Download EGR252S08 Chapter10 Lecture 1 JMB

Statistical Hypothesis Testing
 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.
JMB Ch 10 Lecture 1 8th edition
EGR 252 Spring 2008
Slide 1
Tests of Hypotheses - Graphics I
 We can make a decision about our hypotheses
based on our understanding of probability.
 We can visualize this probability by defining a
rejection region on the probability curve.
 The general location of the rejection region is
determined by the alternate hypothesis.
H0 : μ = _____
H1 : μ < _____
JMB Ch 10 Lecture 1 8th edition
H0 : μ = _____
H1 : μ ≠ _____
EGR 252 Spring 2008
H0 : p = _____
H1 : p > _____
Slide 2
Choosing the Hypotheses
Your turn …
Suppose a coffee vending machine claims it
dispenses an 8-oz cup of coffee. You have been
using the machine for 6 months, but recently it
seems the cup isn’t as full as it used to be. You
plan to conduct a statistical hypothesis test. What
are your hypotheses?
H0 : μ = _____
H1 : μ ≠ _____
?
H0 : μ = _____
H1 : μ < _____
JMB Ch 10 Lecture 1 8th edition
EGR 252 Spring 2008
Slide 3
Potential errors in decision-making
H0 True
H0 False
Do not
reject H0
Correct
Decision
Type II
error
Reject H0
Type I
error
Correct
Decision
 α
 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)
JMB Ch 10 Lecture 1 8th edition
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)
EGR 252 Spring 2008
Slide 4
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 or fail to reject H0, based
on the location of the test statistic.
6. Make an engineering or scientific conclusion.
JMB Ch 10 Lecture 1 8th edition
EGR 252 Spring 2008
Slide 5
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 or fail to reject H0,
based on the P-value.
5. Make an engineering or scientific conclusion.
p = 0.05
P-value
↓
0
JMB Ch 10 Lecture 1 8th edition
0.25
0.50
0.75
1.00
EGR 252 Spring 2008
P-value
Slide 6
Hypothesis Testing Tells Us …
Strong conclusion:
 If our calculated t-value is “outside” tα,ν (approach
1) or we have a small p-value (approach 2), then
we reject H0: μ = μ0 in favor of the alternate
hypothesis.
Weak conclusion:
 If our calculated t-value is “inside” tα,ν (approach 1)
or we have a “large” p-value (approach 2), then
we cannot reject H0: μ = μ0.
Failure to reject H0 does not imply that μ is
equal to the stated value (μ0), only that we do
not have sufficient evidence to support H1.
JMB Ch 10 Lecture 1 8th edition
EGR 252 Spring 2008
Slide 7
Example: Single Sample Test of the Mean Pvalue Approach
A sample of 20 cars driven under varying highway conditions
achieved fuel efficiencies as follows:
Sample mean
x = 34.271 mpg
Sample std dev
s = 2.915 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
JMB Ch 10 Lecture 1 8th edition
EGR 252 Spring 2008
Slide 8
Single Sample Example (cont.)
Approach 2:
X 
T
S/ n
= -1.11842
Find probability from chart or use Excel’s tdist function.
P(x ≤ -1.118) = TDIST (1.118, 19, 1) = 0.139665
p = 0.14
0______________1
Decision: Fail to reject null hypothesis
Conclusion: The mean does not differ significantly from 35 mpg.
JMB Ch 10 Lecture 1 8th edition
EGR 252 Spring 2008
Slide 9
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 “μ< μ0,” tcrit = -1.729
Step 4: tcalc = -1.11842
Step 5: Decision
Fail to reject H0
Step 6: Conclusion: Conclusion: The mean does not differ
significantly from 35 mpg.
JMB Ch 10 Lecture 1 8th edition
EGR 252 Spring 2008
Slide 10