.47
0.50
0.50
5.15
H 0 : D  0
 H 0 : 1   2
H :    2  0
or  0 1
or 

H
:



H
:




0
2
2
H 1 : D  0
 1 1
 1 1
x12
0.3136
0.2500
0.4761
0.3481
0.2209
0.1764
0.2025
0 .2209
0.2500
0.2500
2.709
x2
0.13
0.13
0.18
0.23
0.18
0.31
0.35
0.23
0.31
0.33
2.38
x22
0.0169
0.0169
0.0324
0.0529
0.0324
0.0961
0.1225
0.0529
0.0961
0.1089
0.628
d  x1  x2  0.515  0.238  0.277
a) We assume that the variances are equal. (  12   22 ). So we use the traditional method for this problem.
n  1s12  n2  1s22  9 0.00625   9 0.00684   0.00625   0.00684   0.00655 and

s p2  1
18
2
n1  n2  2
1 
1
 1
1
  0.00625     0.00625 0.2   0.03618
sd  s p2  

10
10


 n1 n2 
df  n1  n2  2  10  10  2  18 .
11
252solnD1 10/18/06
(i) Test Ratio: t 
(Open this document in 'Page Layout' view!)Edited to replace  or  with D .
d  D0
sd

0.277  0
 7.66 . If this test ratio lies between t  , do not reject H 0 .
0.03618
2
Make a diagram with zero in the middle showing 'reject' regions below  t.18
025  2.101 and above
t.18
025  2.101. Since 7.66 falls in a 'reject' region, reject H 0 .
(ii). Critical Value: d CV  D0  t  2 s d  0  2.1010.03618   0.076 . Make a diagram with zero in the
middle showing 'reject' regions below -0.076 and above 0.076. Since d  0.277 falls in a 'reject' region,
reject H 0 .
(iii) Confidence Interval: D  d  t 2 s d  0.277  2.1010.03618   0.277  0.076 or 0.201 to 0.353.
Since zero is not on this interval, reject H 0 .
b) (Optional) We assume that the variances are not equal. ( 12   22 ).
approximation.
So use the Satterthwaite
s12 0.00625
s 2 .00684
s2 s2

 0.000625 , 2 
 0.000684 , so 1  2  0.000625  0.000684  0.001309 ,
n1
10
n2
10
n1 n2
sd 
DF 
s12 s22

 0.001309  0.0362 , d  x1  x2  8.20  7.10  1.10 and
n1 n2
 s12 s22 
  
n

 1 n2 
2

0..001309 2
0.000625 2  0.0006846250 2
 17 .96 . We probably should round down to 17
2
2
 s12 
 s22 
 
 
9
9
n 
 
 1    n2 
n1  1
n2  1
degrees of freedom, but note that, if we use 18 degrees of freedom, our results with this method are the
same as those with the traditional method.
d  D0
0.277  0

 7.66 . If this test ratio lies between t , do not reject H0 .
(i) Test Ratio: t 

sd
0.0362
2
Make a diagram with zero in the middle showing 'reject' regions below  t.17
025  2.110 and above
t.17
025  2.110 . Since 7.66 falls in a 'reject' region, reject
H0 .
(ii). Critical Value: d CV  D0  t  2 s d  0  2.110 0.03618   0.076 . Make a diagram with zero in the
middle showing 'reject' regions below -0.076 and above 0.076. Since d  0.277 falls in a 'reject' region,
reject H 0 .
(iii) Confidence Interval: D  d  t 2 s d  0.277  2.110 0.03618   0.277  0.076 or 0.201 to 0.353.
Since zero is not on this interval, reject H 0 .
Parts not copied ©2003 Roger Even Bove
12