chapter 7 l2
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Chapter 7
Hypothesis Testing Two
PopulationsWeek 10
L3: - Hypothesis Testing Between two means
(2 populations, variance unknown but not equal)- Hypothesis Testing Between two means(2 populations, variance unknown large samples)
- Hypothesis Testing- proportion
( 2 populations) 1September 2013
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Learning Objectives
At the end of the lesson student should beable to
Carry out hypothesis testing fordifference in means for two normalpopulations ( variances unknown)
Carry out hypothesis testing for
difference in means for difference in twoproportion
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Null Hypothesis is the same as before:
TWO V RI NCES RE UNKNOWN ND UNEQU L
Test about the difference
Test statistic:
)/()/(
)()(
2
2
21
2
1
2121
nSnS
xx
T
Case 2:2
2
2
1
But the degree of freedom is given by
1
)/(
1
)/(
)(
2
2
2
2
2
1
2
1
2
1
2
2
2
2
1
2
1
n
ns
n
ns
n
s
n
s
If is not an integer, round down to the nearest integer.
021
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Test about the difference
Case 2:2
2
2
1
021
Critical region:
Alternative Hypothesis Rejection Criteria ( Reject H0)
TEST BOUT TWO NORM L ME NS
WHEN TWO V RI NCES RE UNKNOWN
vv tTortT ,2/,2/ 211: H
vtT ,211: H
vtT ,211: H
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A two sided confidence interval for
when variance is unknown :
)%1(100 21
2
2
2
1
2
1,2/21
21
2
22
1
21
,2/21
n
s
n
stxx
n
s
n
stxx
Confidence interval:
Case 2: 22
2
1
Value of is from the formula earlier5
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6
Example 1:
Two companies manufacture a rubber material intended for use in
an automotive application. 25 samples of material from each
company are tested, and the amount of wear after 1000 cycles are
observed. For company 1, the sample mean and standard deviation
of wear are
and for company 2, we obtain
cycles1000/9.1andcycles1000/12.20 11 mgsmgx
Do the sample data support the claim that the two companiesproduce material with different mean wear? Assume each population
is normally distributed but unequal variances?
cycles1000/9.7andcycles1000/64.11 22 mgsmgx
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7
2. State the null hypothesis H0 and appropriate alternative
hypothesis, H1
0: 210 H 0: 211 H
3. Determine the appropriate test statistic
)/()/(
)()(
2
2
21
2
1
2121
nSnS
xxT
1. Identify the parameter of interest
Parameter of interest; the difference between the true average
Variances unknown and unequal21
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8
5. State the rejection region for the statistic
479.2ifHjectRe 0 T
4. Critical value given 0.01 479.226,01.0, tt v
2 22 2 2 2
1 2
1 2
2 2 2 22 22 2
1 2
1 2
1 2
(1.9) (7.9)
25 2526.77
(1.9) (7.9)
25 25
24 241 1
26
s s
n n
s s
n n
n n
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9
7. Make a decision
Decision: since we reject H0
Enough evidence to support that
479.222.5 T
21
6. Compute the value of the test statistic
22.5
25
)9.7(
25
)9.1(
0)64.1112.20(
22
T
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10
Exercise:
The following data represent the running times of films produced
by 2 motion-picture companies. Test the hypothesis that theaverage running time of films produced by company 2 exceeds the
average running time of films produced by company 1 by 10
minutes against the one-sided alternative that the difference is less
than 10 minutes? Use 0.01 and assume the distributions of
times to be approximately normal with unequal variances.
Time
Company
X1 102 86 98 109 92
X2 81 165 97 134 92 87 114
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2. State the null hypothesis H0 and appropriate alternative
hypothesis, H1
10: 120 H 10: 121 H
3. Determine the appropriate test statistic
)/()/(
)()(
2
2
21
2
1
1212
nSnS
xxT
1. Identify the parameter of interest
Parameter of interest; the difference between the true averageVariances unknown and unequal
12
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5. State the rejection region for the statistic
998.2ifHjectRe 0 T
4. Critical value given 0.01 998.27,01.0, tt v
38.76/)7/333.913(4/)5/8.78(
)7/333.9135/8.78(
1
)/(
1
)/(
)(
22
2
2
2
2
2
2
1
2
1
2
1
2
2
2
2
1
2
1
n
ns
n
ns
n
s
n
s
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7. Make a decision
Decision: since , we fail to reject H0
Not enough evidence that
998.222.0 T
1012
22.07/333.9135/8.78
10)4.97110(
)/()/(
)()(
2
2
21
2
1
1212
nSnS
xxT
6. Compute the value of the test statistic
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Test statistic:
Test about the difference between two means (Large Samples)
0
2
2
2
1
2
1
2121 Hunder)1,0(~)()(
N
n
S
n
S
xxZ
Null Hypothesis:
vs210: H 211211211 :or:or,: HHH
Alternative Hypothesis Rejection Criteria ( Reject H0)
211: H
211: H
211: H
2/2/ zZorzZ
zZ
zZ
Case 3:
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A confidence interval for is)%1(100 21
2
2
2
1
2
1
2/2121
2
2
2
1
2
1
2/21
n
S
n
Szxx
n
S
n
Szxx
The procedures used in hypothesis testing for 2 populationsare the same as in the hypothesis testing for one population.
Conclusion of the test could also be based on the confidence
Interval and the p-value.
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Example 3:
A study wants to investigate if the mean price of a new Proton car has
changed in 2012 compared to 2011. A sample of cars were taken duringthese two years and summary information as given below:
i. Perform the test for the above investigation. What is your conclusion?
ii. Find the two sided 95% CI for the difference between the two means in
2011 and 2012. What is your conclusion comparing with part (i).
Year 2011 2012
Mean car price RM 45,000 RM 48,000
Standard deviation of car
prices
RM 1,500 RM 2,000
Number of cars 150 200
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Solution
1. From the problem context, identify the parameter of interest
21
2. State the null hypothesis H0 and appropriate alternative
hypothesis, H1
210 : H 211: H
3. Determine the appropriate test statistic
)/()/(
)()(
2
2
21
2
1
2121
nSnS
xxZ
To test the difference between the true average large sample
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6. Compute any necessary sample quantities, substitute these into
the equation for the test statistic, and compute the value
,200,150,2000,1500,000,48,45000212121
nnSSXX
Compute the value of the test statistic:
04.16
)200/2000()150/1500(000,48000,45
)/()/()()(
22
2
2
21
2
1
2121
nSnS
xxZ
Find the p - value
P-value = 0)11(2))04.16(1(2 Since p-value < 0.05, we reject the null hypothesis. We conclude that the mean price
of car in 2012 has change significantly compared to 2011.
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From formula:
2
2
2
1
2
1
2/2121
2
2
2
1
2
1
2/21)()()(
n
S
n
Szxx
n
S
n
Szxx
Find a 95% CI on the difference in means.
200
2000150
150096.14800045000200
2000150
150096.1480004500022
21
22
32.263367.336621
Since 1 2 0is not in the interval then we reject H0.
Both propellants are not the same mean burning rate.
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Exercise:
In a test to compare the effectiveness of two drugs designed
to lower cholesterol levels, 75 randomly selected patientswere given drug A and 100 randomly selected patients were
given drug B. Those given drug A reduced their cholesterol
levels by an average of 40mg with standard deviation 12mg,
and those given drug B reduced their cholesterol levels by anaverage 42mg with standard deviation of 15mg.
i. Perform the test for the above investigation. Can you
conclude that the mean reduction using drug B is greater
than that of drug A?ii. Find the two sided 95% CI for the difference between the
two means of reductions using drug A and drug B. What is
your conclusion comparing with part (i).
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TEST FOR TWO PROPORTIONS
Alternative Hypothesis Rejection Criteria ( Reject H0)
Null Hypothesis:210: ppH ):,:(or 210210 ppHppH
2/2/ zZorzZ 211: ppH
zZ211: ppH
211: ppH zZ
Test Problems about two proportions:
Test statistic:
21
21
21
2
22
1
11
n1
n1
21
ppwhen)1,0(NZ
nn
XXp,
n
Xp,
n
Xp;
))(p1(p
ppZ
21
21
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A two sided confidence interval for is:)%1(100 21 pp
ULppLL
n
pp
n
ppzppUL
n
pp
n
ppzppLL
21
2
22
1
112/21
2
22
1
112/21
)1()1()(
)1()1()(
Confidence interval:
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Solution
Step 1 Problem: Test about two proportions large samples
Step 2 H0: pA= pB H1: pA> pB
Example 4:In a study on the effects of sodium restricted diets on hypertension, 24
out of 55 hypertensive patients were on sodium restricted diets, and 36
out of 149 non-hypertensive patients were on sodium restricted diets.
i. Test the hypothesis that the proportion of patients on sodium
restricted diets is higher for hypertensive patients at a=0.05.
ii. What is the P-value for this test?
iii. Construct a two sided 95% CI and comment.
Step 3 Determine the appropriate test statistic
BA
BA
B
B
B
A
A
A
BA
BA
BA
nn
XXp,
n
Xp,
n
Xpwhere
ppifN(0,1)))(p(1p
ppZ
n1
n1
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Step 4 Critical value given 0.05 z= z 0.05 = 1.645
645.1jectRe 0 ZifH
Step 5 Rejection region for the statistic
Step 6 Compute the value of the test statistic:
29.014955
3624
where,9455.4
149
1
55
1)30.01(30.0
24.044.0
24.0,44.0,36,24,149,55
0
pz
ppxxnn BABABA
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Step 7.Make a decision
Since z0> 1.65, then we reject H0. It means that enough
evidence to claim that the proportion of patients on
hypertension is higher than non hypertension patients
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Example 5:
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Solution
Step 1 Problem: Test about two proportions large samples
Step 2 H0: p1= p2 H1: p1> p2
Example 5:
A vote is to be taken among residents of a town and the
surrounding county to determine whether a proposed chemical
plant should be constructed. If 120 of 200 town voters favour the
proposal and 240 of 500 county residents favour it, would youagree that the proportion of town voters favouring the proposal is
higher than the proportion of county voters? Use 0.05
Step 3 Determine the appropriate test statistic
21
21
2
22
1
11
21
n1
n1
21
nn
XXp,
n
Xp,
n
Xpwhere
ppifN(0,1)))(p(1p
ppZ
21
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27
Step 4 Critical value given 0.05 z = z 0.05 = 1.645
645.1jectRe 0 ZifH
Step 5 Rejection region for the statistic
Step 6 Compute the value of the test statistic:
500,200,240,120 2121 nnXX
21
21
2
22
1
11
21
n1
n1
21
nn
XXp,
n
Xp,
n
Xpwhere
ppifN(0,1)))(p(1p
ppZ
21
2.9
500200
240120,
500
240,
200
120 21
Z
ppp
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Step 7.Make a decision
Decision: since , we reject H0 and agree
that the proportion of town voters favouring the
proposal is higher than the proportion of county voters.
645.19.2 Z
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TEST FOR TWO V RI NCES
Alternative Hypothesis Rejection Criteria ( Reject H0)
):
or,:or:vs:
2
2
2
11
2
2
2
11
2
2
2
11
2
2
2
10
H
HHH
2
2
2
11: H
2
2
2
11: H
2
2
2
11: H
Test Problems about two variances:
Test statistic:
29
2
2
2
10
s
s
F
1,1,2/101,1,2/0 2121or
nnnn fFfF
1,1,0 21 nnfF
1,1,0 21 nnfF
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A two sided confidence interval for is:)%1(100 2
2
2
1
Confidence interval:
30
1,1,2/22
2
1
22
2
1
1,1,2/122
2
1
1212
nnnn fs
s
fs
s
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Example 6:
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Solution
Step 1 Problem: Test about two population variances
Step 2
Example 6:A random sample of 12 air pollution index at UTP station produced a
variance 0.0340 while a random sample of another 13 air pollution index
at Tronoh station produced a variance 0.0525.
i. Are the population variances equal?. Use a = 0.05.
ii. Find the 95% two-sided confidence interval on the ratio of two
variances.
Step 3 Determine the appropriate test statistic
2
2
2
11
2
2
2
10 :vs: HH
2
2
21
0s
sF
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Step 4 Critical value a=0.05, fa/2, 12,13 = 3.15
Step 5 Rejection region: Reject IF
Step 6 Compute the value of the test statistic:
32.015.3
11or15.3
13,12,025.0
13,12,975.0013,12,025.00 f
fFfF
0H
648.00525.0
0340.02
2
2
10
s
sF
Step 7.Make a decision
Since 0.32
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A 95% two sided confidence interval for is:222
1
Confidence interval:
33
041.2206.0
)15.3(0525.0
0340.0
15.3
1
0525.0
0340.0
1
2
2
2
1
2
2
2
1
1,1,2/2
2
2
12
2
2
1
1,1,2/
2
2
2
1
1,1,2/2
2
2
1
2
2
2
11,1,2/12
2
2
1
12
12
1212
nn
nn
nnnn
fss
fss
fs
sf
s
s
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