Download EGR252S14_Chapter10Lecture2and3_MDH 2016

Hypothesis Test
Lecture 2
2-Sample Tests
Goodness of Fit
Tests for Independence
EGR252 2015 Ch10 Lec2and3 9th
Slide 1
Hypothesis Testing Basics
 Null hypothesis must be accepted (fail to reject) or
rejected
 Test Statistic: A value which functions as the decision
maker. The decision to “reject” or “fail to reject” is based
on information contained in a sample drawn from the
population of interest.
 Rejection region: If test statistic falls in some interval
which support alternative hypothesis, we reject the null
hypothesis.
 Acceptance Region: It test statistic falls in some interval
which support null hypothesis, we fail to reject the null
hypothesis.
 Critical Value: The point which divide the rejection
region and acceptance
EGR252 2015 Ch10 Lec2and3 9th
Slide 2
Hypothesis Testing Basics
Test statistic; n is large, standard deviation is
known
Test statistic: n is small, standard deviation
is unknown
EGR252 2015 Ch10 Lec2and3 9th
Slide 3
Hypothesis Testing – Approach 1
 Approach 1 - Fixed probability of Type 1 error.
1. State the null and alternative hypotheses.
2. Choose a fixed significance level α.
3. Specify the appropriate test statistic and establish
the critical region based on α. Draw a graphic
representation.
4. Calculate the value of the test statistic based on
the sample data.
5. Make a decision to reject H0 or fail to reject H0,
based on the location of the test statistic.
6. Make an engineering or scientific conclusion.
EGR252 2015 Ch10 Lec2and3 9th
Slide 4
Hypothesis Testing – Approach 2
 Approach 2 - Significance testing based on the
calculated P-value
1. State the null and alternative hypotheses.
2. Choose an appropriate test statistic.
3. Calculate value of test statistic and determine
P-value. Draw a graphic representation.
4. Make a decision to reject H0 or fail to reject H0,
based on the P-value.
5. Make an engineering or scientific conclusion.
p = 0.05
P-value
0
↓
0.25
EGR252 2015 Ch10 Lec2and3 9th
0.50
0.75
1.00
P-value
Slide 5
Two Sample Hypothesis Test
Test relationships between two means
Hypothesis
 Ho: equals to hypothesis
 H1: m1-m2 ≠ d0
 H1: m1-m2 > d0
 H1: m1-m2 < d0
 Test statistic Selection
 Large samples/s1 and s2 known
 Small samples/ s1 = s2
 Small samples/ s1 ≠ s2
 Paired (before and after/ pre-post, etc.)
EGR252 2015 Ch10 Lec2and3 9th
Slide 6
Non-directional - Two-tail Test
m1-m2 ≠ d0
Do Not
Reject H
Reject H
a/2
0
0
Reject H
1-a
–za/2
-ta/2
Critical/Reject Region
z <-za/2 or z >za/2
t <-ta/2 or t >ta/2
EGR252 2015 Ch10 Lec2and3 9th
a/2
za/2
ta/2
0
Slide 7
Directional-Right-tail Test
m1-m2 > d0
Critical/Reject Region
z >za
t >ta
Do Not Reject H
0
Reject H
1-a
EGR252 2015 Ch10 Lec2and3 9th
0
a
za
ta
Slide 8
Directional-Left-tail Test
m1-m2 < d0
Critical/Reject Region
z <-za
t <-ta
Reject H
0
Do Not Reject H
1-a
a
0
–za
-ta
EGR252 2015 Ch10 Lec2and3 9th
Slide 9
Two-Sample Hypothesis Testing
 Define the difference in the two means as:
μ1 - μ2 = d0
where d0 is the actual value of the hypothesized difference
 What are the Hypotheses?
H0: _______________
H1: _______________
or
H1: _______________
or
H1: _______________
EGR252 2015 Ch10 Lec2and3 9th
Slide 10
Two-Sample Hypothesis Testing
A professor has designed an experiment to test the effect of
reading the textbook before attempting to complete a homework
assignment. Four students who read the textbook before
attempting the homework recorded the following times (in hours)
to complete the assignment:
3.1, 2.8, 0.5, 1.9 hours
Five students who did not read the textbook before attempting the
homework recorded the following times to complete the
assignment:
0.9, 1.4, 2.1, 5.3,
***Assume equal variances.
EGR252 2015 Ch10 Lec2and3 9th
4.6 hours
Slide 11
Hypothesis Tests to Conduct
 Lower-tail test (μ1 - μ2 < 0)
 “Fixed α” approach (“Approach 1”) at α = 0.05 level.
 “p-value” approach (“Approach 2”)
 Upper-tail test (μ2 – μ1 > 0)
 “Fixed α” approach at α = 0.05 level.
 “p-value” approach
 Two-tailed test (μ1 - μ2 ≠ 0)
 “Fixed α” approach at α = 0.05 level.
 “p-value” approach
Recall 
t calc
( x1 - x 2 ) - d0

s p 1/ n1  1/ n2
EGR252 2015 Ch10 Lec2and3 9th
Slide 12
Our Example – Hand Calculation
Reading:
n1 = 4 mean x1 = 2.075
s12 = 1.363
n2 = 5 mean x2 = 2.860
s22 = 3.883
No reading:
To conduct the test by hand, we must calculate sp2 .
2
2
(
n
1
)
s

(
n
1
)
s
= 2.803
1
2
2
s p2  1
n1  n2 - 2
and
t calc
( x1 - x 2 ) - d0

s p 1/ n1  1/ n2
EGR252 2015 Ch10 Lec2and3 9th
sp = 1.674
= ????
Slide 13
Lower-tail test (μ1 - μ2 < 0) Why?
Draw the picture:
 Approach 1: df = 7, t0.05,7 = 1.895  tcrit = -1.895
Calculation:
 tcalc = ((2.075-2.860)-0)/(1.674*sqrt(1/4 + 1/5)) =
-0.70
Graphic:
Decision:
Conclusion:
EGR252 2015 Ch10 Lec2and3 9th
Slide 14
Upper-tail test (μ2 – μ1 > 0)
Conclusions
 The data does not support the hypothesis that the
mean time to complete homework is less for
students who read the textbook.
or
 There is no statistically significant difference in the
time required to complete the homework for the
people who read the text ahead of time vs those
who did not.
or
 The data does not support the hypothesis that the
mean completion time is less for readers than for
non-readers.
EGR252 2015 Ch10 Lec2and3 9th
Slide 15
Our Example Using Excel
Reading:
No reading:
s12 = 1.363
s22 = 3.883
n1 = 4 mean x1 = 2.075
n2 = 5 mean x2 = 2.860
If we have reason to believe the population variances are “equal”, we can conduct a t- test
assuming equal variances in Minitab or Excel.
t-Test: Two-Sample Assuming Equal Variances
Read
Mean
Variance
Observations
Pooled Variance
EGR252 2015 Ch10 Lec2and3 9th
DoNotRead
2.075
2.860
1.3625
3.883
4
5
2.8027857
Hypothesized Mean Difference
0
df
7
t Stat
-0.698986
P(T<=t) one-tail
0.2535567
t Critical one-tail
1.8945775
P(T<=t) two-tail
0.5071134
t Critical two-tail
2.3646226
Slide 16
Our Example Using Excel
Reading:
No reading:
n1 = 4
n2 = 5
mean x1 = 2.075
mean x2 = 2.860
s12 = 1.363
s22 = 3.883
What if we do not have reason to believe the population variances are “equal”?
We can conduct a t- test assuming unequal variances in Minitab or Excel.
t-Test: Two-Sample Assuming Equal Variances
Read
Mean
Variance
Observations
Pooled Variance
DoNotRead
t-Test: Two-Sample Assuming Unequal Variances
2.075
2.860
1.3625
3.883
4
5
2.8027857
Read
Mean
DoNotRead
2.075
2.86
1.3625
3.883
Observations
4
5
Variance
Hypothesized Mean Difference
0
Hypothesized Mean Difference
0
df
7
df
7
t Stat
-0.698986
t Stat
P(T<=t) one-tail
0.2535567
P(T<=t) one-tail
0.2409258
t Critical one-tail
1.8945775
t Critical one-tail
1.8945775
P(T<=t) two-tail
0.5071134
P(T<=t) two-tail
0.4818516
t Critical two-tail
2.3646226
t Critical two-tail
2.3646226
EGR252 2015 Ch10 Lec2and3 9th
-0.7426759
Slide 17
Examples
10.30
10.34
10.40 (Excel)
EGR252 2015 Ch10 Lec2and3 9th
Slide 20
Special Case: Paired Sample T-Test
Which designs are paired-sample?
A. Car
Radial Belted
1
**
**
Radial, Belted tires
2
**
**
placed on each car.
3
**
**
4
**
**
B. Person Pre
Post
1
**
**
Pre- and post-test
2
**
**
administered to each
3
**
**
person.
4
**
**
C. Student Test1 Test2
1
**
**
4 scores from test 1,
2
**
**
4 scores from test 2.
3
**
**
4
**
**
EGR252 2015 Ch10 Lec2and3 9th
Slide 21
Example (Paired)
10.43 (Excel)
EGR252 2015 Ch10 Lec2and3 9th
Slide 22
Goodness-of-Fit Tests
Procedures for confirming or refuting
hypotheses about the distributions of random
variables.
Hypotheses:
H0: The population follows a particular distribution.
H1: The population does not follow the distribution.
Examples:
H0: The data come from a normal distribution.
H1: The data do not come from a normal
distribution.
EGR252 2015 Ch10 Lec2and3 9th
Slide 23
Goodness of Fit Tests: Basic Method
Test statistic is χ2
 Draw the picture
 Determine the critical value
χ2 with parameters α, ν = k – 1
Calculate χ2 from the sample
2
(
O
E
)
i
 2 calc   i
Ei
i 1
k
Compare χ2calc to χ2crit
Make a decision about H0
State your conclusion
EGR252 2015 Ch10 Lec2and3 9th
Slide 24
Tests of Independence
 Example:
500 employees were surveyed with respect to
pension plan preferences.
 Hypotheses
H0: Worker Type and Pension Plan are independent.
H1: Worker Type and Pension Plan are not independent.
 Develop a Contingency Table showing the observed values for
the 500 people surveyed.
Pension Plan
Worker Type
#1
#2
#3
Salaried
Hourly
160
40
140
60
40
60
340
160
Total
200
200
100
500
EGR252 2015 Ch10 Lec2and3 9th
Total
Slide 25
Calculation of Expected Values
Worker Type
Salaried
Hourly
Total
Pension Plan
#1
#2
160
140
40
200
60
200
#3
40
60
100
Total
340
160
500
2. Calculate expected probabilities
P(#1 ∩ S) = P(#1)*P(S) = (200/500)*(340/500)=0.272
E(#1 ∩ S) = 0.272 * 500 = 136
#1
#2
#3
S (exp.) 136
?
68
H (exp.)
?
32
64
EGR252 2015 Ch10 Lec2and3 9th
Slide 26
Calculate the Sample-based Statistic
Calculation of the sample-based statistic
(Oi - Ei ) 2
 calc  
Ei
i 1
k
2
= (160-136)^2/(136) + (140-136)^2/(136) +
… (60-32)^2/(32)
= 49.63
EGR252 2015 Ch10 Lec2and3 9th
Slide 27
The Chi-Squared Test of Independence
5. Compare to the critical statistic, χ2α, v
where v = (r – 1)(c – 1) Note: v is the symbol for degrees of freedom
For our example, suppose α = 0.01
χ2 0.01,2 = ___________
χ2 calc = ___________
Decision:
Conclusion:
EGR252 2015 Ch10 Lec2and3 9th
Slide 28