PERTEMUAN 5

PENGUJIAN HIPOTESIS (part 2)

Mata Kuliah : Metode Statistika 2

Politeknik Statistika STIS
 OUTLINE
Prosedur Pengujian Hipotesis untuk:
◼ Beda 2 rata-rata pada sampel independen

◼ Beda 2 rata-rata pada sampel dependen

(data berpasangan)
◼ Beda 2 proporsi

◼ Rasio 2 varians
 Hypothesis Tests for the Difference
Between Two Means

◼ Testing Hypothesis about μ1 – μ2

◼ Use the same situations discussed already:

◼ Standard deviations known or unknown
◼ Sample sizes  30 or not  30
 Hypothesis Tests for
Two Population Proportions
Two Population Means, Independent Samples

Lower tail test: Upper tail test: Two-tailed test:

H0: μ1  μ2 H0: μ1 ≤ μ2 H0: μ1 = μ2

HA: μ1 < μ2 HA: μ1 > μ2 HA: μ1 ≠ μ2
i.e., i.e., i.e.,
H0: μ1 – μ2  0 H0: μ1 – μ2 ≤ 0 H0: μ1 – μ2 = 0
HA: μ1 – μ2 < 0 HA: μ1 – μ2 > 0 HA: μ1 – μ2 ≠ 0
 Hypothesis tests for μ1 – μ2

Population means, independent samples

σ1 and σ2 known Use a z test statistic

Use s to estimate unknown
σ1 and σ2 unknown, σ , approximate with a z
n1 and n2  30 test statistic

σ1 and σ2 unknown, Use s to estimate unknown

n1 or n2 < 30 σ , use a t test statistic
 σ1 and σ2 known

Population means,
independent The test statistic for
samples μ1 – μ2 is:

σ1 and σ2 known * z=
( x − x )− ( μ − μ )
1 2 1 2
2 2
σ1 and σ2 unknown, σ σ2
n1 and n2  30
1
+
n1 n2
σ1 and σ2 unknown,
n1 or n2 < 30
 σ1 and σ2 unknown, large samples

Population means,
independent The test statistic for
samples μ1 – μ2 is:

σ1 and σ2 known
z=
( x − x )− ( μ − μ )
1 2 1 2

σ1 and σ2 unknown,
n1 and n2  30
* s
+
1
2
s2
2

n1 n2
σ1 and σ2 unknown,
n1 or n2 < 30
 σ1 and σ2 unknown, small samples
*Assuming equal variances
Population means,
independent The test statistic for μ1 – μ2 is:

( x − x )− ( μ − μ )
samples

t=
1 2 1 2
σ1 and σ2 known
1 1
sp +
σ1 and σ2 unknown, n1 n 2
n1 and n2  30
Where t/2 or t has (n1 + n2 – 2) d.f.,
σ1 and σ2 unknown,
n1 or n2 < 30
* and

sp =
(n1 − 1)s12 + (n2 − 1)s22
n1 + n2 − 2
 σ1 and σ2 unknown, small samples
*Assuming different variances
Population means,
independent The test statistic for μ1 – μ2 is:

( x − x )− ( μ − μ )
samples

t=
1 2 1 2
σ1 and σ2 known 2 2
s s
1
+ 2
σ1 and σ2 unknown, n1 n 2
n1 and n2  30
Where t/2 or t has d.f.,
2 2
σ1 and σ2 unknown, * (s
s
(1 n1
s
+ 2 n2 ) 2

n ) /(n − 1)+ (s 

v=
n1 or n2 < 30 2
1 1
2
1
2
2 )
2
n2 /(n2 − 1)
 Hypothesis tests for μ1 – μ2
Two Population Means, Independent Samples
Lower tail test: Upper tail test: Two-tailed test:
H0: μ1 – μ2  0 H0: μ1 – μ2 ≤ 0 H0: μ1 – μ2 = 0
HA: μ1 – μ2 < 0 HA: μ1 – μ2 > 0 HA: μ1 – μ2 ≠ 0

  /2 /2

-z z -z/2 z/2

Reject H0 if z < -z Reject H0 if z > z Reject H0 if z < -z/2
or z > z/2
 Pooled sp t Test: Example
You’re a financial analyst for a brokerage firm. Is there a
difference in dividend yield between stocks listed on the
NYSE & NASDAQ? You collect the following data:
NYSE NASDAQ
Number 21 25
Sample mean 3.27 2.53
Sample std dev 1.30 1.16

Assuming equal variances, is

there a difference in average
yield ( = 0.05)?
 Solution
Reject H0 Reject H0
H0: μ1 - μ2 = 0 i.e. (μ1 = μ2)
HA: μ1 - μ2 ≠ 0 i.e. (μ1 ≠ μ2)
 = 0.05 .025 .025

df = 21 + 25 - 2 = 44 -2.0154 0 2.0154 t
Critical Values: t = ± 2.0154
2.040
Test Statistic: Decision:
3.27 − 2.53 Reject H0 at  = 0.05
t= = 2.040
1 1
1.2256 + Conclusion:
21 25 There is evidence of a
difference in means.
 Calculating the Test Statistic
The test statistic is:

t=
(x − x )− (μ − μ ) (3.27 − 2.53) − 0
1 2
= 1 2
= 2.040
1 1 1 1
sp + 1.2256 +
n1 n 2 21 25

sp =
(n1 − 1)s12 + (n2 − 1)s22 =
(21− 1)1.302 + (25 − 1)1.162 = 1.2256
n1 + n2 − 2 21+ 25 − 2
 Hypothesis Testing for
Paired Samples

The test statistic for d is

Paired
samples
d − μd
t=
sd
n
n is the
number n
of pairs
in the
Where t/2 has n - 1 d.f.  i
(d − d) 2

paired and sd is: sd = i=1

sample n −1
 Hypothesis Testing for
Paired Samples
(continued)
Paired Samples

Lower tail test: Upper tail test: Two-tailed test:

H0: μd  0 H0: μd ≤ 0 H0: μd = 0

HA: μd < 0 HA: μd > 0 HA: μd ≠ 0

  /2 /2

-t t -t/2 t/2

Reject H0 if t < -t Reject H0 if t > t Reject H0 if t < -t/2
or t > t/2
Where t has n - 1 d.f.
 Paired Samples Example
◼ Assume you send your salespeople to a “customer
service” training workshop. Is the training effective?
You collect the following data:

Number of Complaints: (2) - (1)  di

Salesperson Before (1) After (2) Difference, di d = n
C.B. 6 4 - 2
= -4.2
T.F. 20 6 -14
M.H. 3 2 - 1
R.K.
M.O.
0
4
0
0
0
- 4 sd =
 i
(d − d) 2

-21 n −1
= 5.67
 Paired Samples: Solution
▪ Has the training made a difference in the number of
complaints (at the 0.01 level)?
Reject Reject
H0: μd = 0
HA: μd  0
/2 /2
 = .01 d = - 4.2 - 4.604 4.604
- 1.66
Critical Value = ± 4.604
d.f. = n - 1 = 4
Decision: Do not reject H0
(t stat is not in the reject region)
Test Statistic:
Conclusion: There is not a
d − μd − 4.2 − 0
t= = = −1.66 significant change in the
sd / n 5.67/ 5 number of complaints.
 Hypothesis Tests for
Two Population Proportions

Population proportions

Lower tail test: Upper tail test: Two-tailed test:

H0: p1  p2 H0: p1 ≤ p2 H0: p1 = p2

HA: p1 < p2 HA: p1 > p2 HA: p1 ≠ p2
i.e., i.e., i.e.,
H0: p1 – p2  0 H0: p1 – p2 ≤ 0 H0: p1 – p2 = 0
HA: p1 – p2 < 0 HA: p1 – p2 > 0 HA: p1 – p2 ≠ 0
 Two Population Proportions
Since we begin by assuming the null
hypothesis is true, we assume p1 = p2
Population
and pool the two 𝑝Ƹ estimates
proportions
The pooled estimate for the
overall proportion is:

n1 pˆ1 + n 2 pˆ 2 x1 + x 2
pˆ = =
n1 + n 2 n1 + n 2
where x1 and x2 are the numbers from
samples 1 and 2 with the characteristic of interest
 Two Population Proportions
(continued)

Population The test statistic for

proportions p1 – p2 is:

z=
( pˆ1 − pˆ 2 ) − ( p1 − p2 )
1 1
pˆ (1 − pˆ )  + 
 n1 n2 
 Hypothesis Tests for
Two Population Proportions
Population proportions
Lower tail test: Upper tail test: Two-tailed test:
H0: p1 – p2  0 H0: p1 – p2 ≤ 0 H0: p1 – p2 = 0
HA: p1 – p2 < 0 HA: p1 – p2 > 0 HA: p1 – p2 ≠ 0

  /2 /2

-z z -z/2 z/2

Reject H0 if z < -z Reject H0 if z > z Reject H0 if z < -z/2
or z > z/2
 Example:
Two population Proportions

Is there a significant difference between the

proportion of men and the proportion of
women who will vote Yes on Proposition A?

◼ In a random sample, 36 of 72 men and 31 of

50 women indicated they would vote Yes

◼ Test at the .05 level of significance

 Example:
Two population Proportions
(continued)
◼ The hypothesis test is:
H0: p1 – p2 = 0 (the two proportions are equal)
HA: p1 – p2 ≠ 0 (there is a significant difference between proportions)

◼ The sample proportions are:

◼ Men: 𝑝Ƹ1 = 36/72 = .50
◼ Women: 𝑝Ƹ 2 = 31/50 = .62

▪ The pooled estimate for the overall proportion is:

x1 + x 2 36 + 31 67
pˆ = = = = .549
n1 + n 2 72 + 50 122
 Example:
Two population Proportions
(continued)
Reject H0 Reject H0
The test statistic for p1 – p2 is:
.025 .025
z=
( pˆ1 − pˆ 2 ) − ( p1 − p2 )
1 1 
pˆ (1 − pˆ ) +  -1.96 1.96
 n1 n 2  -1.31
=
( .50 − .62) − ( 0) = − 1.31
 1 1 Decision: Do not reject H0
.549 (1 − .549)  + 
 72 50 
Conclusion: There is not
significant evidence of a
Critical Values = ±1.96
For  = .05 difference in proportions
who will vote yes between
men and women.
 Hypothesis Tests for
Two Population Variances
Hypothesis Tests for Variances

H0: σ12 – σ22 = 0

Two tailed test
* Tests for Two
HA: σ1 – σ2 ≠ 0
2 2 Population Variances

H0: σ12 – σ22  0 Lower tail test

HA: σ12 – σ22 < 0 F test statistic

H0: σ12 – σ22 ≤ 0 Upper tail test

HA: σ12 – σ22 > 0
 F Test for Difference in Two
Population Variances
Hypothesis Tests for Variances
The F test statistic is:
2
s Tests for Two
F= 1
2 Population Variances
s 2

s12 = Variance of Sample 1

* F test statistic
n1 - 1 = numerator degrees of freedom

s22 = Variance of Sample 2

n2 - 1 = denominator degrees of freedom
 Finding the Critical Value
H0: σ12 – σ22 = 0
H0: σ12 – σ22 ≤ 0 HA: σ12 – σ22 ≠ 0
HA: σ12 – σ22 > 0

 /2
/2
0 F 0 F
Do not Reject H0 Do not Reject H0
reject H0 F reject H0 F/2
◼ rejection region ◼ rejection region for
for a one-tail test a two-tailed test is
(upper tail test) is
s12 s 2
s 2
F = 2  F / 2 or F = 12  F1− / 2
F=  F1
2 s2 s2
s 2
 F Test: An Example

You are a financial analyst for a brokerage firm. You

want to compare dividend yields between stocks listed
on the NYSE & NASDAQ. You collect the following data:
NYSE NASDAQ
Number 21 25
Mean 3.27 2.53
Std dev 1.30 1.16

Is there a difference in the

variances between the NYSE &
NASDAQ at the  = 0.1 level?
 F Test: Example Solution
◼ Form the hypothesis test:
H0: σ21 – σ22 = 0 (there is no difference between variances)
HA: σ21 – σ22 ≠ 0 (there is a difference between variances)

◼ Find the F critical value for  = .1:

◼ Numerator:

◼ df1 = n1 – 1 = 21 – 1 = 20

◼ Denominator:

◼ df2 = n2 – 1 = 25 – 1 = 24

F.95, 20, 24 = 0,48 or F.05, 20, 24 = 2.03

 F Test: Example Solution
(continued)

◼ The test statistic is: H0: σ12 – σ22 = 0

HA: σ12 – σ22 ≠ 0
s12 1.302
F= 2 = 2
= 1.256
s2 1.16
/2 = .05

0
◼Conclusion : do not reject H0 Do not Reject H0
reject H0 F/2 =2.03
◼ There is no evidence of a

difference in variances at  = .1