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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 6 and 7 are in this document. Summary problems D7 and D8 are in the
next document.
-------------------------------------------------------------------------------------------------------------------------------
Problems with 2 Proportions.
 x  45, n1  100
Exercise 10.32 [12.2 in 9th] (12.2 in 8th edition): Assume that  1
. a) At the 5% significance
 x2  25, n2  50
level is there a significant difference between population proportions? b) Construct a 95% confidence
interval for the difference between the two population proportions.
Solution: From Table 3 of the Syllabus Supplement:
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
pcv  p0  z 2  p
Difference
p  p 0
p  p  z 2 sp
H 0 : p  p0
z
between
If p0  0
 p
H 1 : p  p0
p  p1  p2
proportions
 p  p0 q 0  1 n  1 n 
If p  0
p 0  p 01  p 02
p1q1 p2 q 2
q  1 p
s p 
n1

1
n2
or p 0  0
 p 
p01q 01 p02 q 02

n1
n2
Or use s p
2
n p  n2 p2
p0  1 1
n1  n 2
H 0 : p1  p 2
H 0 : p  0
H : p  p 2  0
 x1  45, n1  100
  .01  0 1
Given 
or 
or 
 x2  25, n2  50
H 1 : p1  p 2
H 1 : p  0
H 1 : p1  p 2  0
x
45
x
25
 .45, p2  2 
 .50 . p  p1  p2  .45  .50  .05 .
a) p1  1 
n1 100
n2 50
p0 
x1  x2 45  25
70
n p  n2 p2 100 .45   50 .50 


 0.4667  1 1

q0  1  p0  1  .4667  .5333
n1  n2 100  50 150
n1  n2
100  50
 p  p0q0

1
n1

1
n3

.4667 .5333  1100  150 
.0074667  .08641
z 2  z.005  2.576.
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(Only one of the following methods is needed!)Test Ratio: z 
p  p 0

p
 .05  0
 0.579 . Make a
.08641
diagram showing a 99% 'accept' region between -2.576 and +2.576. Shade the 'reject regions below -2.576
and above 2.576. Since -0.579 lies in the 'accept' region, do not reject H 0 .
or Critical Value: pcv  p0  z  p  0  2.5760.08641  0.233 . Make a diagram showing a
2
99% 'accept' region between -0.223 and +0.223. Shade the 'reject' regions below -0.223 and above 0.223.
Since p  .05 lies in the 'accept' region, do not reject H 0 .
b) p  p  z sp , where p  p1  p2  .45  .50  .05,
2
q1  1  p1  1  .45  .55, q2  1  p2  1  .50  .50,
sp 
p1q1 p2 q2
.45.55  .50 .50 



 .002475  .005000  .007475  0.086458 .
n1
n2
100
50
So p  .05  2.576 0.086458   .05  .223 or -.173 to .273. Note that this interval includes zero and
thus supports the null hypothesis.
Exercise 10.38 [12.7 in 9th] (Not in 8th edition): Of 56 white workers terminated, 29 claimed bias. Of 407
black workers terminated, 126 claimed bias.
a.)At the 5% significance level, is there evidence that white workers are more likely to claim bias
than black workers?
b) Find and interpret the p-value in a).
Solution:
H 0 : p  0
H : p  p 2
H : p  p 2  0
 x  29, n1  56
Given  1
or  0 1
or 
so
  .05  0 1
 x 2  126 , n 2  407
H1 : p  0
H 1 : p1  p 2
H 1 : p1  p 2  0
Population 1 contains white workers and population 2
x
x
29
126
p1  1 
 .5179 , p 2  2 
 .3096 . p  p1  p2  .5179  .3096  .2083
n1 56
n 2 407
a) p0 
x1  x2 29  126 155
n p n p
56 .5179   407 .3096 


 0.3348  1 1 2 2 
n1  n2 56  407 463
n1  n2
56  407
 p  p 0 q 0

1
n1

1
n3

.3348 .6652  156  1 407  .22270 .017857
(Only one of the following methods is needed!)Test Ratio: z 
 .002457   .0052393  .06726
p  p 0
 p

.2083  0
 3.097 . This is a
.06726
one-tail test! Because the alternate hypothesis says that the difference between the proportions is above
zero, this is a right-tail test. Make a diagram with a mean at zero showing a 95% 'accept' region below
z  z .05  1.645 . Shade the 'reject region above 1.645. Since 3.097 lies in the 'reject' region, reject H 0 .
or Critical Value: Since the alternate hypothesis says p  0 , we need a critical value for the difference
between the two proportions that is above zero, so the two-tail formula, pcv  p0  z  p becomes
2
pcv  p0  z  p  0  1.6450.06726  .1106. . Make a diagram showing a 95% 'accept' region
between 0.1106. Shade the 'reject' regions above 0.1106. Since p  .2083 is above .1106 and lies in the
'reject' region, do not reject H 0 . or Confidence Interval: The two sided formula is p  p  z s p .
2
Since the alternate hypothesis is p  0 , a one-sided interval will be p  p  z s p where
2
s p 
p1 q1 p 2 q 2
.5179 .4821  .3096 .6904 



 .0044586  .0004616  .004920  0.07014 . So
n1
n2
56
463
the confidence interval will say p   0.2083  1.645 0.07014   0.2083  .1154  .0929 . The null
hypothesis says p  0 , which is contradicted by p  .0929 . Reject H 0 .
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b) Remember that the alternate hypothesis is p  0 Extreme, in this case means high, since high values of
p will discourage our belief that p  0 . pvalue  Pz  3.097   .5  .4990  .0010
The Instructor’s Solution Manual says (a) There is sufficient evidence to conclude that white workers are
more likely to claim bias than black workers. (b) Using Excel, the p-value is 0.00098. The probability of
obtaining a difference in two sample proportions as large as 0.20828 or larger is 0.00098 when the null
hypothesis is true.
Exercise 10.39 [12.8 in 9th] (Not in 8th edition): 500 African Americans and 500 whites (all with incomes
above $50000) were surveyed with the result that 74% of the African-Americans and 84% of the whites
owned stocks.
a. At the 5% significance level, is there a significant difference between the proportion of African
Americans and the proportion of whites who own stocks?
b. Find and interpret the p-value in a.
c. Set up a 95% confidence interval for the difference between the two proportions.
H : p  p 2
H : p  0
H : p  p 2  0
 p  .74, n1  500
Solution: Given  1
or  0 1
or  0
  .05  0 1
 p 2  .84, n 2  500
H 1 : p1  p 2
H 1 : p  0
H 1 : p1  p 2  0
p  p1  p2  .74  .84  .10 (Population 1 = African American, Population 2 = Whites)
a) p0 
n1 p1  n2 p2 500 .74   500 .84 

 .79 q0  1  p0  1  .79  .21
n1  n2
500  500
 p  p 0 q 0

1
n1

1
n3

.79 .21 1500  1500  .1659 .004
(Only one of the following methods is needed!)Test Ratio: z 
 .02576
p  p 0
p
z 2  z.015  1.960.

 .10  0
 3.882 . Make a
.02576
diagram showing a 95% 'accept' region between -1.960 and +1.960. Shade the 'reject regions below -1.960
and above 1.960. Since -3.882 lies in the lower 'reject' region, reject H 0 .
or Critical Value: pcv  p0  z  p  0  1.9600.02576  0.0505 . Make a diagram showing a
2
95% 'accept' region between -0.0505 and +0.0505. Shade the 'reject' regions below -0.0505 and above
0.0505. Since p  .10 lies in the lower 'reject' region, reject H 0 .
The Instructor’s Solution Manual says that there is sufficient evidence to conclude that a significant
difference exists in the proportion of African Americans and whites who invest in stocks.
b) Since this is a 2-sided test, p  value  2Pz  3.882   2.5  .4999   .0002 . The Instructor’s Solution
Manual should say that the probability of obtaining a test statistic as large in absolute value as 3.8819 or
larger is 0.0002 when the null hypothesis is true, but it apparently got this one wrong.
c) p  p  z sp , where p  p1  p 2  .10, q1  1  p1  1  .74  .26, q 2  1  p 2  1  .84  .16,
2
s p 
p1 q1 p 2 q 2
.74 .26  .84 .16 



 .0003848  .0002688  .0006536  0.02557 .
n1
n2
500
500
So p  .10  1.960 0.02557   .10  .050 or -.150 to .050. Note that this interval does not include zero
and thus contradicts the null hypothesis.
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Exercise 12.32 (New): A random sample of 200 coffee drinkers is taken and their preferences between
Brand A and Brand B are taken before and after an advertising campaign.
After
Before
A B
The results are as follows:
. a) Is there evidence at the 5% significance level that the
A
101 9
 22 68 
B


proportion of coffee drinkers who prefer brand A has risen over the campaign? b) Compute a p-value and
interpret it.
Solution: a) The 2005 version of the outline says the following. In the McNemar Test we compare two
proportions taken from the same sample, which is equivalent to paired samples. Assume that two different
question 2
question 1
yes no
questions are asked of the same group with the following responses.
So, for
x
yes
 11 x12 
x

no
 21 x 22 
example x 21 is the number of people who answered no to question 1 and yes to question 1.
x11  x12  x 21  x 22  n , p1 
z
x12  x 21
x12  x 21
H 0 : p1  p 2
x11  x12
x  x 21
and p 2  11
. If we wish to test 
, let
n
n
H 1 : p1  p 2
. (The test is valid only if x12  x 21  10 .)
Here we have a left-sided test because for us to believe that the proportion that prefer A to rise, we need to
find that x11  x 21  x11  x12 , but this implies x 21  x12 , which means that the numerator of z will be
H 0 : p1  p 2
negative. Our hypotheses are 
. Using our formula for z , we compute
H 1 : p1  p 2
z
x12  x 21
x12  x 21

9  22
9  22

13 2
 2.3349 . For a left – sided test our rejection region is below
31
z .045  1.645 . Since -2.3349 is in the rejection region, we reject the null hypothesis and conclude that the
proportion has risen. b) The p-value is Pz  2.34   .5  .4904  .0096 , which says that if the null
hypothesis is true the probability of getting results as extreme or more extreme than what we actually got is
96 one-hundredths of 1%.
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Problems with 2 Variances.
Exercise 10.40 [10.16 in 9th] (10.15 in 8th edition): Find FU and FL the upper and lower critical values
of the F ratio for the following two-tailed tests: (a)   .10, n1  16, n2  21 , (b)   .05, n1  16, n 2  21 ,
(c)   .02, n1  16, n2  21 , (d)   .01, n1  16, n 2  21 , (e) As  gets smaller (or as the confidence
level gets larger) what happens to the size of the ‘accept’ and ‘reject’ regions?
Solution: For a 2-tail F test, the upper bound is FU  Fn1 1,n2 1 and the lower bound is
2
n1 1,n2 1
FL  F1
1

2
n2 1,n1 1
F
15,20 
FL  F.95
15, 20 and
. For   .10, n1  16, n2  21 , for example, FU  F.05
2
1
20,15
F.05
From the Instructor’s Solution Manual (edited):
These values come from table E.5 in the text. Since my table does not have a df1  15 column, you
would have to average the values in the df1  14 and df1  16 columns to get these numbers.
1
= 0.429
2.33
1
(b)   .05, n1  16, n 2  21
FU = 2.57, FL =
= 0.362
2.76
1
(c)   .02, n1  16, n2  21
FU = 3.09, FL =
= 0.297
3.37
1
(d)   .01, n1  16, n 2  21
FU = 3.50, FL =
= 0.258
3.88
(e) As  gets smaller, the rejection region gets narrower and the acceptance region gets wider. FL
gets smaller and FU gets larger.
(a)   .10, n1  16, n2  21
FU = 2.20, FL =
Exercise 10.43-10.46 [10.19-22 in 9th] (10.18-21 in 8th edition): Assume that you have the following
information: n1  25, n2  25, s12  133.7 and s 22  162.9 . a) What is the value of the F ratio used for
testing equality of variances? b) How many degrees of freedom do you have in the numerator and the
denominator of the f test? c) What are the critical values FU and FL if you do a test for equality of
variances? d) What is your statistical decision?
Solution: a) Given: H 0 :  12   22 , H1 :  12   22 , n1  25, n2  25, s12  133.7 and s 22  162.9.
Fcalc 
s12
s 22

133 .7
 0.826 . b) Since n1  25 and n2  25, the degrees of freedom are df1  n1  1
161 .9
24, 24 .
 25  1  24 and df 2  n 2  1  24 . c) Part of the F table appears below. We find F.025

14
df 2  24 *
0.100
0.050
0.025
0.010
1.80
2.13
2.47
2.93
16
*
1.77
2.09
2.41
2.85
20
24
30
*
1.73
2.03
2.33
2.74
*
1.70
1.98
2.27
2.66
*
1.67
1.94
2.21
2.58
df1
40
50
*
1.64
1.89
2.15
2.49
*
1.62
1.86
2.11
2.44




24, 24  2.27 and
FU  F n1 1,n2 1  F.025
FL  F1n1 1,n2 1 
2
2
75
100
*
1.59
1.82
2.05
2.37
*
1.58
1.80
2.02
2.33
1
1
200
500
*
1.56
1.77
1.98
2.27
*
1.54
1.75
1.95
2.24
10000
*
1.53
1.73
1.94
2.21
1
. d) The

  F 24,24  2.27  0.441
F n2 1,n1 1
.025
2
Instructor’s Solution Manual says that, since Fcalc = 0.826 is between the critical bounds of FU = 2.27 and FL = 0.441,
do not reject H0. There is not enough evidence to conclude that the two population variances are different.
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Exercise 10.47 [10.19 in 9th] (10.18 in 8th edition): Assume that you have the following information:
n1  16 , n 2  13, s12  47.3 and s 22  36.4 , but the populations are very skewed to the right, should you
use the F test to test for equal population variances?
Solution: Given: From the Instructor’s Solution Manual ,
In testing the equality of two population variances, the F-test statistic is very sensitive to the
assumption of normality for each population. If the populations are very right-skewed, the F-test
should not be used.
It really should be noted that the Levene test (to be covered in section F) is a more appropriate test in this
case. However, the entire data sets are needed for the Levene test.
Exercise 10.48 [10.24 in 9th] (10.13? in 8th edition): Again assume that you have the following information:
n1  16 , n 2  13, s12  47.3 and s 22  36.4 and use a 5% significance level. a) Is there evidence of a
difference between  12 and  22 ? b) Test  12   22 . c) Test  12   22 .
Solution: a) H 0 :  12   22 or H 0 :
Fcalc 
s12
s 22

 12
 22
 1. and H1 :  12   22 or H 1 :
47 .3
 1.2995 . Since n1  16 and
36 .4
n 2  13,
 12
 22
 1.
the degrees of freedom are
df1  n1  1
15,12  3.18 and
 16  1  15 and df 2  n2  1  12 . For our upper and lower critical values, we find F.025
15,12 
F.975
1
1
. The first value comes from the table in the text since values of df1  14
12,15  2.96  0.338
F.025
and df1  16 appear in Table 2, but not df1  15 . This is not really a problem, since if we find
14,12  3.21
F.025
and
16,12  3.15
F.025
it is not hard to guess that
15,12 
F.025
14,12  F 16,12
F.025
.025
2
3.21  3.15

 3.18 . The following comes from the Instructor’s Solution Manual.
2
Decision: Since Fcalc = 1.2995 is between the critical bounds of FU = 3.18 and FL = 0.3378, do not
reject H0. There is not enough evidence to conclude that the two population variances are
different.
The Minitab instructions and output for this test on the Normality assumption follow:
MTB > VarTest 16 47.3 13 36.4.
Test for Equal Variances
95% Bonferroni confidence intervals for standard deviations
Sample N Lower StDev Upper
1 16 4.87589 6.87750 11.4026
2 13 4.13630 6.03324 10.7890
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The confidence intervals for the two variances are Bonferroni intervals, which mean that they are
simultaneously valid at the 5% level. The test statistic is the one calculated above and rounded. The p-value
given to the null hypothesis is above our significance level so that we cannot reject the null hypothesis.
b) The statement (  12   22 ) to be evaluated in b) is a strict inequality and thus must be an alternative
hypothesis, so that we have H 0 :  12   22 or H 0 :
 12
 22
 1 and H1 :  12   22 or H 1 :
alternative hypothesis is true, we would expect the ratio, Fcalc 
s12
 12
 22
 1 . If the
to be above one. Remember that
s 22
df1  15 and df 2  12 . Since this is a one-sided test with a significance level of 5%, we compare our
15,12  2.62 . (We can get this from the textbook table or by averaging the values
calculated F ratio, with F.05
14,12 and F 16,12 .) The following comes from the Instructor’s Solution Manual.
of F.05
.05
Decision: Since Fcalc = 1.2995 is less than the critical bound of FU = 2.62, do not reject
H0. There is not enough evidence to conclude that the variance for population 1 is greater
than the variance for population 2.
c) The statement (  12   22 ) to be evaluated in c) is a strict inequality and thus must be an alternative
hypothesis, so that we have H 0 :  12   22 or H 1 :
 12
 22
 1 and H1 :  12   22 or H 1 :
 12
 22
 1. The
Instructor’s Solution Manual implies that if the alternative hypothesis is true, we would expect the ratio,
s2
Fcalc  12 to be below one. Remember that df1  15 and df 2  12 . Since this is a one-sided test with a
s2
significance
15,12 
F.95
level
1
of
5%,
we
compare
our
calculated
F
ratio,
Fcalc 
s12
s 22
 1.2995
with
1
. The following comes from the Instructor’s Solution Manual.
12,15  2.48  0.403
F.05
Decision: Since Fcalc = 1.2995 is greater than the critical bound of FL = 0.403, do not
reject H0. There is not enough evidence to conclude that the variance for population 1 is
less than the variance for population 2.
An alternative way to approach this problem is to note that we can say that our hypotheses are
H 0 :  12   22 , H1 :  12   22 or H 1 :
ratio, Fcalc 
s 22
s12
 22
 12
 1. If the alternative hypothesis is true, we would expect the
to be above one. Remember that df1  15 and df 2  12 . Since this is a one-sided test with a
significance level of 5%, we compare our calculated F ratio, Fcalc 
s 22
s12

1
 0.7695 with
1.2995
15,12  2.48 . Since our calculated F ratio is not above the 5% critical value, we cannot reject the null
F.05
hypothesis.
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Problem D6a: Given: H 0 :  12   22 , H 1 :  12   22 , n1  16 , n 2  21, s12  100 and s 22  400 . Test the
hypotheses using only right tail tests.   .05 . (This is my preferred way of doing problems like this
because the F table is set up to make right tail tests most convenient.)
s 2 100
Solution: The ratio Fcalc  12 
 0.250 has 15 and 20 degrees of freedom. We found that
s 2 400
15,20  2.57 , and we see that our calculated F-ratio does not exceed it. The ratio
F.025
Fcalc 
s 22
s12

400
 4.000 has
100
20,15  2.76, and this second calculated F-ratio does exceed it. Since
20 and 15 degrees of freedom. We found that F.025
one calculated F-ratio is above its bound, we reject the null hypothesis. To get the p-value, compute 2 PF  4.00  on
the computer as follows below.
Welcome to Minitab, press F1 for help.
Results for: notmuch.MTW
MTB > %fareaa
Executing from file: fareaa.MAC
Graphic display of F 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.DF2 must be above 4.
DATA> 20
DATA> 15
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> 4
...working...
F Curve Area
F Curve with numerator DF of 20 and Denominator DF of 15
The Area to the Right of 4 is 0.0043
0.9
0.8
0.7
Density
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0
1
2
3
4
Data A xis
5
6
7
8
Data Display
mode
0.890625
Data Display
std dev
0.631988
So we can say that 2PF  4.00   2.0043   .0086 .
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Exercise 10.49 [10.25 in 9th] (10.24 in 8th edition): A professor claims that there is much more variability
in exam scores of students taking the introductory accounting course as a requirement than for students
taking the same course as part of their accounting major. A random sample of scores of 13 non accounting
majors and a second sample of scores of accounting majors is taken from the recorded scores for one exam.
The results are n1  13, n 2  10, s12  210.2 and s 22  36.5. a) Test the professor’s claim at the   .05
level. b) Interpret the p-value. c) What assumptions are necessary to justify the use of as F test?
Solution: Note that the professor’s conjecture can be written as  1 2   2 2 and must be an alternative
hypothesis.
a) The Instructor’s Solution Manual gives the following hypotheses:
H0:  1 2   2 2 (Where Populations: 1 = nonaccounting students, 2 = accounting students) - The variance
in final exam scores of nonaccounting students is less than or equal to the variance in final exam scores of
accounting students.
H1:  1 2   2 2 - The variance in final exam scores of nonaccounting students is greater than the variance
in final exam scores of accounting students.
We put the variance that should be larger according to the alternative hypothesis on top to get the test ratio
Fcalc 
s1 2
2

210 .2
12,9   3.07 . Since our
= 5.759. This ratio has 12 and 9 degrees of freedom. F.05
36 .5
s2
calculated F-ratio exceeds the table value, we reject the null hypothesis.
12,9   5.11 , we can say that the p-value is above .01. If we check table E.5
b) Since 5.759 is larger than F.01
in the text, we find that the p-value is above .005. The Instructor’s Solution Manual says that using Excel, the
p value = 0.0066.
c) This test assumes that the parent populations are both Normally distributed.
Problem D6: We have the following data for returns on two stocks:
Stock A 7, 8, -5, 9, 11 nA = 5
Stock B 6, 7, 0, 4, 9, 15 nB = 6
a. Find a 95% interval for
 A2
 B2
b. Test the following at a 95% level:
H 0 :  A2   B2
H 1 :  A2   B2
Solution: The formulas for this problem are given in “Confidence limits and Hypothesis Testing for
Variances” in the Syllabus Supplement and summarized on the formula pages.
a) From the data above we can compute s A2  40.00 and s B2  25.36 . The formula given is
s 22
s12
 22 s 22 ( n1 1, n2 1)


F
. If we let Stock A be x 2 , and Stock B be x1 , then we can state that


F n2 1,n1 1  12 s12 2
1
2
DF1  n B  1  5 , DF1  n A  1  4 and
confidence interval formula and get
( 4,5)
F.025
 7.39 , so 1.577 
s A2
s B2
s 22
s12

s A2
s B2

40 .00
 1.577 . Substitute these values in the
25 .36
 A2 s A2 (5, 4)
( 5, 4 )
 9.36 and


F.025 . From the F table, F.025
F4,5   B2 s B2
1
2
 A2
2
1
 2  1.577 9.36  , which becomes 0.213  A2  14 .76 . If we wish an
7.39  B
B
interval for the standard deviations, we can take the square roots to get 0.462 
A
B
 3.842 .
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b) We are testing H 0 :  A2   B2
supplement, test F DF1 , DF2 
s12
s 22
H1 :  A2   B2 or H 0 :
 A2
2
 1 H1 : A2  1 . According to the syllabus
2
B
B
where DF1  n1  1 , DF2  n 2  1 and s12 is the larger of the two
variances according to the alternate hypothesis. Since this is a 1-sided test with   .05 ,
s A2
s B2

40 .00
4,5  5.19 . Since the ratio is smaller than the F, we do not
 1.577 is thus compared to F.05
25 .36
reject H 0 .
10