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3 Introduction When the sample size is small, the estimation and testing procedures of large samples are not appropriate. There are equivalent small sample test and estimation procedures for , the mean of a normal population the difference between two population means 2, the variance of a normal population The ratio of two population variances.
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1
Pertemuan 09 & 10Pengujian Hipotesis
Mata kuliah : A0392 - Statistik EkonomiTahun : 2010
2
Outline Materi :
• Uji hipotesis rata-rata dan proporsi• Uji hipotesis beda dua rata-rata• Uji hipotesis beda dua proporsi
3
Introduction• When the sample size is small, the
estimation and testing procedures of large samples are not appropriate.
• There are equivalent small sample test and estimation procedures for , the mean of a normal populationthe difference between two population
means2, the variance of a normal populationThe ratio of two population variances.
4
The Sampling Distribution of the Sample Mean
• When we take a sample from a normal population, the sample mean has a normal distribution for any sample size n, and
• has a standard normal distribution. • But if is unknown, and we must use s to estimate it, the
resulting statistic is not normalis not normal.
nxz
/
normal!not is / ns
x
x
5
Student’s t Distribution• Fortunately, this statistic does have a sampling
distribution that is well known to statisticians, called the StudentStudent’’s t distribution, s t distribution, with nn-1 -1 degrees of freedom.degrees of freedom.
nsxt/
•We can use this distribution to create estimation testing procedures for the population mean .
6
Properties of Student’s t
• Shape depends on the sample size n or the degrees of degrees of freedom, freedom, nn-1.-1.
• As n increase , variability of t decrease because the estimate s is based on more and more information.
• As n increases the shapes of the t and z distributions become almost identical.
•Mound-shapedMound-shaped and symmetric about 0.•More variable than More variable than zz, with “heavier tails”
Applet
7
Using the t-Table• Table 4 gives the values of t that cut off certain
critical values in the tail of the t distribution.• Index dfdf and the appropriate tail area a a to find
ttaa,,the value of t with area a to its right.
For a random sample of size n = 10, find a value of t that cuts off .025 in the right tail.
Row = df = n –1 = 9
t.025 = 2.262
Column subscript = a = .025
8
Small Sample Inference for a Population Mean
• The basic procedures are the same as those used for large samples. For a test of hypothesis:
.1on with distributi- taon basedregion rejection aor values- using
/
statistic test theusing tailedor two one :H versus:HTest
0
a00
ndfp
nsxt
9
Example
A sprinkler system is designed so that the average time for the sprinklers to activate after being turned on is no more than 15 seconds. A test of 5 systems gave the following times:
17, 31, 12, 17, 13, 25 Is the system working as specified? Test using = .05.
specified) as ng(not worki 15:Hspecified) as (working 15:H
a
0
10
ExampleData:Data: 17, 31, 12, 17, 13, 25First, calculate the sample mean and standard deviation, using your calculator or the formulas in Chapter 2.
387.75
61152477
1
)(
167.196
115
222
nnxx
s
nxx i
11
ExampleData:Data: 17, 31, 12, 17, 13, 25Calculate the test statistic and find the rejection region for =.05.
5161 38.16/387.715167.19
/
:freedom of Degrees :statisticTest
0
ndfns
xt
Rejection Region: Reject H0 if t > 2.015. If the test statistic falls in the rejection region, its p-value will be less than = .05.
12
ConclusionData:Data: 17, 31, 12, 17, 13, 25Compare the observed test statistic to the rejection region, and draw conclusions.
.015.2 if HReject :RegionRejection 38.1 :statisticTest
0
t
t
Conclusion: For our example, t = 1.38 does not fall in the rejection region and H0 is not rejected. There is insufficient evidence to indicate that the average activation time is greater than 15.
15:H 15:H
a
0
13
Approximating the p-value
• You can only approximate the p-value for the test using Table 4.
Since the observed value of t = 1.38 is smaller than t.10 = 1.476,
p-value > .10.
14
The exact p-value
• You can get the exact p-value using some calculators or a computer.
One-Sample T: TimesTest of mu = 15 vs mu > 15
Variable N Mean StDev SE MeanTimes 6 19.17 7.39 3.02
Variable 95.0% Lower Bound T PTimes 13.09 1.38 0.113
p-value = .113 which is greater than .10 as we approximated using Table 4.
Applet
15
Testing the Difference between Two Means
normal. be must spopulation two the small, are sizes sample the Since
and variancesand and means with2 and 1 spopulation from drawn are and size of samples random tindependen 9,Chapter in As
22
21 .21
21
μμ
nn
•To test: •H0:D0 versus Ha:one of three
where D0 is some hypothesized difference, usually 0.
16
Testing the Difference between Two Means
•The test statistic used in Chapter 9
•does not have either a z or a t distribution, and cannot be used for small-sample inference. •We need to make one more assumption, that the population variances, although unknown, are equal.
2
22
1
21
21z
ns
ns
xx
17
Testing the Difference between Two Means
•Instead of estimating each population variance separately, we estimate the common variance with
2)1()1(
21
222
2112
nn
snsns
21
2
021
11
nns
Dxxthas a t distribution with n1+n2-2=(n1-1)+(n2-1) degrees of freedom.
•And the resulting test statistic,
18
Example• Two training procedures are compared by
measuring the time that it takes trainees to assemble a device. A different group of trainees are taught using each method. Is there a difference in the two methods? Use = .01.
Time to Assemble
Method 1 Method 2
Sample size 10 12
Sample mean 35 31
Sample Std Dev
4.9 4.5
0:H 210
21
2
21
11
0:statisticTest
nns
xxt
0:H 21a
19
Example• Solve this problem by approximating the p-
value using Table 4.
Time to Assemble
Method 1 Method 2
Sample size 10 12
Sample mean 35 31
Sample Std Dev 4.9 4.5
99.1121
101942.21
3135:statisticTest
t
Applet
942.2120
)5.4(11)9.4(9
2)1()1(
:Calculate
22
21
222
2112
nn
snsns
20
Example
value)-(21)99.1(
)99.1()99.1( :value-
ptP
tPtPp
.025 < ½( p-value) < .05
.05 < p-value < .10
Since the p-value is greater than = .01, H0 is not rejected. There is insufficient evidence to indicate a difference in the population means.
df = n1 + n2 – 2 = 10 + 12 – 2 = 20
21
Testing the Difference between Two Means
•If the population variances cannot be assumed equal, the test statistic
•has an approximate t distribution with degrees of freedom given above. This is most easily done by computer.
2
22
1
21
21
ns
ns
xxt
1)/(
1)/(
2
22
22
1
21
21
2
2
22
1
21
nns
nns
ns
ns
df
22
The Paired-Difference Test
•Sometimes the assumption of independent samples is intentionally violated, resulting in a matched-pairsmatched-pairs or paired-difference testpaired-difference test.•By designing the experiment in this way, we can eliminate unwanted variability in the experiment by analyzing only the differences,
ddii = = xx11ii – – xx22ii
•to see if there is a difference in the two population means,
23
Example
• One Type A and one Type B tire are randomly assigned to each of the rear wheels of five cars. Compare the average tire wear for types A and B using a test of hypothesis.
Car 1 2 3 4 5Type A 10.6 9.8 12.3 9.7 8.8
Type B 10.2 9.4 11.8 9.1 8.3
0:H 0:H
21a
210
• But the samples are not independent. The pairs of responses are linked because measurements are taken on the same car.
24
The Paired-Difference Test
.1on with distributi- taon basedregion rejection aor value- theUse
. s,difference theofdeviation standard andmean theare and pairs, ofnumber where
/0
statistic test theusing 0 :H test we0:H test To d0210
ndfp
dsdn
nsdt
i
d
d
25
ExampleCar 1 2 3 4 5Type A 10.6 9.8 12.3 9.7 8.8
Type B 10.2 9.4 11.8 9.1 8.3
Difference .4 .4 .5 .6 .5
0:H 0:H
21a
210
.0837
.48 Calculate
1
22
nndd
s
ndd
ii
d
i8.12
5/0837.048.
/0
:statisticTest
ns
dtd
26
ExampleCarCar 1 2 3 4 5Type A 10.6 9.8 12.3 9.7 8.8
Type B 10.2 9.4 11.8 9.1 8.3
Difference .4 .4 .5 .6 .5
Rejection region: Reject H0 if t > 2.776 or t < -2.776.
Conclusion: Since t = 12.8, H0 is rejected. There is a difference in the average tire wear for the two types of tires.
27
Key Concepts
28
The test statistic is a value computed from the sample data, and it is used in making the decision about the rejection of the null hypothesis.
z = p - p
pqn
Test statistic for proportions
Test statistic for proportions
Solution: The preceding example showed that the given claim results in the following null and alternative hypotheses: H0: p = 0.5 and H1: p > 0.5. Because we work under the assumption that the null hypothesis is true with p = 0.5, we get the following test statistic:
npq
z = p – p
= 0.56 - 0.5
(0.5)(0.5) 880
= 3.56
Figure 7-3
31
Inferences about Two Proportions
Assumptions1. We have proportions from two
independent simple random samples.
2. For both samples, the conditions np 5 and nq 5 are satisfied.
32population 2.
The corresponding meanings are attachedto p
2, n
2 , x
2 , p
2. and q
2 , which come from^^
Notation for Two Proportions
For population 1, we let: p
1 = population proportion
n1 = size of the sample x1 = number of successes in the sample
p1 = x1 (the sample proportion)
q1 = 1 – p
1 ^^
n1
^
33
The pooled estimate of p1 and p
2
is denoted by p.
Pooled Estimate of p
1 and p
2
q = 1 – p
=p n1 + n
2
x1 + x
2
34
Test Statistic for Two Proportions
For H0: p1 = p
2 , H0: p1
p2 , H0: p1
p2
H1: p1
p2 , H1: p1
< p2 , H1: p
1> p
2
+z =
( p1
– p2 ) – ( p
1 – p
2 )^ ^
n1
pqn2
pq
35
Test Statistic for Two Proportions
where p1 – p
2 = 0 (assumed in the null hypothesis)
=p1
^ x1
n1
p2
^ x2
n2
=and
and q = 1 – pn1 + n
2
p = x
1 + x
2
For H0: p1 = p
2 , H0: p1
p2 , H0: p1
p2
H1: p1
p2 , H1: p1
< p2 , H1: p
1> p
2
Example:
Example:
200n1
n1= 200
x1 = 24
p1 = x1 = 24 = 0.120^
n2
n2 = 1400x2 = 147p2 = x2 = 147 = 0.105
1400^
H0: p1 = p2, H1: p1 > p2
p = x1 + x2 = 24 + 147 = 0.106875 n1 + n2 200+1400q = 1 – 0.106875 = 0.893125.
Example:
200n1
n1= 200
x1 = 24
p1 = x1 = 24 = 0.120^
n2
n2 = 1400x2 = 147p2 = x2 = 147 = 0.105
1400^
z = (0.120 – 0.105) – 0(0.106875)(0.893125) + (0.106875)(0.893125)
200 1400
z = 0.64
Example:
200n1
n1= 200
x1 = 24
p1 = x1 = 24 = 0.120^
n2
n2 = 1400x2 = 147p2 = x2 = 147 = 0.105
1400^
z = 0.64This is a right-tailed test, so the P-value isthe area to the right of the test statistic z = 0.64. The P-value is 0.2611.Because the P-value of 0.2611 is greater than the significance level of = 0.05, we fail to reject the null hypothesis.
Example: For the sample data listed in Table 8-1, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped.
200n1
n1= 200
x1 = 24
p1 = x1 = 24 = 0.120^
n2
n2 = 1400x2 = 147p2 = x2 = 147 = 0.105
1400^
z = 0.64Because we fail to reject the null hypothesis, we conclude that there is not sufficient evidence to support the claim that the proportion of black drivers stopped by police is greater than that for white drivers. This does not mean that racial profiling has been disproved. The evidence might be strong enough with more data.
Example:
200n1
n1= 200
x1 = 24
p1 = x1 = 24 = 0.120^
n2
n2 = 1400x2 = 147p2 = x2 = 147 = 0.105
1400^
4242
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