Two-Sample Hypothesis Test Examples
Two-Sample Hypothesis Test Examples
(Chapter 11)
1. In a test of the reliability of products produced by two machines, machine A produced 15 defective parts in a run of 280, while machine B produced 10 defective parts in a run of 200. Do these results imply a difference in the reliability of these two machines? (Use = 0.01.)
Step 0 : Step 1 : Step 2 : Step 3 : Step 4 :
Step 5 : Step 6 :
Check Assumptions
( ) nA pA = yA = 15 10 and nA 1- pA = nA - yA = 265 10 ( ) nB pB = yB = 10 10 and nB 1- pB = nB - yB = 190 10
Hypotheses
H0: A - B = 0
Ha: A - B 0
Significance Level
= 0.01
Critical Value(s) and Rejection Region(s)
Critical Value: ?z = ? z0.005 = ?2.58 Reject the null hypothesis if Z ?2.58 or if Z 2.58.
Test Statistic
pc
=
yA + yB nA + nB
=
15 280
+ 10 + 200
=
25 480
Z=
( pA - pB ) - 0
=
15 280
-
10 200
-
0
= 0.1736
pc (1 -
pc
)
1 nA
+
1 nB
25 480
455 480
1 280
+
1 200
p - value = 2 * P(z 0.1736) 2* P(z 0.17) = 2* 0.4325 = 0.8650
Conclusion
Since ?2.58 0.1736 2.58 (p-value 0.8650 > 0.01 = ), we fail to reject
the null hypothesis.
State conclusion in words
At the = 0.01 level of significance, there is not enough evidence to conclude
that there is a difference in the reliability of the two machines.
2. Two sections of a class in statistics were taught by two different methods. Students' scores on a standardized test are shown below. Do the results present evidence of a difference in the effectiveness of the two methods? (Use = 0.01.)
Step 1 : Step 2 : Step 3 :
Step 4 : Step 5 : Step 6 :
Class A
Class B
74
76
78
79
97
75
92
76
79
82
94
93
88
86
78
82
78
100
71
69
93
94
85
84
70
Hypotheses
H0 : ? A - ? B = 0
Ha: ?A - ?B 0
Significance Level
= 0.01
Critical Value(s) and Rejection Region(s)
Since
we
don't
know
the
population
variances
(
2 A
and
2 B
)
and
don't
think
that
they are equal, we'll use the non-pooled t-test.
( ) sA2 =
y
2 A
-
yA nA
2
87960
-
(1022)2
12
nA -1
=
12 - 1
= 83.6061
( ) sB2 =
y
2 B
-
yB nB
2
85841 -
(1051)2
13
nB -1
=
13- 1
= 72.6410
[ ] (( )) (( )) [ ] df = =
sA2 nA + sB2 nB 2
sA2
nA
2
+
sB2
2
nB
nA - 1
nB - 1
=
(83.6061 12) + (72.6410 13) 2
(83.6061 12)2 + (72.6410 13)2
12 - 1
13 - 1
= 22.0147 22
Critical Values: ?t 2,df = = ?t0.005,df = 22 = ?2.82
Reject the null hypothesis if T ?2.82 or if T 2.82.
Test Statistic
yA =
yA = 1022 = 85.1667
nA
12
yB =
yB = 1051 = 80.8462
nB
13
T = (yA - )yB - 0 = (85.1667 - 80.8462) - 0 = 1.2193
sA2 + sB2 nA nB
83.6061 + 72.6410
12
13
p - value = 2 * P(t 1.2193) 2* P(t 1.2) = 2 *0.121 = 0.242
Conclusion
Since ?2.82 1.2193 2.82 (p-value 0.242 > 0.01 = ), we fail to reject the null hypothesis.
State conclusion in words
At the = 0.01 level of significance, there is not enough evidence to conclude that there is a difference in the effectiveness of the two methods.
3. The table below shows the observed pollution indexes of air samples in two areas of a city. Test the hypothesis that the mean pollution indexes are the same for the two areas. (Use = 0.05.)
Step 1 : Step 2 : Step 3 :
Step 4 : Step 5 : Step 6 :
Area A
2.92 4.69 1.88 4.86 5.35 5.81 3.81 5.55
Area B
1.84 3.44 0.95 3.69 4.26 4.95 3.18 4.47
Hypotheses
H0 : ? A - ? B = 0
Ha: ?A - ?B 0
Significance Level
= 0.05
Critical Value(s) and Rejection Region(s)
Since
we
don't
know
the
population
variances
(
2 A
and
2 B
)
but
think
that
they
are not equal (air varies across different areas of the same city due to
industrialization, vegetation, etc.), we'll use the non-pooled t-test.
( ) sA2 =
y
2 A
-
yA nA
2
165.3737
-
(34.87) 2
8
nA -1
=
8-1
= 1.9120
( ) sB2 =
y
2 B
-
yB nB
2
102.4812
-
(26.78)2
8
nB -1
=
8-1
= 1.8336
[ ] [ ] (( )) (( )) df = =
sA2 n A +
sA2
nA
2
+
nA - 1
sB2 n B 2 sB2 nB 2 nB - 1
=
(1.9120 8) + (1.8336 8) 2 (1.9120 8)2 (1.8336 8)2
8-1 + 8-1
= 13.9939 13
Critical Values: ?t 2,df = = ?t0.025,df =13 = ?2.16
Reject the null hypothesis if T ?2.16 or if T 2.16.
Test Statistic
yA =
yA = 34.87 = 4.3588
nA
8
yB =
yB = 26.78 = 3.3475
nB
8
T = (yA - )yB - 0 = (4.3588 - 3.3475) - 0 = 1.4780
sA2 + sB2 nA nB
1.9120 + 1.8336
8
8
p - value = 2 * P(t 1.4780) 2 * P(t 1.5) = 2* 0.079 = 0.158
Conclusion
Since ?2.16 1.4780 2.16 (p-value 0.158 > 0.05 = ), we fail to reject the null hypothesis.
State conclusion in words
At the = 0.05 level of significance, there is not enough evidence to conclude that there is a difference in the mean pollution indexes for the two areas.
4. A closer examination of the records of the air samples in Example 3 reveals that each line of the data actually represents readings on the same day: 2.92 and 1.84 are from day 1, and so forth. Since this affects the validity of the results obtained in Example 10, reanalyze. (Use = 0.05.)
Step 1 : Step 2 : Step 3 : Step 4 :
Step 5 : Step 6 :
Area A
2.92 1.88 5.35 3.81 4.69 4.86 5.81 5.55
Area B
1.84 0.95 4.26 3.18 3.44 3.69 4.95 4.47
yd = A ? B
1.08 0.93 1.09 0.63 1.25 1.17 0.86 1.08
Hypotheses
H0: ?d = 0
Ha: ?d 0
Significance Level
= 0.05
Critical Value(s) and Rejection Region(s)
Since we have paired data and don't know the population variance of the
differences
(
2 d
),
we'll
use
the
paired
t-test.
Critical Values: ?t 2,df = nd -1 = ? t0.025,df =7 = ?2.37
Reject the null hypothesis if T ?2.37 or if T 2.37.
Test Statistic
yd =
yd = 8.09 = 1.0113
nd
8
sd =
2
( ) y
2 d
-
yd nd
nd -1
=
8.45 - (8.09)2
8 = 0.1960 7
T
=
y d sd
-
0
=
1.0113 0.1960
-
0
= 14.5938
nd
8
p - value = 2 * P(t 14.5938) 2 * P(t 4.0) = 2 *0.003 = 0.006
Conclusion
Since 14.5938 2.37 (p-value 0.006 0.05 = ), we shall reject the null hypothesis.
State conclusion in words
At the = 0.05 level of significance, there exists enough evidence to conclude that there is a difference in the mean pollution indexes for the two areas.
5. Eight quantities of effluent from a pulp mill were each divided into ten batches. From each quantity, five randomly selected batches were subjected to a treatment process intended to remove toxic substances. Five fish of the same species were placed in each batch, and the mean number surviving in the five treated and untreated portions of each effluent quantity after five days were recorded and are given below. Test to see if the treatment increased the mean number of surviving fish. (Use = 0.01.)
Quantity No. 1
2
3
4
5
6
7
8
Mean Number Surviving
Untreated
5
Treated
5
yd = U ? T 0
1 1.8 1 3.6 5 2.6 1 5 1.2 4.8 5 5 4.4 2
-4 0.6 -3.8 -1.4 0 -1.8 -1
Step 1 : Step 2 : Step 3 : Step 4 :
Step 5 : Step 6 :
Hypotheses
H0: ?d = 0
Ha: ?d < 0
Significance Level
= 0.01
Critical Value(s) and Rejection Region(s)
Since we have paired data and don't know the population variance of the
differences
(
2 d
),
we'll
use
the
paired
t-test.
Critical Value: -t ,df = nd -1 = -t0.01,df = 7 = -3.00
Reject the null hypothesis if T ?3.00.
Test Statistic
yd =
yd = -11.4 = -1.425
nd
8
sd =
2
( ) y
2 d
-
yd nd
nd -1
=
37 - (-11.4)2
8 = 1.7219 7
T
=
y d sd
-
0
=
-1.425 1.7219
-
0
=
-2.3407
nd
8
p - value = P(t -2.3407) = P(t 2.3407) P(t 2.3) = 0.027
Conclusion
Since ?2.3407 > ?3.00 (p-value 0.027 > 0.01 = ), we fail to reject the null hypothesis.
State conclusion in words
At the = 0.01 level of significance, there is not enough evidence to conclude that the treatment increased the mean number of surviving fish.
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