Download Review of Hypothesis Testing

Chapter 10, sections 1 and 4
Two-sample Hypothesis Testing
• Test hypotheses for the difference between
two independent population means (standard
deviations known or unknown)
• Use the F table to find critical F values
• Complete an F test for the difference
between two variances
Difference Between Two Means
Goal: Test hypothesis or form a confidence interval for the
difference between two population means, μ1 – μ2.
Assumptions:
• Different data sources-- populations are Unrelated and
Independent
• two samples are randomly and independently drawn from
these populations.
• Sample selected from one population has no effect on the
sample selected from the other population
• population distributions are normal or both sample sizes
are  30
Sample 1 : X 1, S1, n1
Population 1 : 1,  1
Sample 2 : X 2, S2 , n2
Population 2 :  2,  2
Possible Hypotheses Are:
Two-tail test:
Upper-tail test:
H0: μ1 = μ2
H1: μ1 ≠ μ2
H0: μ1 ≤ μ2
H1: μ1 > μ2
i.e.,
H0: μ1 – μ2 = 0
H1: μ1 – μ2 ≠ 0
i.e.,
H0: μ1 – μ2 ≤ 0
H1: μ1 – μ2 > 0
Lower-tail test:
H0: μ1  μ2
H1: μ1 < μ2
i.e.,
H0: μ1 – μ2  0
H1: μ1 – μ2 < 0
• Population standard deviations are known, σ1
and σ2 known, use Z test.
•
The test statistic for μ1 – μ2 , based on sample
sample means, ( X  X ) is:

X
Z
1
1
2
 X   μ  μ
2
1
σ
σ

n
n
2

2
2
1
2
1
2
Standard Error of the pooled population, σ
• The Confidence interval for μ1 – μ2 is:

( 1   2)  X1  X 2

2
2
σ1
σ2
Z

n1
n2
X1X 2
σ1 and σ2 Unknown
Assumptions:
 Samples are randomly and independently drawn
 Populations are normally distributed or both sample sizes are at least
30
 Population variances are unknown but assumed equal, σ1 and σ2
unknown, but known to be equal
 The population variances are assumed equal, so use the two sample
standard deviations and pool them to estimate σ
 The test Statistic for μ1 – μ2 has a t distribution with a degree
of freedom of (n1+n2-2):

X
t
1

 X 2   μ1  μ2 
1 1 
S p   
 n1 n2 
2
• Where:
2
Sp

n

1
 1 S 1  n2  1 S 2
(n1  1 )  ( n2  1)
2
2
• The confidence interval for μ1 – μ2 is:


(  1   2 )  X 1  X 2  t n1  n 2 - 2
1
1 
S   
 n1 n 2 
2
p
• Example:
• Last week you were given a sample of 69 beer, 54 are U.S.-made and
15 are foreign-made. Additional information were provided with
respect to price, calories, and percent alcohol content.
• Let’s assume the two populations are unrelated, independent, and
approximately normally distributed with equal variance
• Assume that the two samples are independently drawn.
Samples Information:
Price($) Calories %Alcohol Content
NON U.S. Sample
Mean
Std Deviation
Sample size
Variance
5.86
0.90
15
$0.82
138.60
27.61
15
$762.40
4.13
1.74
15
$3.04
Mean
Std Deviation
Sample size
Variance
4.71
1.48
54
2.18
143.39
30.67
54
940.73
4.50
1.53
54
2.35
U.S. Sample
• Questions:
1. Is there evidence of a difference in mean calories
of us and non-U.S. beers?
2. What is the 95% confidence interval for the
difference in mean calories?
3. Are conclusions in 1 and 2 consistent?
4. U.S. beers have about 10% more alcohol than
non-U.S. beers.
5. Is the assumption of equal population variances,
that you used for 1 and 4 a valid assumption?
6. Is there evidence that there is less variation in
price of imported beers than price of domestic
beers
Hypothesis Tests for Variances
• Test of two population variances
• Hypotheses:
H0: σ12 = σ22
H1: σ12 ≠ σ22
H0: σ12 / σ22=1
H1: σ12 / σ22≠1
H0: σ12  σ22
H1: σ12 < σ22
H0: σ12 / σ22 1
H1: σ12 / σ22 <1
Lower-tail test
H0: σ12 ≤ σ22
H1: σ12 > σ22
H0: σ12 / σ22 ≤1
H1: σ12 / σ22 >1
Upper-tail test
Two-tail test
• The test statistic from samples is
2
F(n
1
 1 ), ( n 2  1 ), 
S1
 2
S2
S12 = Variance of Sample 1
n1 - 1 = numerator degrees of freedom
S22 = Variance of Sample 2
n2 - 1 = denominator degrees of freedom
• F-Distribution can take values from 0 to
infinity
• It is a right-skewed distribution
1.
Finding the critical lower and upper tail values
2.
Find FU from the F table for n1 – 1 numerator and n2 – 1
denominator degrees of freedom
3.
Find FL using the formula:
•
FL 
1
FU*
Where FU* is from the F table with n2 – 1 numerator and n1 – 1
denominator degrees of freedom (i.e., switch the d.f. from FU)
/2
/2
Do not
reject H0
0
H0: σ12 = σ22
H1: σ12 ≠ σ22
F
FL
FU