1 / 50100%
Chapter 13
13.1a Equal-variances estimator
+± α21
2
p2/21 n
1
n
1
st)xx(
= (524 – 469)
±
2.009
+
+ +
25
1
25
1
22525
131)125(129)125( 22
= 55
±
73.87
b Equal-variances estimator
+± α21
2
p2/21 n
1
n
1
st)xx(
= (524 – 469)
±
2.009
+
+ +
25
1
25
1
22525
260)125(255)125( 22
= 55
±
146.33
c The interval widens.
d Equal-variances estimator
+± α21
2
p2/21 n
1
n
1
st)xx(
= (524 – 469)
±
1.972
+
+ +
100
1
100
1
2100100
131)1100(129)1100( 22
= 55
±
36.26
e The interval narrows.
13.2
)(:H 210 µµ
= 0
)(:H 211 µµ
0
a Equal-variances test statistic
Rejection region:
= 22,025.
t
–2.074 or
=>
να ,2/
tt
=
22,025.
t
2.074
+
µµ
=
21
2
p
2121
n
1
n
1
s
)()xx(
t
=
+
+ +
12
1
12
1
21212
16)112(18)112(
0)7174(
22
= .43, p-value = .6703. There is not enough
evidence to infer that the population means differ.
b Equal-variances test statistic
Rejection region:
=<
να
,2/
tt
= 22,025.
t
–2.074 or
=>
να ,2/
tt
=
22,025.
t
2.074
+
µµ
=
21
2
p
2121
n
1
n
1
s
)()xx(
t
=
+
+ +
12
1
12
1
21212
198)112(210)112(
0)7174(
22
= .04, p-value = .9716. There is not enough
evidence to infer that the population means differ.
c The value of the test statistic decreases and the p-value increases.
25
d Equal-variances test statistic
Rejection region:
= 298,025.
t
–1.960 or
=>
να ,2/
tt
=
298,025.
t
1.960
+
µµ
=
21
2
p
2121
n
1
n
1
s
)()xx(
t
=
+
+ +
150
1
150
1
2150150
16)1150(18)1150(
0)7174(
22
= 1.53, p-value = .1282. There is not
enough evidence to infer that the population means differ.
e The value of the test statistic increases and the p-value decreases.
f Rejection region:
=<
να
,2/
tt
= 22,025.
t
–2.074 or
=> να ,2/
tt
=
22,025.
t
2.074
+
µµ
=
21
2
p
2121
n
1
n
1
s
)()xx(
t
=
+
+ +
12
1
12
1
21212
16)112(18)112(
0)7176(
22
= .72, p-value = .4796. There is not enough
evidence to infer that the population means differ.
g The value of the test statistic increases and the p-value decreases.
13.3 a Unequal-variances estimator
1n
)n/s(
1n
)n/s(
)n/sn/s(
2
2
2
2
2
1
2
1
2
1
2
2
2
21
2
1
+
+
=ν
= 64.8 (rounded to 65, approximated by
70=ν
)
+± α2
2
2
1
2
1
2/21 n
s
n
s
t)xx(
= (63 – 60)
±
1.667
+45
7
50
18 22
= 3
±
4.59
b Unequal-variances estimator
1n
)n/s(
1n
)n/s(
)n/sn/s(
2
2
2
2
2
1
2
1
2
1
2
2
2
21
2
1
+
+
=ν
= 63.1 (rounded to 63, approximated by
60=ν
)
+± α2
2
2
1
2
1
2/21 n
s
n
s
t)xx(
= (63 – 60)
±
1.671
+45
15
50
41 22
= 3
±
10.38
c The interval widens.
d Unequal-variances estimator
1n
)n/s(
1n
)n/s(
)n/sn/s(
2
2
2
2
2
1
2
1
2
1
2
2
2
21
2
1
+
+
=ν
= 131 (approximated by
140=ν
)
+± α2
2
2
1
2
1
2/21 n
s
n
s
t)xx(
= (63 – 60)
±
1.656
+90
7
100
18 22
= 3
±
3.22
26
e The interval narrows.
13.4
)(:H 210 µµ
= 0
)(:H 211 µµ
>
0
a Unequal-variances test statistic
1n
)n/s(
1n
)n/s(
)n/sn/s(
2
2
2
2
2
1
2
1
2
1
2
2
2
21
2
1
+
+
=ν
= 200.4 (rounded to 200)
Rejection region:
653.1ttt 200,05., ==> να
+
µµ
=
2
2
2
1
2
1
2121
n
s
n
s
)()xx(
t
=
+
150
54
150
128
0)405412(
22
= .62, p-value = .2689. There is not enough evidence to infer that
1
µ
is greater than
2
µ
.
b Unequal-variances test statistic
1n
)n/s(
1n
)n/s(
)n/sn/s(
2
2
2
2
2
1
2
1
2
1
2
2
2
21
2
1
+
+
=ν
= 223.1 (rounded to 223)
Rejection region:
645.1ttt 223,05., => να
+
µµ
=
2
2
2
1
2
1
2121
n
s
n
s
)()xx(
t
=
+
150
16
150
31
0)405412(
22
= 2.46, p-value = .0074. There is enough evidence to infer that
1
µ
is
greater than
2
µ
.
c The value of the test statistic increases and the p-value decreases.
d Unequal-variances test statistic
1n
)n/s(
1n
)n/s(
)n/sn/s(
2
2
2
2
2
1
2
1
2
1
2
2
2
21
2
1
+
+
=ν
= 25.6 (rounded to 26)
Rejection region:
706.1ttt
26,05.,
==>
να
+
µµ
=
2
2
2
1
2
1
2121
n
s
n
s
)()xx(
t
=
+
20
54
20
128
0)405412(
22
= .23, p-value = .4118. There is not enough evidence to infer that
1
µ
is greater than
2
µ
.
e The value of the test statistic decreases and the p-value increases.
27
f Unequal-variances test statistic
Rejection region:
653.1ttt 200,05., ==> να
+
µµ
=
2
2
2
1
2
1
2121
n
s
n
s
)()xx(
t
=
+
150
54
150
128
0)405409(
22
= .35, p-value = .3624. There is not enough evidence to infer that
1
µ
is greater than
2
µ
.
g The value of the test statistic decreases and the p-value increases.
13.5 a Equal-variances t-test degrees of freedom = 28, Unequal-variances t-test degrees of freedom =26.4
b Equal-variances t-test degrees of freedom = 24, Unequal-variances t-test degrees of freedom = 10.7
c Equal-variances t-test degrees of freedom = 98 , Unequal-variances t-test degrees of freedom = 91.2
d Equal-variances t-test degrees of freedom = 103, Unequal-variances t-test degrees of freedom = 78.5
13.6 a In all cases the equal-variances t-test degrees of freedom is greater than the unequal-variances t-test degrees
of freedom.
13.7
)(:H 210 µµ
= 0
)(:H 211 µµ
< 0
Two-tail F test: F = 1.02, p-value = .9856; use equal-variances test statistic
Rejection region:
=<
να
,
tt
372.1t 10,10. =
+
µµ
=
21
2
p
2121
n
1
n
1
s
)()xx(
t
=
,43.
6
1
6
1
266
4.6653)16(5.6767)16(
0)83.38150.361( =
+
+ +
p-value = .3382. The manager should
choose to use cameras.
13.8
)(:H 210 µµ
= 0
)(:H 211 µµ
< 0
Two-tail F test: F = 1.02, p-value = .9823; use equal-variances test statistic
Rejection region:
=<
να
,
tt
330.1t 18,10. =
+
µµ
=
21
2
p
2121
n
1
n
1
s
)()xx(
t
=
,04.2
10
1
10
1
21010
79.5)110(88.5)110(
0)30.710.5( =
+
+ +
p-value = .0283. There is enough
evidence to infer that there are fewer errors when the yellow ball is used.
28
13.9
)(:H 210 µµ
= 0
)(:H 211 µµ
0
Two-tail F test: F = .11, p-value = .0106; use unequal-variances test statistic
1n
)n/s(
1n
)n/s(
)n/sn/s(
2
2
2
2
2
1
2
1
2
1
2
2
2
21
2
1
+
+
=ν
= 8.59 (rounded to 9)
Rejection region:
833.1t 9,05. =
or
=>
να ,2/
tt
833.1t 9,05. =
+
µµ
=
2
2
2
1
2
1
2121
n
s
n
s
)()xx(
t
=
,24.
8
71.367
8
21.42
0)50.1625.18( =
+
p-value = .8123. There is not enough evidence to infer that
the two groups differ in the number of pictures printed.
13.10
)(:H 210 µµ
= 0
)(:H 211 µµ
0
Two-tail F test: F = 1.04, p-value = .9873; use equal-variances test statistic
Rejection region:
771.1t 13,05. =
or
=>
να ,2/
tt
771.1t 13,05. =
+
µµ
=
21
2
p
2121
n
1
n
1
s
)()xx(
t
=
,59.1
6
1
9
1
269
778,725)16(196,755)19(
0)093,4372,3( =
+
+ +
p-value = .1368. There is not
enough evidence to infer a difference between the two types of vacation expenses.
13.11a
)(:H 210 µµ
= 0
)(:H 211 µµ
> 0
Two-tail F test: F = 1.59, p-value = .3050; use equal-variances test statistic
Rejection region:
684.1ttt
38,05.,
=>
να
+
µµ
=
21
2
p
2121
n
1
n
1
s
)()xx(
t
=
+
+ +
25
1
15
1
22515
35.3)125(23.4)115(
0)36.3193.36(
22
= 4.61, p-value = 0. There is enough
evidence to infer that Tastee is superior.
b
+± α21
2
p2/21 n
1
n
1
st)xx(
= (36.93 – 31.36)
±
2.021
+
+ +
25
1
15
1
22515
35.3)125(23.4)115( 22
= 5.57
±
2.43; LCL = 3.14, UCL = 8.00
c The histograms are somewhat bell shaped. The weight gains may be normally distributed.
29
13.12
)(:H 210 µµ
= 0
µµ )(:H 211
0
Two-tail F test: F = .99, p-value = .9571; use equal-variances test statistic
Rejection region:
960.1t 238,025. =
or
=>
να ,2/
tt
960.1t 238,025. =
+
µµ
=
21
2
p
2121
n
1
n
1
s
)()xx(
t
=
,55.1
120
1
120
1
2120120
45.4)1120(43.4)1120(
0)12.901.10(
22 =
+
+ +
p-value = .1204. There is not
enough evidence to infer that oat bran is different from other cereals in terms of cholesterol reduction?
13.13 a
)(:H 210 µµ
= 0
µµ )(:H 211
0
Two-tail F test: F = .50, p-value = 0; use unequal-variances test statistic
1n
)n/s(
1n
)n/s(
)n/sn/s(
2
2
2
2
2
1
2
1
2
1
2
2
2
21
2
1
+
+
=ν
= 449
Rejection region:
960.1ttt
449,025.,2/
=<
να
or
960.1ttt
449,025.,2/
=>
να
+
µµ
=
2
2
2
1
2
1
2121
n
s
n
s
)()xx(
t
=
+
250
32.43
250
77.30
0)96.5299.58(
22
= 1.79, p-value = .0734. There is not enough evidence to
conclude that a difference in mean listening times exist between the two populations.
b
+± α2
2
2
1
2
1
2/21 n
s
n
s
t)xx(
= (58.99 –52.96)
±
1.960
+250
32.43
250
77.30 22
= 6.03
±
6.59;
LCL = –.56, UCL = 12.62
c The histograms are bell shaped.
13.14 a
)(:H 210 µµ
= 0
)(:H 211 µµ
> 0
Two-tail F test: F = 1.01, p-value = .9619; use equal-variances test statistic
Rejection region:
=>
να
,
tt
645.1t 282,05.
+
µµ
=
21
2
p
2121
n
1
n
1
s
)()xx(
t
=
,88.2
159
1
125
1
2159125
99.6)1159(02.7)1125(
0)40.5781.59(
22 =
+
+ +
p-value = .0021. There is
enough evidence to infer that the cruise ships are attracting younger customers.
30
b
+± α21
2
p2/21 n
1
n
1
st)xx(
= (59.81 – 57.40)
±
2.576
+
+ +
159
1
125
1
2159125
99.6)1159(02.7)1125( 22
= 2.41
±
2.16;
LCL = .25, UCL = 4.57
13.15a
)(:H 210 µµ
= 0
µµ )(:H 211
0
Two-tail F test: F = .98, p-value = .9254; use equal-variances test statistic
Rejection region:
972.1ttt 198,025.,2/ =< να
or
972.1ttt 198,025.,2/ => να
+
µµ
=
21
2
p
2121
n
1
n
1
s
)()xx(
t
=
+
+ +
100
1
100
1
2100100
90.2)1100(87.2)1100(
0)66.923.10(
22
= 1.40, p-value = .1640. There is not
enough evidence to infer that the distance males and females drive differs.
b
+± α21
2
p2/21 n
1
n
1
st)xx(
= (10.23 – 9.66)
±
1.972
+
+ +
100
1
100
1
2100100
90.2)1100(87.2)1100( 22
= .57
±
.80; LCL = –.23, UCL = 1.37
c The histograms are bell shaped.
13.16
)(:H 210 µµ
= 0
>µµ )(:H 211
0
Two-tail F test: F = .98, p-value = .9520; use equal-variances test statistic
Rejection region:
671.1ttt 58,05., => να
+
µµ
=
21
2
p
2121
n
1
n
1
s
)()xx(
t
=
,94.
30
1
30
1
23030
93.21)130(69.21)130(
0)20.11050.115(
22 =
+
+ +
p-value = .1753. There is not
enough evidence to retain supplier A - switch to supplier B.
13.17
)(:H 210 µµ
= 0
µµ )(:H 211
0
Two-tail F test: F = .92, p-value = .5000; use equal-variances test statistic
Rejection region:
960.1ttt 594,025.,2/ =< να
or
960.1ttt 594,025.,2/ => να
31
+
µµ
=
21
2
p
2121
n
1
n
1
s
)()xx(
t
=
,16.
290
1
306
1
2290306
58.5)1290(36.5)1306(
0)49.556.5(
22 =
+
+ +
p-value = .8759. There is no
evidence of a difference in job tenures between men and women.
13.18a
)(:H 210 µµ
= 0
µµ )(:H 211
0
Two-tail F test: F = 5.18, p-value = .0019; use unequal-variances test statistic
1n
)n/s(
1n
)n/s(
)n/sn/s(
2
2
2
2
2
1
2
1
2
1
2
2
2
21
2
1
+
+
=ν
= 33.9 (rounded to 34)
Rejection region:
724.2ttt 34,005.,2/ =< να
or
724.2ttt 34,005.,2/ => να
+
µµ
=
2
2
2
1
2
1
2121
n
s
n
s
)()xx(
t
=
,94.2
16
03.9
24
54.20
0)44.5642.70(
22 =
+
p-value = .0060. There is enough evidence to conclude that
the two packages differ in the amount of time needed to learn how to use them.
b
+± α2
2
2
1
2
1
2/21 n
s
n
s
t)xx(
= (70.42 –56.44)
±
2.030
+16
03.9
24
54.20 22
= 13.98
±
9.67;
LCL = 4.31, UCL = 23.65
c The amount of time is required to be normally distributed.
d The histograms are somewhat bell shaped.
13.19a
)(:H 210 µµ
= 0
)(:H 211 µµ
< 0
Two-tail F test: F = 5.18, p-value = .0019; use unequal-variances test statistic
1n
)n/s(
1n
)n/s(
)n/sn/s(
2
2
2
2
2
1
2
1
2
1
2
2
2
21
2
1
+
+
=ν
= 276.5 (rounded to 277)
Rejection region:
326.2ttt 277,01., =< να
+
µµ
=
2
2
2
1
2
1
2121
n
s
n
s
)()xx(
t
=
+
200
09.3
200
39.1
0)80.702.5(
22
= –11.60, p-value = 0. There is enough evidence to infer that the
amount of time wasted in unsuccessful firms exceeds that of successful firms.
32
b
+± α2
2
2
1
2
1
2/21 n
s
n
s
t)xx(
= (5.02 – 7.80)
±
1.960
+200
09.3
200
39.1 22
= –2.78
±
.47;
LCL = –3.25, UCL = –2.31. Workers in unsuccessful companies waste on average between 2.31 and 3.25 hours per
week more than workers in successful companies.
13.20
)(:H 210 µµ
= 0
>µµ )(:H 211
0
Two-tail F test: F = .73, p-value = .0699; use equal-variances test statistic
Rejection region:
645.1ttt 268,05., => να
+
µµ
=
21
2
p
2121
n
1
n
1
s
)()xx(
t
=
+
+ +
145
1
125
1
2145125
053).1145(045).1125(
0)601.646(.
22
= 7.54, p-value = 0. There is enough
evidence to conclude that the reaction time of drivers using cell phones is slower that for non-cell phone users.
13.21
)(:H 210 µµ
= 0
µµ )(:H 211
0
Two-tail F test: F = 1.10, p-value = .6406; use equal-variances test statistic
Rejection region:
973.1ttt 183,025.,2/ =< να
or
973.1ttt 183,025.,2/ => να
+
µµ
=
21
2
2121
11
)()(
nn
s
xx
t
p
=
+
+ +
90
1
95
1
29095
045).190(048).195(
0)662.654(.
22
= –1.17, p-value = .2444. There is not
enough evidence to infer that the type of discussion affects reaction times.
13.22
)(:H 210 µµ
= 0
)(:H 211 µµ
> 0
Two-tail F test: F = .97, p-value = .9054; use equal-variances test statistic
Rejection region:
656.1ttt 143,05., => να
+
µµ
=
21
2
p
2121
n
1
n
1
s
)()xx(
t
=
+
+ +
81
1
64
1
28164
61.1)181(59.1)164(
0)94.518.6(
22
= .90, p-value = .1858. There is not enough
evidence to infer that people spend more time researching for a financial planner than they do for a stock broker.
13.23
)(:H 210 µµ
= 0
33
)(:H 211 µµ
< 0
Two-tail F test: F = .74, p-value = .0446; use unequal-variances test statistic
1n
)n/s(
1n
)n/s(
)n/sn/s(
2
2
2
2
2
1
2
1
2
1
2
2
2
21
2
1
+
+
=ν
= 373
Rejection region:
645.1ttt 373,05., =< να
+
µµ
=
2
2
2
1
2
1
2121
n
s
n
s
)()xx(
t
=
+
202
85.6
173
90.5
0)80.6671.63(
22
= –4.69, p-value = 0. There is enough evidence to infer that
students without textbooks outperform those with textbooks.
13.24
)(:H 210 µµ
= 0
µµ )(:H 211
0
Two-tail F test: F = .85, p-value = .2494; use equal-variances test statistic
Rejection region:
960.1ttt 413,025.,2/ =< να
or
960.1ttt 413,025.,2/ => να
+
µµ
=
21
2
p
2121
n
1
n
1
s
)()xx(
t
=
+
+ +
202
1
213
1
2202213
64.23)1202(82.21)1213(
0)43.15485.149(
22
= –2.05, p-value = .0412. There is
enough evidence to conclude that there are differences in service times between the two chains.
13.25
)(:H 210 µµ
= 0
µµ )(:H 211
0
Two-tail F test: F = .71, p-value = .1214; use equal-variances test statistic
Rejection region:
975.1ttt 168,025.,2/ =< να
or
975.1ttt 168,025.,2/ => να
+
µµ
=
21
2
p
2121
n
1
n
1
s
)()xx(
t
=
+
+ +
91
1
79
1
29179
64.3)191(06.3)179(
0)67.5105.53(
22
= 2.65, p-value = .0087. There is enough
evidence to conclude that the two types of specialties differ in the time devoted to patients.
13.26 a
)(: 210 µµH
= 0
µµ )(: 211
H
0
Two-tail F test: F = 1.51, p-value = .0402; use unequal-variances test statistic
34
1n
)n/s(
1n
)n/s(
)n/sn/s(
2
2
2
2
2
1
2
1
2
1
2
2
2
21
2
1
+
+
=ν
= 190
Rejection region:
973.1ttt 190,025.,2/ =< να
or
973.1ttt 190,025.,2/ => να
+
µµ
=
2
2
2
1
2
1
2121
n
s
n
s
)()xx(
t
=
+
100
00.26
100
99.31
0)14.12693.130(
22
= 1.16, p-value = .2467. There is not enough evidence to infer
that differences exist between the two types of customers.
13.27
)(:H 210 µµ
= 0
)(:H 211 µµ
> 0
Two-tail F test: F = 1.07, p-value = .8792; use equal-variances test statistic
Rejection region:
684.1ttt
38,05.,
=>
να
+
µµ
=
21
2
p
2121
n
1
n
1
s
)()xx(
t
=
+
+ +
20
1
20
1
22020
06.15)120(60.15)120(
0)20.6960.73(
22
= .91, p-value = .1849. There is not
enough evidence to infer that the new design tire lasts longer than the existing design tire.
13.28
0)(:H 210 =µµ
)(:H 211 µµ
< 0
Two-tail F test: F = .95, p-value = .8252; use equal-variances test statistic
Rejection region:
653.1ttt 178,05., => να
+
µµ
=
21
2
p
2121
n
1
n
1
s
)()xx(
t
=
+
+ +
90
1
90
1
22020
755,10)190(506,10)190(
0)563,63245,60
22
= −2.09
t =
09.2
, p-value = .0189. There is enough evidence to conclude that commission salespeople outperform fixed-
salary salespersons
13.29
0)(:H 210 =µµ
)(:H 211 µµ
0
Two-tail F test: F = .88, p-value = .4709; use equal-variances test statistic
Rejection region:
645.1ttt 429,025.,2/ => να
and
645.1ttt 429,025.,2/ = να
35
+
µµ
=
21
2
p
2121
n
1
n
1
s
)()xx(
t
=
+
+ +
338
1
93
1
233893
69.52)1338(45.49)193(
0)86.66197.633
22
= −4.58
t = −4.58, p-value = 0. There is enough evidence to conclude there is a difference in scores between those who have
and those who do not have accidents in a three-year period.
13.30
)(:H 210 µµ
= 0
)(:H 211 µµ
> 0
Two-tail F test: F = .41, p-value = 0; use unequal-variances test statistic
1n
)n/s(
1n
)n/s(
)n/sn/s(
2
2
2
2
2
1
2
1
2
1
2
2
2
21
2
1
+
+
=ν
= 222
Rejection region:
645.1ttt 222,05., => να
+
µµ
=
2
2
2
1
2
1
2121
n
s
n
s
)()xx(
t
=
+
130
42.4
130
84.2
0)27.1120.14(
22
= 6.28, p-value = 0. There is enough evidence to conclude that
bottles of wine with metal caps are perceived to be cheaper.
13.31 The data are observational. Experimental data could be produced by randomly assigning babies to either
Tastee or the competitor’s product.
13.32 Assuming that the volunteers were randomly assigned to eat either oat bran or another grain cereal the data are
experimental.
13.33 The data are observational. It is not possible to conduct a controlled experiment.
13.34 a The data are observational because the students decided themselves which package each would use.
b The professor can randomly assign one of the two packages to each student.
c Better students tend to choose Program B and better students learn how to use computer software more quickly.
13.35 a The data are observational.
b A powerful dictator can randomly order medical students to either general/family practice or pediatrics.
13.36a Let students select the section they wish to attend and compare test results.
b Randomly assign students to either section and compare test results.
36
13.37 Randomly assign patients with the disease to receive either the new drug or a placebo.
13.38a Randomly select finance and marketing MBA graduates and determine their starting salaries.
b Randomly assign some MBA students to major in finance and others to major in marketing. Compare starting
salaries after they graduate.
c Better students may be attracted to finance and better students draw higher starting salaries.
13.39a The data are observational because to obtain experimental data would entail randomly assigning some people
to smoke and others not to smoke.
b It is possible that some people smoke because of a genetic defect (Genetics have been associated with
alcoholism.), which may also be linked to lung cancer.
c In our society the experiment described in part a is impossible.
13.40
D0 :H µ
= 0
D1 :H µ
< 0
Rejection region:
895.1ttt 7,05., ==< να
,22.3
8/17.4
075.4
n/s
x
t
DD
D
D=
=
µ
=
p-value = .0073. There is enough evidence to infer that the Brand A is better
than Brand B.
13.41
D0 :H µ
= 0
D1 :H µ
< 0
Rejection region:
895.1ttt 7,05., ==< να
,20.2
8/225.
0175.
n/s
x
t
DD
D
D=
=
µ
=
p-value = .0320. There is enough evidence to infer that ABS is better.
13.42
D0 :H µ
= 0
D1 :H µ
> 0
Rejection region:
943.1ttt 6,05., ==> να
,98.1
7/48.2
086.1
n/s
x
t
DD
D
D=
=
µ
=
p-value = .0473. There is enough evidence to infer that the camera reduces the
number of red-light runners.
13.43a
D0 :H µ
= 0
<µD1 :H
0
37
Rejection region:
796.1ttt 11,05., ==< να
=
=
µ
=12/02.3
000.1
n/s
x
t
DD
D
D
–1.15, p-value = .1375. There is not enough evidence to infer that the new fertilizer is
better.
b
D
D
2/D n
s
tx α
±
=
92.100.1
12
02.3
201.200.1 ±=±
; LCL = –2.92, UCL = .92
c The differences are required to be normally distributed
d No, the histogram is bimodal.
e The data are experimental.
f The experimental design should be independent samples.
13.44 a
D0 :H µ
= 0
>µD1 :H
0
Rejection region:
796.1ttt 11,05., ==> να
=
=
µ
=12/88.5
008.3
n/s
x
t
DD
D
D
1.82, p-value = .0484. There is enough evidence to infer that companies with exercise
programs have lower medical expenses.
b
D
D
2/D n
s
tx α
±
=
74.308.3
12
88.5
201.208.3 ±=±
; LCL = –.66, UCL = 6.82
c Yes because medical expenses will vary by the month of the year.
13.45
D0 :H µ
= 0
>µD1 :H
0
Rejection region:
656.1ttt 149,05., => να
=
=
µ
=150/1.99
04.12
n/s
x
t
DD
D
D
1.53, p-value = .0638. There is not enough evidence to infer that mortgage payments
have increases in the past 5 years.
13.46
D0 :H µ
= 0
µD1 :H
0
Rejection region:
009.2ttt 49,025.,2/ =< να
or
009.2ttt 49,025.,2/ => να
38
=
=
µ
=50/22.2
016.1
n/s
x
t
DD
D
D
–3.70, p-value = .0006. There is enough evidence to infer that waiters and waitresses
earn different amounts in tips.
13.47 a
D
D
2/D n
s
tx α
±
=
16.875.19
40
63.30
684.175.19 ±=±
; LCL = 11.59, UCL = 27.91
b
D0 :H µ
= 0
>µD1 :H
0
Rejection region:
684.1ttt
39,05.,
=>
να
=
=
µ
=40/63.30
075.19
n/s
x
t
DD
D
D
4.08, p-value = .0001. There is enough evidence to conclude that companies that
advertise in the Yellow Pages have higher sales than companies that do not.
c The histogram of the differences is bell shaped.
d No, because we expect a great deal of variation between stores.
13.48a
D0 :H µ
= 0
>µD1 :H
0
Rejection region:
345.1ttt
14,10.,
==>
να
,92.16
15/14.13
040.57
n/s
x
t
DD
D
D=
=
µ
=
p-value = 0. There is enough evidence to conclude that heating costs for
insulated homes is less than that for uninsulated homes.
b
D
D
2/D n
s
tx α
±
=
28.740.57
15
14.13
145.240.57 ±=±
; LCL = 50.12, UCL = 64.68
c Differences are required to be normally distributed.
13.49
D0 :H µ
= 0
µD1 :H
0
Rejection region:
014.2ttt 44,025.,2/ =< να
or
014.2ttt 44,025.,2/ => να
=
=
µ
=45/16.317
094.42
n/s
x
t
DD
D
D
–.91, p-value = .3687. There is not enough evidence to infer men and women spend
different amounts on health care.
39
13.50
D0 :H µ
= 0
<µD1 :H
0
Rejection region:
654.1ttt 169,05., =< να
=
=
µ
=170/94.1568
035.183
n/s
x
t
DD
D
D
–1.52, p-value = .0647. There is not enough to infer stock holdings have
decreased.
13.51
D0 :H µ
= 0
D1 :H µ
> 0
Rejection region:
690.1ttt
37,05.,
=>
να
=
=
µ
=38/1634.
00422.
n/s
x
t
DD
D
D
1.59, p-value = .0599. There is not enough evidence to conclude that ratios are higher
this year.
13.52
D0 :H µ
= 0
D1 :H µ
> 0
Rejection region:
676.1ttt
54,05.,
=>
να
=
=
µ
=55/92.1854
085.520
n/s
x
t
DD
D
D
2.08, p-value = .0210. There is enough evidence to infer that company 1’s
calculated tax payable is higher than company 2’s.
13.53
D0 :H µ
= 0
D1 :H µ
> 0
Rejection region:
729.1ttt
19,05.,
==>
να
=
=
µ
=20/22.7
055.4
n/s
x
t
DD
D
D
2.82, p-value = .0055. There is enough evidence to that the new design tire lasts longer
than the existing design.
13.54 The matched pairs experiment reduced the variation caused by different drivers.
13.55
D0 :H µ
= 0
D1 :H µ
> 0
Rejection region:
711.1ttt
24,05.,
==>
να
40
=
=
µ
=25/851,22
04587
n/s
x
t
DD
D
D
1.00, p-value = .1628. There is not enough evidence to infer that finance majors
attract higher salary offers than do marketing majors.
13.56 Salary offers and undergraduate GPA are not as strongly linked as are salary offers and MBA GPA.
13.57 a
D0 :H µ
= 0
<µD1 :H
0
Rejection region:
684.1ttt
41,05.,
=<
να
=
=
µ
=42/95.1
010.
n/s
x
t
DD
D
D
–.33, p-value = .3704. There is not enough evidence to infer that for companies where
an offspring takes the helm there is a decrease in operating income.
b
D0 :H µ
= 0
D1 :H µ
> 0
Rejection region:
660.1ttt 97,05., ==> να
=
=
µ
=98/83.2
024.1
n/s
x
t
DD
D
D
4.34, p-value = 0. There is enough evidence to infer that when an outsider becomes
CEO the operating income increases.
13.58 a
1/:H 2
2
2
10 =σσ
1/:H 2
2
2
11 σσ
Rejection region:
88.1FFF 29,29,05.,,2/ 21 => ννα
or
53.88.1/1F/1F/1FF 29,29,05.,,2/,,2/1 1221 ===< νναννα
F =
2
2
2
1s/s
= 350/700 =.50, p-value = .0669. There is enough evidence to conclude that the population variances
differ.
b Rejection region:
98.2FFF 14,14,025.,,2/ 21 ==> ννα
or
34.98.2/1F/1F/1FF 14,14,025.,,2/,,2/1 1221 ====< νναννα
F =
2
2
2
1s/s
= 350/700 =.50, p-value = .2071. There is not enough evidence to conclude that the population variances
differ.
c The value of the test statistic is unchanged and in this exercise the conclusion changed as well..
13.59 a LCL =
21 ,,2/
2
2
2
1
F
1
s
s
ννα
=
03.4
1
19
28
= .366, UCL =
12 ,,2/
2
2
2
1F
s
sννα
=
03.4
19
28
= 5.939
41
b LCL =
21 ,,2/
2
2
2
1
F
1
s
s
ννα
=
27.2
1
19
28
= .649, UCL =
12 ,,2/
2
2
2
1F
s
sννα
=
27.2
19
28
= 3.345
c The interval narrows.
13.60
1/:H 2
2
2
10 =σσ
1/:H 2
2
2
11 σσ
Rejection region:
78.3FFF 10,9,025.,,2/ 21 ==> ννα
or
25.96.3/1F/1F/1FF 9,10,025.,,2/,,2/1 1221 ====< νναννα
F =
2
2
2
1s/s
= .0000057/.0000114 =.50, p-value = .3179. There is not enough evidence to conclude that the two
machines differ in their consistency of fills.
13.61
1/:H 2
2
2
10 =σσ
1/:H 2
2
2
11 <σσ
Rejection region:
314.18.3/1F/1F/1FF 9,9,05.,,,,1 1221 ====< νναννα
F =
2
2
2
1s/s
= .1854/.1893 =.98, p-value = .4879. There is not enough evidence to infer that the second method is
more consistent than the first method.
13.62
1/:H 2
2
2
10 =σσ
1/:H 2
2
2
11 σσ
Rejection region:
72.3FFF 10,10,025.,,2/ 21 ==> ννα
or
269.72.3/1F/1F/1FF 10,10,025.,,2/,,2/1 1221 ====< νναννα
F =
2
2
2
1s/s
= 193.67/60.00 = 3.23, p-value = .0784. There is not enough evidence to infer that the variances of the
marks differ between the two sections.
13.63
1/:H 2
2
2
10 =σσ
1/:H 2
2
2
11 >σσ
Rejection region:
39.1FFF 99,99,05.,, 21 => ννα
F =
2
2
2
1s/s
= 19.38/12.70 = 1.53, p-value = .0183. There is enough evidence to infer that limiting the minimum and
maximum speeds reduces the variation in speeds.
13.64
1/:H 2
2
2
10 =σσ
1/:H 2
2
2
11 σσ
Rejection region:
48.1FFF 99,99,025.,,2/ 21 => ννα
or
68.48.1/1F/1F/1FF 99,99,025.,,2/,,2/1 1221 ===< νναννα
42
F =
2
2
2
1s/s
= 41,309/19,850 = 2.08, p-value = .0003. There is enough evidence to conclude that the variances differ.
13.65
2
2
2
10 /:H σσ
= 1
2
2
2
11 /:H σσ
< 1
Rejection region:
63.60.1/1F/1F/1FF 51,51,05.,,,,1 1221 ===< νναννα
F =
2
2
2
1/ss
= .0261/.0875 = .298, p-value = 0. There is enough evidence to infer that portfolio 2 is riskier than
portfolio 1.
13.66
2
2
2
10 /:H σσ
= 1
σσ 2
2
2
11 /:H
1
Rejection region:
39.1FFF 99,99,05.,,2/ 21 => ννα
or
72.39.1/1F/1F/1FF 99,99,05.,,,,1 1221 ===< νναννα
F =
2
2
2
1/ss
= 3.35/10.95 = .31, p-value = 0. There is enough evidence to conclude that the variance in service times
differ.
13.67
)pp(:H 210
= 0
)pp(:H 211
0
a
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
+
100
1
100
1
)425.1(425.
)40.45(.
= .72, p-value = 2P(Z > .72) = 2(1 – .7642)
= .4716.
b
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
+
400
1
400
1
)425.1(425.
)40.45(.
= 1.43, p-value = 2P(Z > 1.43) = 2(1 – .9236)
= .1528.
c The p-value decreases.
13.68
)pp(:H 210
= 0
)pp(:H 211
0
a
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
,07.1
225
1
225
1
)575.1(575.
)55.60(. =
+
p-value = 2P(Z > 1.07) = 2(1 – .8577)
= .2846
43
b
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
,01.2
225
1
225
1
)925.1(925.
)90.95(. =
+
p-value = 2P(Z > 2.01) = 2(1 – .9778)
= .0444.
c. The p-value decreases.
d
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
,01.2
225
1
225
1
)075.1(075.
)05.10(. =
+
p-value = 2P(Z > 2.01) = 2(1 – .94778)
= .0444.
e. The p-value decreases.
13.69 a
2
22
1
11
2/21 n
)p
ˆ
1(p
ˆ
n
)p
ˆ
1(p
ˆ
z)p
ˆ
p
ˆ
(
+
± α
= (.18–.22)
100
)22.1(22.
100
)18.1(18.
645.1
+
±
= –.040
±
.0929
b
2
22
1
11
2/21 n
)p
ˆ
1(p
ˆ
n
)p
ˆ
1(p
ˆ
z)p
ˆ
p
ˆ
(
+
± α
= (.48–.52)
100
)52.1(52.
100
)48.1(48.
645.1
+
±
= –.040
±
.1162
c The interval widens.
13.70
)pp(:H 210
= 0
> )pp(:H 211
0
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
,70.1
178
1
229
1
)177.1(177.
)140.205(. =
+
p-value = P(Z > 1.70) = 1 – .9554 = .0446.
There is enough evidence to conclude that those who paid the regular price are more likely to buy an extended
warranty.
13.71
)pp(:H 210
= 0
0)pp(:H 211
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
,69.1
209
1
83
1
)209.1(209.
)234.145(. =
+
p-value = 2P(Z < −1.69) = 2(.0455)
= .0910.
There is not enough evidence to conclude that new and old accounts are different with respect to overdue accounts.
44
13.72
)pp(:H 210
= 0
> )pp(:H 211
0
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
,74.1
804
1
562
1
)0132.1(0132.
)0087.0196(. =
+
p-value = P(Z > 1.74) = 1 – .9591 = .0409.
There is enough evidence to conclude that those who score under 600 are more likely to default than those who
score 60 or more.
13.73a
)pp(:H 210
= 0
> )pp(:H 211
0
Rejection region:
05.
zzz => α
= 1.645
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
+
800
1
1100
1
)518.1(518.
)46.56(.
= 4.31, p-value = 0. There is enough evidence to infer that
the leader’s popularity has decreased.
b
)pp(:H 210
= .05
> )pp(:H 211
.05
Rejection region:
05.
zzz => α
= 1.645
2
22
1
11
2121
n
)p
ˆ
1(p
ˆ
n
)p
ˆ
1(p
ˆ
)pp()p
ˆ
p
ˆ
(
z
+
=
=
800
)46.1(46.
1100
)56.1(56.
05.)46.56(.
+
= 2.16, p-value = P(Z > 2.16) = 1 – .9846
= .0154.
There is enough evidence to infer that the leader’s popularity has decreased by more than 5%.
c
2
22
1
11
2/21
n
)p
ˆ
1(p
ˆ
n
)p
ˆ
1(p
ˆ
z)p
ˆ
p
ˆ
(
+
±
α
=
800
)46.1(46.
1100
)56.1(56.
96.1)46.56(.
+
±
= .10
±
.045
13.74
)pp(:H 210
= –.08
< )pp(:H 211
–.08
Rejection region:
01.
zzz =< α
= –2.33
2
22
1
11
2121
n
)p
ˆ
1(p
ˆ
n
)p
ˆ
1(p
ˆ
)pp()p
ˆ
p
ˆ
(
z
+
=
=
300
)28.1(28.
300
)11.1(11.
)08.()28.11(.
+
= –2.85, p-value =P(Z < –2.85) = 1 – .9978
= .0022.
There is enough evidence to conclude that management should adopt process 1.
45
13.75
)pp(:H 210
= 0
> )pp(:H 211
0
Rejection region:
05.
zzz => α
= 1.645
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
+
1109
1
1604
1
)068.1(068.
)064.071(.
= .71, p-value = P(Z > .71) = 1 – .7611
= .2389.
There is not enough evidence to infer that the claim is false.
13.76 a
)pp(:H 210
= 0
< )pp(:H 211
0
Rejection region:
05.
zzz =< α
= –1.645
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
+
6281
1
6281
1
)104.1(104.
)115.093(.
= –4.04, p-value = 0. There is enough evidence to infer
that Plavix is effective.
13.77 a
)pp(:H 210
= 0
< )pp(:H 211
0
Rejection region:
01.
zzz =< α
= –2.33
0095.
000,11
104
p
ˆ1==
0172.
000,11
189
p
ˆ2==
01335.
000,22
189104
p
ˆ=
+
=
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
+
000,11
1
000,11
1
)01335.1(01335.
)0172.0095(.
= –4.98, p-value = 0. There is enough evidence
to infer that aspirin is effective in reducing the incidence of heart attacks.
13.78
)pp(:H 210
= 0
> )pp(:H 211
0
Rejection region:
645.1zzz 05. ==> α
0985.
000,11
084,1
p
ˆ1==
0906.
000,11
997
p
ˆ2==
0946.
000,22
997084,1
p
ˆ=
+
=
46
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
,00.2
000,11
1
000,11
1
)0946.1(0946.
)0906.0985(. =
+
p-value = P(Z > 2.00) = 1 – .9772 = .0228.
There is enough evidence to infer that aspirin leads to more cataracts.
13.79
)pp(:H 210
= 0
< )pp(:H 211
0
Rejection region:
01.
zzz =< α
= –2.33
0289.
594,2
75
p
ˆ1==
0509.
594,2
132
p
ˆ2==
0399.
594,2594,2
13275
p
ˆ=
+
+
=
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
,04.4
594,2
1
594,2
1
)0399.1(0399.
)0509.0289(. =
+
p-value = 0.
There is enough evidence to infer that Letrozole works.
13.80
)pp(:H 210
= 0
< )pp(:H 211
0
Rejection region:
645.1zzz 05. ==< α
2228.
395
88
p
ˆ1==
2586.
406
105
p
ˆ2==
2409.
406395
10588
p
ˆ=
+
+
=
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
,19.1
406
1
395
1
)2409.1(2409.
)2586.2228(. =
+
p-value = P(Z < –1.19) = .1170. There is not
enough evidence to infer that exercise training reduces mortality.
13.81
)pp(:H 210
= 0
)pp(:H 211
0
a
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
,75.
294
1
350
1
)8975.1(8975.
8878.9057(. =
+
p-value = 2P(Z > .75) = 2(1 – .7734)
= .4532.
There is not enough evidence to infer that the two populations of car owners differ in their satisfaction levels.
13.82
)pp(:H 210
= 0
47
)pp(:H 211
> 0
Rejection region:
10.
zzz => α
= 1.28
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
+
162
1
38
1
)11.1(11.
)0741.2632(.
= 3.35, p-value = 0. There is enough evidence to conclude that
smokers have a higher incidence of heart diseases than nonsmokers.
b
2
22
1
11
2/21 n
)p
ˆ
1(p
ˆ
n
)p
ˆ
1(p
ˆ
z)p
ˆ
p
ˆ
(
+
± α
= (.2632–0741)
162
)0741.1(0741.
38
)2632.1(2632.
645.1
+
±
=.1891
±
.1223; LCL = .0668, UCL = .3114
13.83a
)pp(:H 210
= 0
< )pp(:H 211
0
Rejection region:
10.
zzz =< α
= −1.28
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
,90.
500
1
400
1
)842.1(842.
)852.830(. =
+
p-value = P(Z < −.90) = .1841. There is not enough
evidence to infer that there has been an increase in belief in the greenhouse effect.
b
2
22
1
11
2/21 n
)p
ˆ
1(p
ˆ
n
)p
ˆ
1(p
ˆ
z)p
ˆ
p
ˆ
(
+
± α
= (.830–.852)
500
)852.1(852.
400
)830.1(830.
645.1
+
±
= −.022
±
.040;
LCL = −.062 and UCL = .018.
13.84
)pp(:H 210
= 0
> )pp(:H 211
0
Rejection region:
05.
zzz => α
= 1.645
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
+
462
1
477
1
)3504.1(3504.
)3420.3585(.
= .53, p-value = P(Z > .53) = 1 – .7019
= .2981.
There is not enough evidence to infer that the use of illicit drugs in the United States has increased in the past
decade.
13.85
)pp(:H 210
= –.02
48
< )pp(:H 211
–.02
Rejection region:
05.
zzz =< α
= –1.645
2
22
1
11
2121
n
)p
ˆ
1(p
ˆ
n
)p
ˆ
1(p
ˆ
)pp()p
ˆ
p
ˆ
(
z
+
=
=
200
)11.1(11.
200
)055.1(055.
)02.()11.055(.
+
= –1.28, p-value = P(Z < –1.28) = .1003. There is not
enough evidence to choose machine A.
13.86
)pp(:H 210
= 0
> )pp(:H 211
0
Rejection region:
05.
zzz => α
= 1.645
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
+
619
1
231
1
)1035.1(1035.
)0905.1385(.
z = 2.04, p-value = P(Z > 2.04) = 1 – .9793
= .0207.
There is enough evidence to conclude that health conscious adults are more likely to buy Special X.
13.87a
)pp(:H 210
= 0
> )pp(:H 211
0
Rejection region:
05.
zzz => α
= 1.645
4172.
163
68
p
ˆ1==
2685.
108
29
p
ˆ2==
3579.
108163
2968
p
ˆ=
+
+
=
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
,50.2
108
1
163
1
)3579.1(3579.
)2685.4172(. =
+
p-value = P(Z > 2.50) = 1 – .9938
= .0062.
There is enough evidence to conclude that members of segment 1 are more likely to use the service than members of
segment 4.
b
)pp(:H 210
= 0
)pp(:H 211
0
Rejection region:
05.2/ zzz =< α
= –1.96 or
05.2/ zzz => α
= 1.96
3704.
54
20
p
ˆ1==
4348.
23
10
p
ˆ2==
3896.
2354
1020
p
ˆ=
+
+
=
49
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
,53.
23
1
54
1
)3896.1(3896.
4348.3704(. =
+
p-value = 2P(Z < –.53) = 2(.2981)
= .5962.
There is not enough evidence to infer that retired persons and spouses that work in the home differ in their use of
services such as Quik Lube.
13.88
)pp(:H 210
= 0
)pp(:H 211
0
Rejection region:
025.2/ zzz =< α
= –1.96 or
025.2/ zzz => α
= 1.96
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
=
+
316
1
382
1
)1132.1(1132.
)1297.0995(.
= –1.25, p-value = 2P(Z < –1.25) = 2(.1056) =
.2112.
There is not enough evidence to infer differences between the two sources.
13.89
)(:H 210 µµ
= 0
)(:H 211 µµ
> 0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Two-Sample Assuming Equal Variances
During Before
Mean 5746.07 5372.13
Variance 167289 194772
Observations 15 24
Pooled Variance 184373
Hypothesized Mean Difference 0
df 37
t Stat 2.65
P(T<=t) one-tail 0.0059
t Critical one-tail 1.6871
P(T<=t) two-tail 0.0119
t Critical two-tail 2.0262
t = 2.65, p-value = .0059. There is enough evidence to conclude that the campaign is successful.
13.90 Gross sales must increase by 50/.20 = $250 to pay for ads.
)(:H 210 µµ
= 250
)(:H 211 µµ
> 250
50
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Two-Sample Assuming Equal Variances
During Before
Mean 5746.07 5372.13
Variance 167289 194772
Observations 15 24
Pooled Variance 184373
Hypothesized Mean Difference 250
df 37
t Stat 0.88
P(T<=t) one-tail 0.1931
t Critical one-tail 1.6871
P(T<=t) two-tail 0.3862
t Critical two-tail 2.0262
t = .88, p-value = .1931. There is not enough evidence to conclude that the ads are profitable.
13.91
)(:H 210 µµ
= 0
)(:H 211 µµ
< 0
1
2
3
4
5
6
7
8
9
10
11
12
13
A B C
t-Test: Two-Sample Assuming Unequal Variances
Group 1 Group 2
Mean 4.94 9.48
Variance 11.19 20.29
Observations 68 193
Hypothesized Mean Difference 0
df 158
t Stat -8.73
P(T<=t) one-tail 0.0000
t Critical one-tail 1.6546
P(T<=t) two-tail 0.0000
t Critical two-tail 1.9751
t = –8.73, p-value = 0. There is enough evidence to conclude that men and women who suffer heart attacks vacation
less than those who do not suffer heart attacks.
13.92
D0 :H µ
= 0
<µD1 :H
0
51
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Paired Two Sample for Means
Drug Placebo
Mean 18.43 22.03
Variance 30.39 66.37
Observations 100 100
Pearson Correlation 0.69
Hypothesized Mean Difference 0
df 99
t Stat -6.09
P(T<=t) one-tail 0.0000
t Critical one-tail 1.6604
P(T<=t) two-tail 0.0000
t Critical two-tail 1.9842
t = –6.09, p-value = 0. There is enough evidence to infer that the new drug is effective.
13.93
)(:H 210 µµ
= 0
µµ )(:H 211
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Two-Sample Assuming Equal Variances
Vendor Delivered
Mean 19.50 20.03
Variance 14.35 14.97
Observations 205 155
Pooled Variance 14.62
Hypothesized Mean Difference 0
df 358
t Stat -1.29
P(T<=t) one-tail 0.0996
t Critical one-tail 1.6491
P(T<=t) two-tail 0.1993
t Critical two-tail 1.9666
t = –1.29, p-value = .1993. There is not enough evidence of a difference in reading time between the two groups.
13.94
)pp(:H 210
= 0
< )pp(:H 211
0
52
1
2
3
4
5
6
7
8
9
10
11
A B C D
z-Test: Two Proportions
Last Year This Year
Sample Proportions 0.6758 0.7539
Observations 327 382
Hypothesized Difference 0
z Stat -2.30
P(Z<=z) one tail 0.0106
z Critical one-tail 1.6449
P(Z<=z) two-tail 0.0212
z Critical two-tail 1.96
z = –2.30, p-value = .0106. There is enough evidence to infer an increase in seatbelt use.
13.95a
)(:H 210 µµ
= 0
)(:H 211 µµ
< 0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Two-Sample Assuming Equal Variances
5 Years This Year
Mean 32.42 33.72
Variance 36.92 45.52
Observations 200 200
Pooled Variance 41.22
Hypothesized Mean Difference 0
df 398
t Stat -2.02
P(T<=t) one-tail 0.0218
t Critical one-tail 1.2837
P(T<=t) two-tail 0.0436
t Critical two-tail 1.6487
t = –2.02, p-value = .0218. There is enough evidence to infer that housing cost a percentage of total income has
increased.
b The histograms are be bell shaped.
13.96a
)(:H 210 µµ
= 0
µµ )(:H 211
0
53
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Two-Sample Assuming Equal Variances
Male Female
Mean 39.75 49.00
Variance 803.88 733.16
Observations 20 20
Pooled Variance 768.52
Hypothesized Mean Difference 0
df 38
t Stat -1.06
P(T<=t) one-tail 0.1490
t Critical one-tail 1.3042
P(T<=t) two-tail 0.2980
t Critical two-tail 1.6860
t = –1.06, p-value = .2980. There is not enough evidence to conclude that men and women differ in the amount of
time spent reading magazines.
b
)(:H 210 µµ
= 0
)(:H 211 µµ
< 0
1
2
3
4
5
6
7
8
9
10
11
12
13
A B C
t-Test: Two-Sample Assuming Unequal Variances
Low High
Mean 33.10 56.84
Variance 278.69 1047.81
Observations 21.00 19
Hypothesized Mean Difference 0.00
df 26.00
t Stat -2.87
P(T<=t) one-tail 0.0040
t Critical one-tail 1.3150
P(T<=t) two-tail 0.0080
t Critical two-tail 1.7056
t = –2.87, p-value = .0040. There is enough evidence to conclude that high-income individuals devote more time to
reading magazines than do low-income individuals.
13.97a
D0 :H µ
= 0
µD1 :H
0
54
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Paired Two Sample for Means
Female Male
Mean 55.68 56.40
Variance 105.64 116.75
Observations 25 25
Pearson Correlation 0.96
Hypothesized Mean Difference 0
df 24
t Stat -1.13
P(T<=t) one-tail 0.1355
t Critical one-tail 1.3178
P(T<=t) two-tail 0.2710
t Critical two-tail 1.7109
t = –1.13, p-value = .2710. There is no evidence to infer that gender is a factor.
b A large variation within each gender group was expected.
c The histogram of the differences is somewhat bell shaped.
13.98
)pp(:H 210
= 0
> )pp(:H 211
0
1
2
3
4
5
6
7
8
9
10
11
A B C D
z-Test: Two Proportions
This Year 3 Years Ago
Sample Proportions 0.4351 0.3558
Observations 393 385
Hypothesized Difference 0
z Stat 2.26
P(Z<=z) one tail 0.0119
z Critical one-tail 1.2816
P(Z<=z) two-tail 0.0238
z Critical two-tail 1.6449
z = 2.26, p-value = .0119. There is enough evidence to infer that Americans have become more distrustful of
television and newspaper reporting this year than they were three years ago.
13.99
)(:H 210 µµ
= 25
)(:H 211 µµ
> 25
55
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Two-Sample Assuming Equal Variances
A nondefectives A nondefectives
Mean 230.13 200.92
Variance 79.51 59.04
Observations 24 24
Pooled Variance 69.27
Hypothesized Mean Difference 25
df 46
t Stat 1.75
P(T<=t) one-tail 0.0433
t Critical one-tail 1.6787
P(T<=t) two-tail 0.0865
t Critical two-tail 2.0129
t = 1.75, p-value = .0433. There is enough evidence to conclude that machine A should be purchased.
13.100
)pp(:H 210
= 0
)pp(:H 211
0
The totals in columns A through D are 5788, 265, 5154, and 332, respectively.
1
2
3
4
5
6
7
A B C D E
z-Test of the Difference Between Two Proportions (Case 1)
Sample 1 Sample 2 z Stat -4.28
Sample proportion 0.045800 0.064400 P(Z<=z) one-tail 0.0000
Sample size 5788 5154 z Critical one-tail 1.6449
Alpha 0.05 P(Z<=z) two-tail 0.0000
z Critical two-tail 1.9600
z = –4.28, p-value = 0. There is enough evidence to infer that the defective rate differs between the two machines.
13.101
D0 :H µ
= 0
>µD1 :H
0
56
Dry Cleaner
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Paired Two Sample for Means
Dry C Before Dry C After
Mean 168.00 165.50
Variance 351.38 321.96
Observations 14 14
Pearson Correlation 0.86
Hypothesized Mean Difference 0
df 13
t Stat 0.96
P(T<=t) one-tail 0.1780
t Critical one-tail 1.7709
P(T<=t) two-tail 0.3559
t Critical two-tail 2.1604
t = .96, p-value = .1780. There is not enough evidence to conclude that the dry cleaner sales have decreased.
Doughnut shop
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Paired Two Sample for Means
Donut Before Donut After
Mean 308.14 295.29
Variance 809.67 812.07
Observations 14 14
Pearson Correlation 0.86
Hypothesized Mean Difference 0
df 13
t Stat 3.24
P(T<=t) one-tail 0.0032
t Critical one-tail 1.7709
P(T<=t) two-tail 0.0065
t Critical two-tail 2.1604
t = 3.24, p-value = .0032. There is enough evidence to conclude that the doughnut shop sales have decreased.
57
Convenience store
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Paired Two Sample for Means
Convenience Before Convenience After
Mean 374.64 348.14
Variance 2270.40 2941.82
Observations 14 14
Pearson Correlation 0.97
Hypothesized Mean Difference 0
df 13
t Stat 7.34
P(T<=t) one-tail 0.0000
t Critical one-tail 1.7709
P(T<=t) two-tail 0.0000
t Critical two-tail 2.1604
t = 7.34, p-value = 0. There is enough evidence to conclude that the convenience store sales have decreased.
13.102a
)(:H 210 µµ
= 0
<µµ )(:H 211
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Two-Sample Assuming Equal Variances
20-year-old 40-year-old
Mean 125.74 129.93
Variance 31.90 31.95
Observations 26 24
Pooled Variance 31.92
Hypothesized Mean Difference 0
df 48
t Stat -2.62
P(T<=t) one-tail 0.0059
t Critical one-tail 1.2994
P(T<=t) two-tail 0.0119
t Critical two-tail 1.6772
t = –2.62, p-value = .0059. There is enough evidence to infer that 40-year-old men have more iron in their bodies
than do 20-year-old men.
b
)(:H 210 µµ
= 0
<µµ )(:H 211
0
58
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Two-Sample Assuming Equal Variances
20-year-old 40-year-old
Mean 134.02 141.11
Variance 36.15 39.47
Observations 26 24
Pooled Variance 37.74
Hypothesized Mean Difference 0
df 48
t Stat -4.08
P(T<=t) one-tail 0.0001
t Critical one-tail 1.2994
P(T<=t) two-tail 0.0002
t Critical two-tail 1.6772
t = –4.08, p-value = .0001. There is enough evidence to infer that 40-year-old women have more iron in their bodies
than do 20-year-old women.
13.103a
)pp(:H 210
= 0
> )pp(:H 211
0
1
2
3
4
5
6
7
8
9
10
11
A B C D
z-Test: Two Proportions
Depressed Not Depressed
Sample Proportions 0.2879 0.2004
Observations 132 1058
Hypothesized Difference 0
z Stat 2.33
P(Z<=z) one tail 0.0100
z Critical one-tail 2.3263
P(Z<=z) two-tail 0.0200
z Critical two-tail 2.5758
z = 2.33, p-value = .0100. There is enough evidence to infer that men who are clinically depressed are more likely to
die from heart diseases.
b No, we cannot establish a causal relationship.
13.104 a
)(:H 210 µµ
= 0
>µµ )(:H 211
0
59
1
2
3
4
5
6
7
8
9
10
11
12
13
A B C
t-Test: Two-Sample Assuming Unequal Variances
Exercise Drug
Mean 13.52 9.92
Variance 5.76 13.16
Observations 25 25
Hypothesized Mean Difference 0
df 42
t Stat 4.14
P(T<=t) one-tail 0.0001
t Critical one-tail 2.4185
P(T<=t) two-tail 0.0002
t Critical two-tail 2.6981
t = 4.14, p-value = .0001. There is enough evidence that exercise is more effective than medication in reducing
hypertension.
b
1
2
3
4
5
6
7
8
A B C D E F
t-Estimate of the Difference Between Two Means (Unequal-Variances)
Sample 1 Sample 2 Confidence Interval Estimate
Mean 13.52 9.92 3.60 1.76
Variance 5.76 13.16 Lower confidence limit 1.84
Sample size 25 25 Upper confidence limit 5.36
Degrees of freedom 41.63
Confidence level 0.95
±
LCL = 1.84, UCL = 5.36
c The histograms are bell shaped.
13.105
)(:H 210 µµ
= 0
<µµ )(:H 211
0
1
2
3
4
5
6
7
8
9
10
11
12
13
A B C
t-Test: Two-Sample Assuming Unequal Variances
Group 1 Group 2
Mean 7.46 8.46
Variance 25.06 12.98
Observations 50 50
Hypothesized Mean Difference 0
df 89
t Stat -1.14
P(T<=t) one-tail 0.1288
t Critical one-tail 1.6622
P(T<=t) two-tail 0.2575
t Critical two-tail 1.9870
60
t = –1.14, p-value = .1288. There is not enough evidence to conclude that people who exercise moderately more
frequently lose weight faster
13.106
D0 :H µ
= 0
<µD1 :H
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Paired Two Sample for Means
Group 1 Group 2
Mean 7.53 8.57
Variance 29.77 43.37
Observations 50 50
Pearson Correlation 0.89
Hypothesized Mean Difference 0
df 49
t Stat -2.40
P(T<=t) one-tail 0.0100
t Critical one-tail 1.6766
P(T<=t) two-tail 0.0201
t Critical two-tail 2.0096
t = –2.40, p-value = .0100. There is enough evidence to conclude that people who exercise moderately more
frequently lose weight faster
13.107
)pp(:H 210
= 0
> )pp(:H 211
0
1
2
3
4
5
6
7
8
9
10
11
A B C D
z-Test: Two Proportions
Special K Other
Sample Proportions 0.635 0.53
Observations 200 200
Hypothesized Difference 0
z Stat 2.13
P(Z<=z) one tail 0.0166
z Critical one-tail 1.6449
P(Z<=z) two-tail 0.0332
z Critical two-tail 1.9600
z = 2.13, p-value = .0166. There is enough evidence to conclude that Special K buyers like the ad more than non-
buyers.
13.108
)pp(:H 210
= 0
> )pp(:H 211
0
61
1
2
3
4
5
6
7
8
9
10
11
A B C D
z-Test: Two Proportions
Special K Other
Sample Proportions 0.575 0.515
Observations 200 200
Hypothesized Difference 0
z Stat 1.20
P(Z<=z) one tail 0.1141
z Critical one-tail 1.6449
P(Z<=z) two-tail 0.2282
z Critical two-tail 1.9600
z = 1.20, p-value = .1141. There is not enough evidence to conclude that Special K buyers are more likely to think
the ad is effective.
13.109
)(:H 210 µµ
= 0
<µµ )(:H 211
0
1
2
3
4
5
6
7
8
9
10
11
12
13
A B C
t-Test: Two-Sample Assuming Unequal Variances
Small space Large space
Mean 1245.7 1915.8
Variance 23812 65566
Observations 25 25
Hypothesized Mean Difference 0
df 39
t Stat -11.21
P(T<=t) one-tail 0.0000
t Critical one-tail 1.6849
P(T<=t) two-tail 0.0000
t Critical two-tail 2.0227
t = –11.21, p-value = 0. There is enough evidence to infer that students write in such a way as to fill the allotted
space.
13.110
)(:H 210 µµ
= 0
>µµ )(:H 211
0
62
1
2
3
4
5
6
7
8
9
10
11
12
13
A B C
t-Test: Two-Sample Assuming Unequal Variances
Computer No Computer
Mean 69,933 48,246
Variance 63,359,040 101,588,525
Observations 89 61
Hypothesized Mean Difference 0
df 109
t Stat 14.07
P(T<=t) one-tail 0.0000
t Critical one-tail 1.2894
P(T<=t) two-tail 0.0000
t Critical two-tail 1.6590
t = 14.07, p-value = 0. There is enough evidence to conclude that single-person businesses that use a PC earn more.
13.111
)pp(:H 210
= 0
> )pp(:H 211
0
1
2
3
4
5
6
7
8
9
10
11
A B C D
z-Test: Two Proportions
New Older
Sample Proportions 0.948 0.920
Observations 250 250
Hypothesized Difference 0
z Stat 1.26
P(Z<=z) one tail 0.1037
z Critical one-tail 1.2816
P(Z<=z) two-tail 0.2074
z Critical two-tail 1.6449
z = 1.26, p-value = .1037. There is not enough evidence to conclude that the new company is better.
13.112
)(:H 210 µµ
= 0
)(:H 211 µµ
<
0
63
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Two-Sample Assuming Equal Variances
Supplement Placebo
Mean 19.02 21.85
Variance 41.34 25.49
Observations 48 48
Pooled Variance 33.41
Hypothesized Mean Difference 0
df 94
t Stat -2.40
P(T<=t) one-tail 0.0092
t Critical one-tail 1.6612
P(T<=t) two-tail 0.0183
t Critical two-tail 1.9855
t = –2.40, p-value = .0092. There is enough evidence to infer that taking vitamin and mineral supplements daily
increases the body's immune system?
13.113
)pp(:H 210
= 0
< )pp(:H 211
0
1
2
3
4
5
6
7
8
9
10
11
A B C D
z-Test: Two Proportions
$100-Limit $3000-Limit
Sample Proportions 0.5234 0.5551
Observations 491 490
Hypothesized Difference 0
z Stat -1.00
P(Z<=z) one tail 0.1598
z Critical one-tail 1.2816
P(Z<=z) two-tail 0.3196
z Critical two-tail 1.6449
z = –1.00, p-value = .1598. There is not enough evidence to infer that the dealer at the more expensive table is
cheating.
13.114
)(:H 210 µµ
= 0
µµ )(:H 211
0
64
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Two-Sample Assuming Equal Variances
Female Male
Mean 7.27 7.02
Variance 2.57 2.85
Observations 103 97
Pooled Variance 2.71
Hypothesized Mean Difference 0
df 198
t Stat 1.08
P(T<=t) one-tail 0.1410
t Critical one-tail 1.6526
P(T<=t) two-tail 0.2820
t Critical two-tail 1.9720
t = 1.08, p-value = .2820. There is not enough evidence to conclude that female and male high school students differ
in the amount of time spent at part-time jobs.
13.115
)(:H 210 µµ
= 0
>µµ )(:H 211
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Two-Sample Assuming Equal Variances
City Suburb
Mean 2.42 1.97
Variance 1.08 0.77
Observations 70 78
Pooled Variance 0.92
Hypothesized Mean Difference 0
df 146
t Stat 2.85
P(T<=t) one-tail 0.0025
t Critical one-tail 1.6554
P(T<=t) two-tail 0.0050
t Critical two-tail 1.9763
t = 2.85, p-value = .0025. There is enough evidence to infer that city households discard more newspaper than do
suburban households.
13.116
)(:H 210 µµ
= 0
>µµ )(:H 211
0
65
1
2
3
4
5
6
7
8
9
10
11
12
13
A B C
t-Test: Two-Sample Assuming Unequal Variances
Teenagers 20-to-30
Mean 18.18 14.30
Variance 357.32 130.79
Observations 176 154
Hypothesized Mean Difference 0
df 293
t Stat 2.28
P(T<=t) one-tail 0.0115
t Critical one-tail 1.6501
P(T<=t) two-tail 0.0230
t Critical two-tail 1.9681
t = 2.28, p-value = .0115.There is enough evidence to infer that teenagers see more movies than do twenty to thirty
year olds.
13.117
)pp(:H 210
= 0
< )pp(:H 211
0
1
2
3
4
5
6
7
8
9
10
11
A B C D
z-Test: Two Proportions
No HS HS
Sample Proportions 0.127 0.358
Observations 63 257
Hypothesized Difference 0
z Stat -3.54
P(Z<=z) one tail 0.0002
z Critical one-tail 1.6449
P(Z<=z) two-tail 0.0004
z Critical two-tail 1.96
z = –3.54, p-value = .0002. There is enough evidence to conclude that Californians who did not complete high
school are less likely to take a course in the university’s evening program.
13.118
)(:H 210 µµ
= 0
>µµ )(:H 211
0
66
1
2
3
4
5
6
7
8
9
10
11
12
13
A B C
t-Test: Two-Sample Assuming Unequal Variances
Group 1 Groups 2-4
Mean 11.58 10.60
Variance 9.28 21.41
Observations 269 981
Hypothesized Mean Difference 0
df 644
t Stat 4.15
P(T<=t) one-tail 0.0000
t Critical one-tail 1.6472
P(T<=t) two-tail 0.0000
t Critical two-tail 1.9637
t = 4.15, p-value = 0. There is enough evidence to conclude that on average the market segment concerned about
eating healthy food (group 1) outspends the other market segments.
13.119
)(:H 210 µµ
= 0
<µµ )(:H 211
0
1
2
3
4
5
6
7
8
9
10
11
12
13
A B C
t-Test: Two-Sample Assuming Unequal Variances
Sale-CDs Sale-fax
Mean 59.04 65.57
Variance 425.4 849.7
Observations 122 144
Hypothesized Mean Difference 0
df 256
t Stat -2.13
P(T<=t) one-tail 0.0171
t Critical one-tail 1.6508
P(T<=t) two-tail 0.0341
t Critical two-tail 1.9693
t = –2.13, p-value = .0171. There is enough evidence to conclude that those who buy the fax/copier outspend those
who buy the package of CD-ROMS.
13.120
)(:H 210 µµ
= 0
<µµ )(:H 211
0
67
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Two-Sample Assuming Equal Variances
No Yes
Mean 91,467 97,836
Variance 461,917,705 401,930,840
Observations 466 55
Pooled Variance 455,676,297
Hypothesized Mean Difference 0
df 519
t Stat -2.09
P(T<=t) one-tail 0.0184
t Critical one-tail 1.6478
P(T<=t) two-tail 0.0369
t Critical two-tail 1.9645
t = –2.09, p-value = .0184. There is enough evidence to infer that professors aged 55 to 64 who plan to retire early
have higher salaries than those who don’t plan to retire early.
13.121
)pp(:H 210
= 0
)pp(:H 211
0
The data must first be unstacked. Success = 2
1
2
3
4
5
6
7
8
9
10
11
A B C D
z-Test: Two Proportions
Female Male
Sample Proportions 0.5945 0.6059
Observations 762 746
Hypothesized Difference 0
z Stat -0.45
P(Z<=z) one tail 0.3256
z Critical one-tail 1.6449
P(Z<=z) two-tail 0.6512
z Critical two-tail 1.96
z = −.45, p-value = .6512. There is not enough evidence to conclude that men and women differ in their choices of
Christmas trees.
Case 13.1To test whether gender bias exists, we test to determine whether the proportions of loans denied is greater
for women (
1
p
) than for men (
2
p
).
0pp:H 210 =
0pp:H 211 >
68
1
2
3
4
5
6
7
A B C D E
z-Test of the Difference Between Two Proportions (Case 1)
Sample 1 Sample 2 z Stat 0.98
Sample proportion 0.1217 0.0933 P(Z<=z) one-tail 0.1633
Sample size 115 1050 z Critical one-tail 1.6449
Alpha 0.05 P(Z<=z) two-tail 0.3266
z Critical two-tail 1.9600
z = .98, p-value = .1633. There is not enough evidence to infer that loan requests by women are denied more
frequently than loan requests by men.
Interest rates (points above prime)
)(:H 210 µµ
= 0
>µµ )(:H 211
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Two-Sample Assuming Equal Variances
W Rate M Rate
Mean 1.55 1.28
Variance 0.41 0.45
Observations 101 952
Pooled Variance 0.44
Hypothesized Mean Difference 0
df 1051
t Stat 3.85
P(T<=t) one-tail 6.28E-05
t Critical one-tail 1.6463
P(T<=t) two-tail 0.0001
t Critical two-tail 1.9622
t = 3.85, p-value = 0. There is enough evidence to infer that the mean rate of women’s loans is greater than the mean
rate for men’s loans.
Case 13.2 Arrival times t-tests of
D
µ
D0 :H µ
= 1
D1 :H µ
> 1
DD
DD
n/s
x
tµ
=
69
Cambridge:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Paired Two Sample for Means
C Ambulance C Fire
Mean 12.86 10.64
Variance 9.71 3.39
Observations 280 280
Pearson Correlation 0.28
Hypothesized Mean Difference 1
df 279
t Stat 6.50
P(T<=t) one-tail 0.0000
t Critical one-tail 1.6503
P(T<=t) two-tail 0.0000
t Critical two-tail 1.9685
t = 6.50, p-value = 0. There is sufficient evidence to infer that in Cambridge fire trucks arrive at the scene more than
one minute sooner than ambulances.
Kitchener:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Paired Two Sample for Means
K Ambulance K Fire
Mean 9.79 8.55
Variance 10.73 3.26
Observations 506 506
Pearson Correlation 0.55
Hypothesized Mean Difference 1
df 505
t Stat 1.98
P(T<=t) one-tail 0.0243
t Critical one-tail 1.6479
P(T<=t) two-tail 0.0487
t Critical two-tail 1.9647
t = 1.98, p-value = .0243. There is sufficient evidence to infer that in Kitchener fire trucks arrive at the scene more
than one minute sooner than ambulances.
70
Waterloo:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Paired Two Sample for Means
W Ambulance W Fire
Mean 12.86 12.17
Variance 11.01 19.00
Observations 150 150
Pearson Correlation 0.80
Hypothesized Mean Difference 1
df 149
t Stat -1.46
P(T<=t) one-tail 0.0733
t Critical one-tail 1.6551
P(T<=t) two-tail 0.1466
t Critical two-tail 1.9760
t = –1.46, p-value = 1– .0733 = .9267. There is no evidence to infer that in Waterloo fire trucks arrive at the scene
more than one minute sooner than ambulances.
Frequency of arrivals in less than 8 minutes: z-tests of
21
pp
(case 1) (Data were recoded so that 2 = yes, 1 = no)
0)pp(:H 210 =
)pp(:H 211
< 0
+
=
21
21
n
1
n
1
)p
ˆ
1(p
ˆ
)p
ˆ
p
ˆ
(
z
Cambridge
1
2
3
4
5
6
7
8
9
10
11
A B C D
z-Test: Two Proportions
C Ambulance C Fire
Sample Proportions 0.0536 0.0857
Observations 280 280
Hypothesized Difference 0
z Stat -1.49
P(Z<=z) one tail 0.0676
z Critical one-tail 1.6449
P(Z<=z) two-tail 0.1352
z Critical two-tail 1.9600
z = –1.49, p-value = .0676. There is not enough evidence to infer that in Cambridge fire trucks arrive at the scene in
less than 8 minutes more frequently than do ambulances.
Kitchener:
71
1
2
3
4
5
6
7
8
9
10
11
A B C D
z-Test: Two Proportions
K Ambulance K Fire
Sample Proportions 0.2866 0.3992
Observations 506 506
Hypothesized Difference 0
z Stat -3.77
P(Z<=z) one tail 0.0001
z Critical one-tail 1.6449
P(Z<=z) two-tail 0.0002
z Critical two-tail 1.9600
z = –3.77, p-value = .0001. There is enough evidence to infer that in Kitchener fire trucks arrive at the scene in less
than 8 minutes more frequently than do ambulances.
Waterloo:
1
2
3
4
5
6
7
8
9
10
11
A B C D
z-Test: Two Proportions
W Ambulance W Fire
Sample Proportions 0.1000 0.1667
Observations 150 150
Hypothesized Difference 0
z Stat -1.70
P(Z<=z) one tail 0.0447
z Critical one-tail 1.6449
P(Z<=z) two-tail 0.0894
z Critical two-tail 1.9600
z = –1.70, p-value = .0447. There is enough evidence to infer that in Waterloo fire trucks arrive at the scene in less
than 8 minutes more frequently than ambulances.
Case 13.3 Length of ownership:
)(:H 210 µµ
= 0
µµ )(:H 211
0
72
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A B C
t-Test: Two-Sample Assuming Equal Variances
Would buy Would not buy
Mean 5.98 7.11
Variance 8.88 14.87
Observations 43 187
Pooled Variance 13.77
Hypothesized Mean Difference 0
df 228
t Stat -1.81
P(T<=t) one-tail 0.0359
t Critical one-tail 1.65
P(T<=t) two-tail 0.0717
t Critical two-tail 1.97
t = −1.81, p-value = .0717. There is some but not enough evidence to infer that the years of ownership differ
between those who would and those who would not buy a generator.
Compare homeowners who do not consider the resale value of their home versus homeowners who do
The data were sorted. Success = buy
0)pp(:H 210 =
< )pp(:H 211
0
1
2
3
4
5
6
7
8
9
10
11
A B C D
z-Test: Two Proportions
Do not consider resale Consider resale
Sample Proportions 0.1262 0.2362
Observations 103 127
Hypothesized Difference 0
z Stat -2.13
P(Z<=z) one tail 0.0167
z Critical one-tail 1.6449
P(Z<=z) two-tail 0.0334
z Critical two-tail 1.96
z = −2.13, p-value = .0167. There is enough evidence to infer that the proportion of those who would buy a
generator is greater for homeowners who consider resale value than homeowners who do not consider resale value.
County
0)pp(:H 210 =
)pp(:H 211
0
The data were sorted. Success = buy
73
1
2
3
4
5
6
7
8
9
10
11
A B C D
z-Test: Two Proportions
Miami-Dade Other
Sample Proportions 0.1791 0.1979
Observations 134 96
Hypothesized Difference 0
z Stat -0.36
P(Z<=z) one tail 0.3591
z Critical one-tail 1.6449
P(Z<=z) two-tail 0.7182
z Critical two-tail 1.96
z = −.36, p-value = .7182. There is not enough evidence to conclude that the proportion of those who would buy a
generator differ between Miami-Dade county and the other counties.
d. Advertising should focus on all counties with an emphasis on newer developments. Advertising should stress how
a generator would increase the resale value of a home.
74
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