1. Statistical Hypothesis Testing Review
A statistical hypothesis is an assertion concerning one or more populations.
In statistics, a hypothesis test is conducted on a set of two mutually exclusive statements:
H0 : null hypothesis
H1 : alternate hypothesis
Example
H0 : μ = 17
H1 : μ ≠ 17
We sometimes refer to the null hypothesis as the “equals” hypothesis.
Draw critical region, critical value.
EGR Ch10 Lec2and3 9th edition rev3
EGR 252 F06 Ch. 10 8th edition

2. Potential errors in decision-makingα
Probability of committing a Type I error
Probability of rejecting the null hypothesis given that the null hypothesis is true
P (reject H0 | H0 is true) β
Probability of committing a Type II error
Power of the test = 1 - β
(probability of rejecting the null hypothesis given that the alternate is true.)
Power = P (reject H0 | H1 is true)
Power of the test = 1- β
EGR Ch10 Lec2and3 9th edition rev3
EGR 252 F06 Ch. 10 8th edition

3. 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 edition rev3
EGR 252 F06 Ch. 10 8th edition

4. 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 edition rev3
EGR 252 F06 Ch. 10 8th edition

5. Example: Single Sample Test of the Mean P-value Approach
A sample of 20 cars driven under varying highway conditions achieved fuel efficiencies as follows:
Sample mean x = mpg
Sample std dev s = mpg
Test the hypothesis that the population mean equals 35.0 mpg vs. μ < 35.
Step 1: State the hypotheses.
H0: μ = 35
H1: μ < 35
Step 2: Determine the appropriate test statistic.
σ unknown, n = 20 Therefore, use t distribution
H0 : μ = 35
H1 : μ < 35
n = 20
use t-distribution
EGR Ch10 Lec2and3 9th edition rev3
EGR 252 F06 Ch. 10 8th edition

6. Single Sample Example (cont.)
Approach 2: =
Find probability from chart or use Excel’s tdist function.
P(x ≤ ) = TDIST (1.118, 19, 1) =
p = 0.14
0______________ Decision: Fail to reject null hypothesis
Conclusion: The mean is not significantly less than 35 mpg.
t = =( )/(2.915/(SQRT(20)))
P = =TDIST(1.118,19,1)
draw the graphs.
NOTE: if we look at table A.4, pg. 672, the α value associated to (by symmetry) falls between 0.15 and 0.10.
13.86% of the area under the curve lies to the left of t = Judge that H0 is plausible (fail to reject) and conclude that μ does not differ significantly from 35.
EGR Ch10 Lec2and3 9th edition rev3
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7. Example (concl.) Approach 1: Predetermined significance level (alpha)
Step 1: Use same hypotheses.
Step 2: Let’s set alpha at 0.05.
Step 3: Determine the critical value of t that separates the “reject H0 region” from the “do not reject H0 region”.
t, n-1 = t0.05,19 = 1.729
Since H1 specifies “< ” we declare tcrit =
Step 4: Using the equation, we calculate tcalc =
Step 5: Decision Fail to reject H0
Step 6: Conclusion: The mean is not significantly less than 35 mpg.
(one-sided or one-tailed test)
t0.05,19 = 1.729, tcrit =
t =
draw the picture …
EGR Ch10 Lec2and3 9th edition rev3
EGR 252 F06 Ch. 10 8th edition

8. Your turn … same data, different hypotheses
A sample of 20 cars driven under varying highway conditions achieved fuel efficiencies as follows:
Sample mean = mpg
Sample std dev (s) = mpg
Test the hypothesis that the population mean equals 35.0 mpg vs. μ ≠ 35 at an α level of Be sure to draw the picture.
Step 1
Step 2
Step 3
Step 4
Step 5
Step 6 (Conclusion will be different.)
t0.025,19 =
2.093
EGR Ch10 Lec2and3 9th edition rev3
EGR 252 F06 Ch. 10 8th edition

9. 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
EGR Ch10 Lec2and3 9th edition rev3

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)
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11. 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 edition rev3
EGR 252 F06 Ch. 10 8th edition

12. Your turn … Recall  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 edition rev3
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 edition rev3
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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 edition rev3
EGR 252 F06 Ch. 10 8th edition

15. Upper-tail test (μ2 – μ1 > 0) Conclusions
The data do 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 do 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 edition rev3
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
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 edition rev3
EGR 252 F06 Ch. 10 8th edition

17. Another Example: Low Carb Meals
Suppose we want to test the difference in carbohydrate content between two “low-carb” meals. Random samples of the two meals are tested in the lab and the carbohydrate content per serving (in grams) is recorded, with the following results:
n1 = 15 x1 = s12 = 11
n2 = 10 x2 = s22 = 23
tcalc = ______________________
ν = ______________
(using equation in table 10.3)
tcalc = =( )/(SQRT(11/15+23/10)) =
df from equation in table 10.2, pg. 313
ν = ≈ 15
EGR Ch10 Lec2and3 9th edition rev3
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18. Example (cont.) What are our options for hypotheses?
H0: μ1 - μ2 = or H0: μ1 - μ2 = 0
H1: μ1 - μ2 > 0 H1: μ1 - μ2 ≠ 0
At an α level of 0.05,
One-tailed test, t0.05, 15 = 1.753
Two-tailed test, t0.025, 15 = 2.131
How are our conclusions affected?
Our data don’t support a conclusion that the mean carb content of the two meals are different at an alpha level of (What is H1 ?)
Our data do support a conclusion that Meal 1 has more average carbs than Meal 2 at an alpha level of (What is H1 ?)
H0: μ1 - μ2 = 0 vs H0: μ1 - μ2 = 0
H1: μ1 - μ2 > 0 H1: μ1 - μ2 ≠ 0
t0.05, 15 = 1.753
t0.025, 15 = 2.131
Our data don’t support a conclusion that the two meals are different at an alpha level of .05
Our data do support a conclusion that meal 1 has more carbs than meal 2 at an alpha level of .05
note,
1-sided p-value =TDIST(1.895,15,1) =
2-sided p-value =TDIST(1.895,15,2) =
EGR Ch10 Lec2and3 9th edition rev3
EGR 252 F06 Ch. 10 8th edition

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 edition rev3
EGR 252 F06 Ch. 10 8th edition

20. Sheer Strength Example*
An article in the Journal of Strain Analysis compares several methods for predicting the shear strength of steel plate girders. Data for two of these methods, when applied to nine specific girders, are shown in the table on the next slide. We would like to determine if there is any difference, on average, between the two methods.
Procedure: We will conduct a paired-sample t-test at the 0.05 significance level to determine if there is a difference between the two methods.
* adapted from Montgomery & Runger, Applied Statistics and Probability for Engineers.
difference scores, d
0.119
0.159
etc.
EGR Ch10 Lec2and3 9th edition rev3
EGR 252 F06 Ch. 10 8th edition

21. Sheer Strength Example Data
Girder
Karlsruhe Method
Lehigh
Method
Difference (d) 1
1.186
1.061
0.125 2
1.151
0.992
0.159 3
1.322
1.063
0.259 4
1.339
1.062
0.277 5
1.200
1.065
0.135 6
1.402
1.178
0.224 7
1.365
1.037
0.328 8
1.537
1.086
0.451 9
1.559
1.052
0.507
EGR Ch10 Lec2and3 9th edition rev3
EGR 252 F06 Ch. 10 8th edition

22. Sheer Strength Example Calculations
Hypotheses:
H0: μD = 0
H1: μD ≠ t0.025,8 = Why 8?
Calculation of difference scores (d), mean and standard deviation, and tcalc …
d =
sd =
tcalc = ( d – d0 ) = ( ) =
sd / sqrt(n) ( / 3)
t0.025,8 = 2.306
difference scores – previous page
tcalc = (dbar-d0)
sd/sqrt(n)
EGR Ch10 Lec2and3 9th edition rev3
EGR 252 F06 Ch. 10 8th edition

23. What does this mean? Draw the graphic: Decision: Conclusion:
Graphic: t-test with tcrit = (lower boundary) and 2.306(upper boundary) and tcalc = 6.05
Decision: reject H0
Conclusion: The two methods produce different results, on average.
Since we subtracted Lehigh scores from Karlsruhe scores, and the resulting difference scores were positive, we have evidence that the Karlsruhe method yields larger strength predictions.
EGR Ch10 Lec2and3 9th edition rev3
EGR 252 F06 Ch. 10 8th edition

24. 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 edition rev3

25. 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
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26. 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 edition rev3
EGR 252 F06 Ch. 10 8th edition

27. 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 edition rev3
EGR 252 F06 Ch. 10 8th edition

28. 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 edition rev3
EGR 252 F06 Ch. 10 8th edition

29. 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 edition rev3
EGR 252 F06 Ch. 10 8th edition

30. The Chi-Squared Test in Minitab 15
Chi-Square Test: pp1, pp2, pp3
Expected counts are printed below observed counts
Chi-Square contributions are printed below expected counts
pp pp pp Total
Total
Test statistic: Chi-Sq calc = , DF = 2, P-Value = Reject Ho.
Conclude that worker and plan are not independent.
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 edition rev3
EGR 252 F06 Ch. 10 8th edition