1. Lecture 2 2-Sample Tests Goodness of Fit Tests for Independence
Hypothesis Test
Lecture 2
2-Sample Tests
Goodness of Fit
Tests for Independence
EGR Ch10 Lec2and3 9th

2. 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
EGR Ch10 Lec2and3 9th

3. Hypothesis Testing Basics
Test statistic; n is large, standard deviation is known
Test statistic: n is small, standard deviation is unknown
EGR Ch10 Lec2and3 9th

4. Hypothesis Testing – Approach 1
Approach 1 - Fixed probability of Type 1 error.
State the null and alternative hypotheses.
Choose a fixed significance level α.
Specify the appropriate test statistic and establish the critical region based on α. Draw a graphic representation.
Calculate the value of the test statistic based on the sample data.
Make a decision to reject H0 or fail to reject H0, based on the location of the test statistic.
Make an engineering or scientific conclusion.
recall our question about the amount of coffee in the cup …
1. H0 : μ = 8 oz.
H1 : μ < 8 oz.
2. α = 0.05
3. zα =
5. if zcalc < , reject H0
if zcalc > , fail to reject H0
6. e.g., coffee in the cup is significantly less than 8 oz.
or coffee in the cup is not significantly less than 8 oz.
EGR Ch10 Lec2and3 9th
EGR 252 F06 Ch. 10 8th edition

5. Hypothesis Testing – Approach 2
Approach 2 - Significance testing based on the calculated P-value
State the null and alternative hypotheses.
Choose an appropriate test statistic.
Calculate value of test statistic and determine P-value. Draw a graphic representation.
Make a decision to reject H0 or fail to reject H0, based on the P-value.
Make an engineering or scientific conclusion.
recall our question about the amount of coffee in the cup …
1. H0 : μ = 8 oz.
H1 : μ < 8 oz.
2. if variance known or n large, z-test (assume z in this case)
3. from zcalc, determine p-value from table A.3 or as given by statistical software packages.
4. P = 0, H0 rejected / not plausible (e.g., coffee in the cup is significantly less than 8 oz.)
P = 1, H0 is not rejected (coffee in the cup is not significantly less than 8 oz.)
Note: Approach 1 is the classical method. Approach 2 is gaining acceptance, partly because of the increasing availability of statistical software packages. The conclusion based on a P-value requires judgment. The smaller the P-value, the less plausible is the null hypothesis.
p = ↓
P-value
0.25
0.50
0.75
1.00
P-value
EGR Ch10 Lec2and3 9th
EGR 252 F06 Ch. 10 8th edition

6. 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.)
EGR Ch10 Lec2and3 9th

7. Non-directional - Two-tail Test
m1-m2 ≠ d0
Critical/Reject Region
z <-za/2 or z >za/2
t <-ta/2 or t >ta/2
-ta/2
ta/2
EGR Ch10 Lec2and3 9th

8. Directional-Right-tail Test
m1-m2 > d0
Critical/Reject Region
z >za
t >ta ta
EGR Ch10 Lec2and3 9th

9. Directional-Left-tail Test
m1-m2 < d0
Critical/Reject Region
z <-za
t <-ta
-ta
EGR Ch10 Lec2and3 9th

10. 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
NOTE: d0 is often 0 (there is, statistically speaking, no difference in the means)
H0: μ1 - μ2 = 0
H1: μ1 - μ2 < 0 (note: compare lower to higher for lower-tail test)
H1: μ1 - μ2 ≠ 0
H1: μ1 – μ2 > 0 (note: compare higher to lower for upper-tail test)
EGR Ch10 Lec2and3 9th
EGR 252 F06 Ch. 10 8th edition

11. 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, 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, , hours
***Assume equal variances.
EGR Ch10 Lec2and3 9th

12. 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)
Recall 
Note that we have to compare higher – lower mean to conduct an upper-tail test
EGR Ch10 Lec2and3 9th
EGR 252 F06 Ch. 10 8th edition

13. Our Example – Hand Calculation
Reading:
n1 = 4 mean x1 = s12 = 1.363
No reading:
n2 = 5 mean x2 = s22 = 3.883
To conduct the test by hand, we must calculate sp2 .
= sp = 1.674
and = ????
t-test
sp2 = (3(1.363)+4(3.883))/(4+5-2) = , s = 1.674
EGR Ch10 Lec2and3 9th
EGR 252 F06 Ch. 10 8th edition

14. Lower-tail test (μ1 - μ2 < 0) Why?
Draw the picture:
Approach 1: df = 7, t0.05,7 =  tcrit =
Calculation:
tcalc = (( )-0)/(1.674*sqrt(1/4 + 1/5)) =
-0.70
Graphic:
Decision:
Conclusion:
tcalc = (( )-0)/(1.674*sqrt(1/4 + 1/5)) = -0.70
Approach 1:
df = 7, t0.05,7 =  tcrit =
Approach 2:
=TDIST(0.7,7,1) =
Decision: fail to reject H0
EGR Ch10 Lec2and3 9th
EGR 252 F06 Ch. 10 8th edition

15. 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.
The data does not support the hypothesis that the mean completion time is less for readers than for non-readers.
tcalc = (( )-0)/(1.674*sqrt(1/4 – 1/5)) = 0.70
Approach 1:
df = 7, t0.5,7 =  tcrit = 1.895
Approach 2:
=TDIST(0.7,7,1) =
Decision: fail to reject H0
Conclusion: the data do not support the hypothesis that the mean time to complete homework is more for students who do not read the textbook
EGR Ch10 Lec2and3 9th
EGR 252 F06 Ch. 10 8th edition

16. Our Example Using Excel
Reading: n1 = 4 mean x1 = s12 = 1.363
No reading: n2 = 5 mean x2 = s22 = 3.883
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
DoNotRead
Mean
2.075
2.860
Variance
1.3625
3.883
Observations 4 5
Pooled Variance
Hypothesized Mean Difference df 7
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
t-test
sp2 = (3(1.363)+4(3.883))/(4+5-2) = , s = 1.674
EGR Ch10 Lec2and3 9th
EGR 252 F06 Ch. 10 8th edition

17. Our Example Using Excel
Reading: n1 = 4 mean x1 = s12 = 1.363
No reading: n2 = 5 mean x2 = 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
DoNotRead
Mean
2.075
2.860
Variance
1.3625
3.883
Observations 4 5
Pooled Variance
Hypothesized Mean Difference df 7
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
t-Test: Two-Sample Assuming Unequal Variances
Read
DoNotRead
Mean
2.075
2.86
Variance
1.3625
3.883
Observations 4 5
Hypothesized Mean Difference df 7
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
t-test
sp2 = (3(1.363)+4(3.883))/(4+5-2) = , s = 1.674
EGR Ch10 Lec2and3 9th
EGR 252 F06 Ch. 10 8th edition

18. Examples
10.30
10.34
10.40 (Excel)
EGR Ch10 Lec2and3 9th

19. Special Case: Paired Sample T-Test
Which designs are paired-sample?
Car Radial Belted
1 ** ** Radial, Belted tires
2 ** ** placed on each car.
3 ** **
4 ** **
Person Pre Post
1 ** ** Pre- and post-test
2 ** ** administered to each
3 ** ** person.
Student Test1 Test2
1 ** ** 4 scores from test 1,
2 ** ** 4 scores from test 2.
paired-sample
paired sample
maybe – if we have information that the test1 and test2 scores can be matched to a particular individual for every subject in the study, it is a paired-sample experiment; otherwise it is a 2-sample experiment.
EGR Ch10 Lec2and3 9th
EGR 252 F06 Ch. 10 8th edition

20. Example (Paired)
10.43 (Excel)
EGR Ch10 Lec2and3 9th

21. 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.
EGR Ch10 Lec2and3 9th

22. 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
Compare χ2calc to χ2crit
Make a decision about H0
State your conclusion
draw χ2 graph
k = number of “cells” (note: some texts use k-1-h where h is the number of parameters in the distribution being tested – e.g., 1 for Poisson, 2 for normal)
Table A.5, pg
show calc and crit on the drawing
EGR Ch10 Lec2and3 9th
EGR 252 F06 Ch. 10 8th edition

23. Tests of Independence Salaried 160 140 40 340 Hourly 60 200 100 500
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.
Worker Type
Pension Plan
Total #1 #2 #3
Salaried
160
140 40
340
Hourly 60
200
100
500
EGR Ch10 Lec2and3 9th
EGR 252 F06 Ch. 10 8th edition

24. Calculation of Expected Values
Worker Type
Pension Plan
Total #1 #2 #3
Salaried
160
140 40
340
Hourly 60
200
100
500
2. Calculate expected probabilities
P(#1 ∩ S) = P(#1)*P(S) = (200/500)*(340/500)=0.272
E(#1 ∩ S) = * 500 = 136
P(#1 ∩ S) =P(#1)*P(S) = (200/500)*(340/500)= E(#1 ∩ S) = 0.272*500 = 136
P(#1 ∩ H) = P(#1)*P(H) = (200/500)*(160/500)= E(#1 ∩ H) = 64
#1 #2 #3
S (exp)
H(exp) #1 #2 #3
S (exp.)
136 ? 68
H (exp.) 64 32
EGR Ch10 Lec2and3 9th
EGR 252 F06 Ch. 10 8th edition

25. Calculate the Sample-based Statistic
Calculation of the sample-based statistic
= ( )^2/(136) + ( )^2/(136) +
… (60-32)^2/(32)
= 49.63
( )^2/136 + ( )^2/136 + (40-68)^2/68 + (40-64)^2/64 + (60-64)^2/64 + (60-32)^2/32
= 49.63
EGR Ch10 Lec2and3 9th
EGR 252 F06 Ch. 10 8th edition

26. 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:
2013
χ20.01,2__ = (from Table A.5, pp 740)
χ2calc> χ2crit so reject the null hypothesis and
conclude that worker and plan are not independent
EGR Ch10 Lec2and3 9th
EGR 252 F06 Ch. 10 8th edition