Chapter 11:
Hypothesis Testing II
11.1 n = 25 paired observations with sample means of 50 and 60 for populations
1 and 2. Can you reject the null hypothesis at an alpha of .05 if
a. sd = 20, H 0 : 1  2  0; H 1 : 1  2  0;
10  0
= 2.500, p-value is between .010 and .005.
20 25
Reject H 0 at alpha of .05
t
Paired T-Test and CI
N
Mean
StDev SE Mean
Difference 25 10.0000 20.0000
4.0000
95% lower bound for mean difference: 3.1565
T-Test of mean difference = 0 (vs > 0): T-Value = 2.50
b. sd = 30,
t
H
0
P-Value = 0.010
: 1  2  0; H 1 : 1  2  0;
10  0
= 1.67, p-value = .054. Do not reject
30 25
H
0
at alpha of .05
Paired T-Test and CI
N
Mean
StDev SE Mean
Difference 25 10.0000 30.0000
6.0000
95% lower bound for mean difference: -0.2653
T-Test of mean difference = 0 (vs > 0): T-Value = 1.67
c. sd = 15,
t
H
0
P-Value = 0.054
: 1  2  0; H 1 : 1  2  0;
10  0
= 3.33, p-value = .001. Reject
15 25
H
0
at alpha of .05
Paired T-Test and CI
N
Mean
StDev SE Mean
Difference 25 10.0000 15.0000
3.0000
95% lower bound for mean difference: 4.8674
T-Test of mean difference = 0 (vs > 0): T-Value = 3.33
d. sd = 40,
t
H
0
P-Value = 0.001
: 1  2  0; H 1 : 1  2  0;
10  0
= 1.25, p-value = .112. Do not reject
40 25
H
0
at alpha of .05
Paired T-Test and CI
N
Mean
StDev SE Mean
Difference 25 10.0000 40.0000
8.0000
95% lower bound for mean difference: -3.6871
T-Test of mean difference = 0 (vs > 0): T-Value = 1.25
P-Value = 0.112
Chapter 11: Hypothesis Testing II
213
11.2 n = 25 paired observations with sample means of 50 and 56 for populations
1 and 2. Can you reject the null hypothesis at an alpha of .05 if
a. sd = 20, H 0 : 1  2  0; H 1 : 1  2  0;
t
(6)  0
= -1.50, p-value = .073. Do not reject
20 25
H
0
at alpha of .05
Paired T-test and CI
N
Mean
StDev SE Mean
Difference 25 -6.00000 20.00000 4.00000
95% upper bound for mean difference: 0.84353
T-Test of mean difference = 0 (vs < 0): T-Value = -1.50
P-Value = 0.073
b. sd = 30, H 0 : 1  2  0; H 1 : 1  2  0;
t
(6)  0
= -1.00, p-value = .164. Do not reject
30 25
H
0
at alpha of .05
Paired T-Test and CI
N
Mean
StDev SE Mean
Difference 25 -6.00000 30.00000 6.00000
95% upper bound for mean difference: 4.26529
T-Test of mean difference = 0 (vs < 0): T-Value = -1.00
c. sd = 15,
t
P-Value = 0.164
H 0 : 1  2  0; H 1 : 1  2  0;
(6)  0
= -2.00, p-value = .028. Reject
15 25
H
0
at alpha of .05
Paired T-Test and CI
N
Mean
StDev SE Mean
Difference 25 -6.00000 15.00000 3.00000
95% upper bound for mean difference: -0.86735
T-Test of mean difference = 0 (vs < 0): T-Value = -2.00
d. sd = 40,
t
P-Value = 0.028
H 0 : 1  2  0; H 1 : 1  2  0;
(6)  0
= -.75, p-value = .230. Do not reject
40 25
H
0
at alpha of .05
Paired T-Test and CI
N
Mean
StDev SE Mean
Difference 25 -6.00000 40.00000 8.00000
95% upper bound for mean difference: 7.68706
T-Test of mean difference = 0 (vs < 0): T-Value = -0.75
11.3
H
0
:  x   y  0; H 1 :  x   y  0;
.0518  0
= 2.04, p-value = .043. Reject
.3055 145
4.3%
t
P-Value = 0.230
H
0
at alpha levels in excess of
214
Statistics for Business & Economics, 6th Edition
Paired T-Test and CI
N
Mean
StDev
SE Mean
Difference 145 0.051800 0.305500 0.025370
95% CI for mean difference: (0.001654, 0.101946)
T-Test of mean difference = 0 (vs not = 0): T-Value = 2.04
11.4
H
0
:  x   y  0; H 1 :  x   y  0;
85.8  71.5
z
(19.13) 2 /151  (12.2) 2 /108
Reject
11.5
H
0
0
at all common levels of alpha
1.91  .21
(1.32) 2 /125  (.53) 2 / 86
Reject
H
H
0
= 7.334.
:  x   y  0; H 1 :  x   y  0;
z
11.6
P-Value = 0.043
H
0
= 12.96.
at all common levels of alpha
:  x   y  0; H 1 :  x   y  0;
2.71  2.79
z
= -1.0207,
(.64) /114  (.56) 2 /123
p-value = 2[1-FZ(1.02)] = 2[1-.8461] = .3078
Therefore, reject H 0 at levels of alpha in excess of 30.78%
11.7
H
2
0
:  x   y  0; H 1 :  x   y  0;
(nx  1) sx  (ny  1) s y
2
sp 
2
t
nx  ny  2
X  Y  D0
sp
2
nx

sp
2
ny
=
2
=
35(22.93) 2  35(27.56) 2
= 642.66925
36  36  2
36.21  47.56
= -1.8995
642.66925 642.66925

36
36
t70 (1.8995)=.0308; p-value = 2(.0308) = .0616.
Reject H 0 at levels in excess of 6.16%
11.8
Assuming both populations are normal with equal variances:
H 0 :  x   y  0; H 1 :  x   y  0;
69(6.14) 2  50(4.29) 2
= 29.592247
70  51  2
3.97  2.86
X  Y  D0
=
= 1.108
t
2
2
29.592247
29.592247
sp
s

 p
70
51
n
n
s2 p 
x
y
Chapter 11: Hypothesis Testing II
Therefore, do not reject
11.9
H
0
t
0
at the 10% alpha level since 1.108 < 1.645 = t(119,.05)
9254  8167
= 1.275
3632605 3632605

10
10
at the 10% alpha level since 1.275 < 1.33 = t(18,.1)
9(2107) 2  9(1681) 2
= 3,632,605, t 
10  10  2
Therefore, do not reject
H
0
:  x   y  0; H 1 :  x   y  0;
s2 p 
11.10
H
215
H
0
:  x   y  0; H 1 :  x   y  0;
1475  0
= 2.239, p-value = .0301. Reject
1862.985 8
H
0
at levels in excess of 3%
Paired T-Test and CI: Male, Female
Paired T for Male - Female
N
Mean
StDev SE Mean
Male
8 46437.5
2680.1
947.5
Female
8 44962.5
2968.4
1049.5
Difference 8 1475.00 1862.99
658.66
95% lower bound for mean difference: 227.11
T-Test of mean difference = 0 (vs > 0): T-Value = 2.24
11.11
H
t
11.12 a.
0
:  x   y  0; H 1 :  x   y  0; reject
4.5  0
= 2.665. Reject
4.1352 6
H
0
H
0
H
0
P-Value = 0.030
if t(5,.05) > 2.015
at the 5% level
: Px  Py  0; H 1 : Px  Py  0;
.42  .50
(.4636)(1  .4636) (.4636)(1  .4636)

500
600
= -2.65 p-value = .004. Therefore, reject H 0 at all common levels of alpha
b.
pˆ o 
500(.42)  600(.50)
= .4636, z 
500  600
H
: Px  Py  0; H 1 : Px  Py  0;
0
500(.60)  600(.64)
= .6218,
500  600
.60  .64
z
= -1.36
(.6218)(1  .6218) (.6218)(1  .6218)

500
600
p-value = .0869. Therefore, reject H 0 at .10, but do not reject at the .05 level
pˆ o 
216
c.
Statistics for Business & Economics, 6th Edition
H
0
: Px  Py  0; H 1 : Px  Py  0;
500(.42)  600(.49)
= .4582,
500  600
.42  .49
z
= -2.32
(.4582)(1  .4582) (.4582)(1  .4582)

500
600
p-value = .0102.
Therefore, reject H 0 at the .05 level, but do not reject at the .01 level
pˆ o 
d.
H
0
: Px  Py  0; H 1 : Px  Py  0;
.25  .34
= -3.25
(.299)(1  .299) (.299)(1  .299)

500
600
p-value = .0006. Therefore, reject H 0 at all common levels of alpha
pˆ o 
e.
500(.25)  600(.34)
= .299, z 
500  600
H
0
: Px  Py  0; H 1 : Px  Py  0;
500(.39)  600(.42)
= .4064,
500  600
.39  .42
z
= -1.01
(.4064)(1  .4064) (.4064)(1  .4064)

500
600
p-value = .1562. Therefore, do not reject H 0 at any common level of alpha
pˆ o 
11.13
H
0
: Px  Py  0; H 1 : Px  Py  0;
900(.60)  900(.66)
= .63,
900  900
.60  .66
z
= -2.63
(.63)(1  .63) (.63)(1  .63)

900
900
p-value = .0043. Therefore, reject H 0 at all common levels of alpha
pˆ o 
11.14
H
0
pˆ o 
: Px  Py  0; H 1 : Px  Py  0;
.384  .52
= -6.97
(.44)(.56) (.44)(.56)

1556
1108
at all common levels of alpha
1556(.384)  1108(.52)
= .44, z 
1556  1108
Reject
H
0
Chapter 11: Hypothesis Testing II
11.15
H
0
: Px  Py  0; H 1 : Px  Py  0; reject
H
0
217
if |z.025| > 1.96
368(.25)  116(.319)
= .266
368  116
.25  .319
z
= -1.466. Do not reject
(.266)(.734) (.266)(.734)

368
116
pˆ o 
11.16
H
0
: Px  Py  0; H 1 : Px  Py  0; reject
H
0
H
0
at the 5% level
if |z.025| > 1.96
78  208
= .36714
175  604
.446  .344
z
= 2.465. Reject
(.36714)(.63286) (.36714)(.63286)

175
604
pˆ o 
11.17
H
0
H
0
at the 5% level
: Px  Py  0; H 1 : Px  Py  0;
191  145
= .614
381  166
.501  .873
z
= -8.216.
(.614)(.386) (.614)(.386)

381
166
Reject H 0 at all common levels of alpha
pˆ o 
11.18
H
0
: Px  Py  0; H 1 : Px  Py  0; reject
H
0
if |z.05| > 1.645
: Px  Py  0; H 1 : Px  Py  0; reject
H
0
if z.01 < -2.33
138  128
= .554
240  240
.575  .533
z
= .926.
(.554)(.446) (.554)(.446)

240
240
Do not reject H 0 at the 5% level
pˆ o 
11.19
H
0
480  790
= .577
1200  1000
.4  .79
z
= -18.44. Reject
(.577)(.423) (.577)(.423)

1200
1000
pˆ o 
H
0
at the 1% level
218
Statistics for Business & Economics, 6th Edition
11.20 a.
2
2
2
H 0 :   100; H 1 :   100;  

2
(24,.025)
 39.36, 
Therefore, reject
2
(24,.010)
H
(n  1) s 2


2
24(165)
= 39.6,
100
 42.98
at the 2.5% level but not
0
at the 1% level of significance.
2
2
2
H 0 :   100; H 1 :   100;  
b.

2
(28,.025)
 44.46, 
Therefore, reject
2
(28,.010)
H

2
(24,.050)
 36.42, 
Therefore, reject
(24,.025)
0

2
(30,.100)
 40.26, 
H
0
2
(40,.100)
 
(n  1) s 2

2


(n  1) s 2

H
0
11.23
2
H
0
24(159)
= 38.16,
100

37(67)
= 24.79,
100
if  2(7,.10) > 12.02
7(933.982)
= 13.0757, Therefore, reject
500
9(5.1556)
= 20.6224. Reject
2.25
H
0
H
0
H
0
at the 10% level
if  2(9,.05) > 16.92
at the 5% level
:  2  300; H 1 :  2  300;
29(480)
2 
= 46.4, p-value = .0214. Reject
300
H

at any common level of significance.
11.22 a. s2 = 5.1556
b. H 0 :  2  2.25; H 1 :  2  2.25; reject
2 
2
 51.81
:  2  500; H 1 :  2  500; reject
2
(n  1) s 2
at the 5% level but not at the 2.5% level of significance.
Therefore, do not reject
11.21
28(165)
= 46.2,
100
 39.36
2
2
2
H 0 :   100; H 1 :   100;  
d.

at the 2.5% level but not at the 1% level of significance.
0
2
H

2
 48.28
2
2
2
H 0 :   100; H 1 :   100;  
c.
(n  1) s 2
0
H
0
at the 5% level
11.24 The hypothesis test assumes that the population values are normally distributed
2
H 0 :   2.0; H 1 :   2.0; reject H 0 if  (19,.05) > 30.14
2 
19(2.36)2
= 26.4556. Do not reject
(2)2
H
0
at the 5% level
Chapter 11: Hypothesis Testing II
11.25
H
0
219
:   18.2; H 1 :   18.2;
24(15.3)2
= 16.961.
 
(18.2)2
Do not reject H 0 at the 10% level since  2 >15.66 =  2(24,.10)
2
11.26 a.
H
0
:  2x   2 y ; H 1 :  2x   2 y
H
F = 125/51 = 2.451. Reject
H
b.
0
:
2
x
  y ; H 1 :
2
2
F = 235/125 = 1.88. Reject
H
c.
0
H
0

H
0
H
0
2
y
at the 5% level since 1.88 > 1.69  F(43,44,.05)
:  2x   2 y ; H 1 :  2x   2 y
F = 134/51 = 2.627. Reject
d.
x
at the 1% level since 2.451 > 2.11  F(44,40,.01)
0
at the 1% level since 2.627 > 2.11  F(47,40,.01)
:  2x   2 y ; H 1 :  2x   2 y
F = 167/88 = 1.90. Reject
H
0
at the 5% level since 1.90 > 1.79  F(24,38,.05)
:  2x   2 y ; H 1 :  2x   2 y
F = 1614.208/451.770 = 3.573.
Reject H 0 at the 1% level since 3.573 > 2.41F(29,29,.01)
11.27
H
11.28
H
0
0
:  2 x   2 y ; H 1 :  2 x   2 y ; reject
F = 114.09/16.08 = 7.095. Reject
H
0
H
0
if F(3,6,.05) > 4.76
at the 5% level
11.29
:  2x   2 y ; H 1 :  2x   2 y ;
F=(27.56)2/(22.93)2=1.44.
Do not reject H 0 at the 10% level since 1.44<1.84F(35,35,.05)
11.30
:  2x   2 y ; H 1 :  2x   2 y ;
F = (2107)2/(1681)2 = 1.57
Therefore, do not reject H 0 at the 10% level since 1.57 < 3.18  F(9,9,.05)
11.31
:  2x   2 y ; H 1 :  2x   2 y ;
F = (24.4)2/(20.2)2 = 1.46.
Do not reject H 0 at the 5% level since 1.46 < 9.28  F(3,3,.05)
H
H
H
0
0
0
11.32 No. The probability of rejecting the null hypothesis given that it is true is 5%.
220
Statistics for Business & Economics, 6th Edition
H :
11.33 a.
0
 1.6; H 1 :   1.6; reject
H
0
if |z.05|> 1.645
1.615  1.6
= 1.20, p-value =2[1-FZ(1.2)]= 2302.
.05 16
Do not reject H 0 at the 10% level
z
H
b.
0
:   .05; H 1 :   .05; reject
15(.086)2
= 44.376. Reject
 
(.05)2
2
H
H
if  2(15,.10)  22.31
0
at the 10% level
0
11.34 a. Assume that the population is normally distributed
One-Sample T: Grams:
Test of mu = 5 vs mu not = 5
Variable
N
Mean
Grams:11-34
12
4.9725
Variable
Grams:11-34
x  4.9725; s  .0936 ,
t
(
StDev
0.0936
95.0% CI
4.9130, 5.0320)
H :
0
SE Mean
0.0270
T
-1.02
P
0.331
 5; H 1 :   5; reject
4.9725  5
= -1.018. Do not reject
.0936 12
H
0
H
0
if |t(11, .025| > 2.201
at the 5% level
b. Assume that the population is normally distributed
2
H 0 :   .025; H 1 :   .025; reject H 0 if  (11,.05)  19.68
2 
11(.0936)2
= 154.19. Therefore, reject
(.025)2
H
0
at the 5% level
11.35 a. x  333/ 9  37; s  312 8 = 6.245
H :
 40; H 1 :   40; reject
H
0
t
37  40
= -1.44. Do not reject
6.245 9
H
0
b.
H
0
0
:   6; H 1 :   6; reject
H
8(6.245)2
= 8.67. Do not reject
 
(6)2
2
11.36
H
0
at the 5% level
if  2(8,.10)  13.36
H
0
at the 10% level
:  x   y  0; H 1 :  x   y  0;
(nx  1) sx  (ny  1) s y
2
sp 
2
0
if t8,.05 < -1.86
nx  ny  2
2
=
33(2.21) 2  85(1.69) 2
= 3.4525
34  85  2
Chapter 11: Hypothesis Testing II
X  Y  D0
t
sp
2
nx

sp
=
2
ny
2.21  1.47
= 1.966
3.4525 3.4525

34
86
p-value is between (.025, .010) x 2 = .05 and .02.
Reject H 0 at levels in excess of 5%
H :
11.37 a.
 4; H 1 :   4; reject
0
H
0
if t.05 > 1.671
4.4  4
= 2.574. Reject H 0 at the 5% level
1.3 70
b. H 0 :  x   y  0; H 1 :  x   y  0; reject H 0 if t .05 < -1.645
t
(nx  1) sx  (ny  1) s y
2
sp 
2
t
X  Y  D0
2
nx
Reject
H
0

H
sp
=
2
ny
0
69(1.3)2  105(1.4)2
= 1.853
70  106  2
4.4  5.3
= -4.293.
1.853 1.853

70
106
at levels in excess of 5%
:  x   y  0; H 1 :  x   y  0; reject
(nx  1) sx  (ny  1) s y
2
sp 
2
t
=
nx  ny  2
sp
11.38
2
2
nx  ny  2
X  Y  D0
sp
2
nx

sp
=
2
ny
H
Do not reject
0
H
0
if |t .05| > 1.645
43(18.20) 2  67(18.94) 2
=
= 347.980
44  68  2
35.02  36.34
=-.210.
347.98 347.98

44
68
at levels in excess of 5%
11.39 Presuming the populations are normally distributed with equal variances, the
samples must be independent random samples:
H 0 :  x   y  0; H 1 :  x   y  0; reject H 0 if t(6,.01) < -3.143
(nx  1) sx  (ny  1) s y
2
sp 
2
t
nx  ny  2
X  Y  D0
sp
2
nx
Reject

H
sp
0
2
ny
=
2
=
3(24.4) 2  3(14.6) 2
= 106.58
442
78  114.7
= -5.027.
106.58 106.58

4
4
at levels in excess of 1%
221
222
Statistics for Business & Economics, 6th Edition
11.40 Assuming the populations are normally distributed with equal variances and
independent random samples:
Magazine A: X  10.968; sx  2.647 , Magazine B: Y  6.738; sy  1.636
H
0
:  x   y  0; H 1 :  x   y  0; reject
(nx  1) sx  (ny  1) s y
2
sp 
2
t
2
nx  ny  2
X  Y  D0
sp
2
nx
Reject

H
sp
=
2
ny
0
H
0
if t(10,.05) > 1.812
5(2.647) 2  5(1.636) 2
=
= 4.8416
662
10.968  6.738
= 3.330.
4.8416 4.8416

6
6
at levels in excess of 5%
11.41 Assume that the populations are normally distributed with equal variances and
independent random samples:
Magazine A: X  7.045; sx  2.1819 , Magazine B: Y  6.777; s y  2.85
H
0
:  x   y  0; H 1 :  x   y  0;
(nx  1) sx  (ny  1) s y
2
sp 
2
t
2
=
nx  ny  2
X  Y  D0
sp
2
nx

sp
=
2
ny
5(2.1819)2  5(2.85)2
= 6.4416
662
7.045  6.777
= .183. Do not reject
6.4416 6.4416

6
6
H
0
at any common
level of alpha
11.42
H
0
:  x   y  0; H 1 :  x   y  0; Sample sizes greater than 100, use the z-test.
2.83  3.0
z
= -2.30, p-value = 1 – FZ(2.3) = 1 - .9893 = .0107
(.89)2 (.67) 2

202
291
Therefore, reject H 0 at levels of alpha in excess of 1.07%
11.43
H
0
:  x   y  0; H 1 :  x   y  0; . Sample sizes less than 100, use the t-test
(nx  1) sx  (ny  1) s y
2
sp 
2
t
2
nx  ny  2
X  Y  D0
sp
2
nx

sp
2
ny
=
=
82(.649)2  53(.425) 2
= .32675
83  54  2
6.543  6.733
= -1.901. p-value is between (.05 and .025)
.32675 .32675

83
54
x 2 = .10 and .05. Reject
H
0
at any alpha of .10 or higher.
Chapter 11: Hypothesis Testing II
H
11.44 a.
0
: P  .5; H 1 : P  .5; reject
H
0
223
if z.05 < -1.645
.455  .5
= -1.2. Do not reject H 0 at the 5% level
(.5)(.5) /178
b. H 0 : Px  Py  0; H 1 : Px  Py  0; reject Ho if |z.025| > 1.96
z
.5068  .455
= .932
1
1
(.478)(.522)(

)
148 178
Therefore, do not reject H 0 at the 5% level
75  81
= .478, z 
148  178
pˆ o 
11.45
H
0
:  x   y  0; H 1 :  x   y  0; reject Ho if t(44,.05) < -1.684
(nx  1) sx  (ny  1) s y
2
sp 
2
t
nx  ny  2
X  Y  D0
sp
2
nx
Reject
11.46
H
0
pˆ o 

H
sp
0
H
0
2
=
ny
=
22(.055)2  22(.058)2
= .00319
23  23  2
.058  .146
= -5.284.
.00319 .00319

23
23
at any common level of alpha
: Px  Py  0; H 1 : Px  Py  0; reject Ho if |z.01| < -2.33
.164  .239
= -1.19.
1
1
(.211)(.789)( 
)
67 113
at the 1% level
11  27
=.211, z 
67  113
Do not reject
11.47
2
H
0
: Px  Py  0; H 1 : Px  Py  0;
pˆ o 
47  40
= .630435, z 
69  69
H
: Px  Py  0; H 1 : Px  Py  0;
.6812  .5797
1
1
(.630435)(.369565)(  )
69 69
2[1-FZ(1.24)] = 2[1-.8925] = .1075.
Reject H 0 at levels of alpha in excess of 10.75%
11.48
0
.564  .691
= -1.653,
1
1
(.617)(.383)(  )
94 68
p-value = 1–FZ(1.65)]=.0495
Therefore, reject H 0 at levels of alpha in excess of 4.95%
pˆ o 
53  47
= .617, z 
94  68
= 1.235, p-value =
224
11.49
Statistics for Business & Economics, 6th Edition
H 0 :  x   y ; H 1 :  x   y ; sx  2.64665, s y  1.63561, F 
(2.647)2/(1.63656)2 = 2.618. Do not reject
H
0
s2x
=
s2 y
at the 5% level, 2.618 < 5.05 
F(5,5,.05)
11.50
H 0 :  x   y ; H 1 :  x   y ; sx  4.16314, s y  4.05421 , F 
(4.16314)2/(4.05421)2 = 1.0545. Do not reject
H
0
s2x
=
s2 y
at the 5% level, 1.0545 < 2.98
 F(10,10,.05). There is insufficient evidence to suggest that the population
variances differ between the two forecasting analysts.
11.51 a.
H
0
:  x   y  0; H 1 :  x   y  0;
df = n1 + n2 – 2 = 27 + 27 – 2 = 52; t52,.05 = 1.675
2
2
(n  1) s1  (n2  1) s2
(27  1)100  (27  1)150
s2 p  1

 125
n1  n2  2
52
x2  x1
64  60
tcalc 

 1.99
2
2
125 125
s p s p


60 64
n1
n2
At the .05 level of significance, reject Ho and accept the alternative that the mean
output per hectare is significantly greater with the new procedure.
b. 95% acceptance interval:
2
150
1
s
F26,26,.025  2.20 , P(
 1.50 , because F
 22  2.20)  .95 , Fcalc 
100
2.20 s1
calc is within the acceptance interval, there is not sufficient evidence against
the null hypothesis that the sample variances are not significantly different
from each other.
11.52 a.
: P2  P1  0; H 1 : P2  P1  0; reject
0
pˆ o 
258  260
= .3453, z 
800  700
H
H
: P2  P1  0; H 1 : P2  P1  0; reject
H
0
H
0
if |z.015| > 2.17
.3714  .3225
= 1.987
(.3453)(.6547) (.3453)(.6547)

800
700
Therefore, reject H 0 at the 5% level, but do not reject at the 3% level
b.
0
if |z.03| > 1.88
Chapter 11: Hypothesis Testing II
225
.3714  .3225
= 1.987
(.3453)(.6547) (.3453)(.6547)

800
700
Therefore, reject H 0 at the 3% level
pˆ o 
258  260
= .3453, z 
800  700
11.53 Assume that the population of matched differences are normally distributed
H 0 :  x   y  0; H 1 :  x   y  0; reject H 0 if |t(9,.05)| > 1.833
x of the matched differences = 1.13, s of the matched differences = 1.612
1.13  0
= 2.22, p-value =.054.
t
1.612 10
Reject H 0 at the 10%, but not the 5% level
Paired T-Test and CI: VARIETY A, VARIETY B
Paired T for VARIETY A - VARIETY B
N
Mean
StDev SE Mean
VARIETY A
10 11.9300
2.9265
0.9254
VARIETY B
10 10.8000
2.5237
0.7981
Difference 10 1.13000 1.61180 0.50969
95% CI for mean difference: (-0.02301, 2.28301)
T-Test of mean difference = 0 (vs not = 0): T-Value = 2.22
P-Value = 0.054
a. The box plots of the raw data show similar medians and interquartile ranges
for both brands. However, brand 2 is dominated by three outliers that is
skewing the brand 2 data to the right:
1000
900
800
700
saleb2
11.54
600
500
400
300
200
100
0
saleb2
saleb4
The descriptive statistics shows the effect of the extreme outliers on brand 2 sales
– note the sizeable standard deviation of brand 2:
226
Statistics for Business & Economics, 6th Edition
Descriptive Statistics: saleb2, saleb4
Variable
saleb2
saleb4
N
52
52
Mean
181.2
140.29
Median
127.0
125.50
TrMean
155.7
136.80
Variable
saleb2
saleb4
Minimum
59.0
55.00
Maximum
971.0
305.00
Q1
94.8
101.25
Q3
203.3
182.75
StDev
154.9
60.84
SE Mean
21.5
8.44
The matched pairs t-test on the original data shows a significant difference between the
weekly sales with brand 2 found to be significantly larger than brand 4 at the .05 level:
Paired T-Test and CI: saleb2, saleb4
Paired T for saleb2 - saleb4
Variable
N
Mean
StDev
SE Mean
saleb2
52
181.2
154.9
21.5
saleb4
52
140.3
60.8
8.4
Difference
52
40.9
169.5
23.5
95% lower bound for mean difference: 1.5
T-Test of mean difference = 0 (vs > 0): T-Value = 1.74
P-Value = 0.044
b. However, with only the largest outlier removed from the data of brand 2, the
difference between the two brands becomes insignificant at the .05 level:
Paired T-Test and CI: saleb2_1, saleb4 (with outlier removed)
Paired T for saleb2_1 - saleb4
N
Mean
StDev
SE Mean
saleb2_1
51
165.7
108.5
15.2
saleb4
51
140.8
61.3
8.6
Difference
51
24.9
125.7
17.6
95% lower bound for mean difference: -4.6
T-Test of mean difference = 0 (vs > 0): T-Value = 1.42
11.55 a.
H
0
P-Value = 0.081
:  x   y  0; H 1 :  x   y  0;
Results for: Ole.MTW
Two-Sample T-Test and CI: Olesales, Carlsale
Two-sample T for Olesales vs Carlsale
N
Mean
StDev
SE Mean
Olesales 156
3791
5364
429
Carlsale 156
2412
4249
340
Difference = mu Olesales - mu Carlsale
Estimate for difference: 1379
95% lower bound for difference: 475
T-Test of difference = 0 (vs >): T-Value = 2.52
310
Both use Pooled StDev = 4839
Reject Ho at the .01 level of significance
P-Value = 0.006
DF =
Chapter 11: Hypothesis Testing II
b.
H
0
:  x   y  0; H 1 :  x   y  0;
Two-Sample T-Test and CI: Oleprice, Carlpric
Two-sample T for Oleprice vs Carlpric
N
Mean
StDev
SE Mean
Oleprice 156
0.819
0.139
0.011
Carlpric 156
0.819
0.120
0.0096
Difference = mu Oleprice - mu Carlpric
Estimate for difference: -0.0007
95% CI for difference: (-0.0297, 0.0283)
T-Test of difference = 0 (vs not =): T-Value = -0.05
0.962 DF = 310
Both use Pooled StDev = 0.130
P-Value =
Do not reject Ho at any common level of significance. Note that the 95%
confidence interval contains 0, therefore, no evidence of a difference.
11.56 Flour A:  A  8,  2 A  .04 . Flour B:  B  8,  2 B  .06
Mix X M  X A  X B , M  8  8  16
 2 M  .04  .06  2(.40) .04 .06  .1392
 2 M .1392
 2 xM 

 (.186)2  .0346,  xM  .186
4
4
z.005  2.575 , Acceptance interval: 16  2.575(.186) = 16  .48
The control limits will be at 16.48 and 15.52
227