Lecture 9
Fundamentals of Hypothesis
Testing (1):
One-Sample Tests
K. H. Hamed, Cairo University
Lecture 9 – Slide 1
Outline
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•
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•
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Hypothesis testing methodology
Z test for the mean (σ known)
P-value approach to hypothesis testing
Connection to confidence interval estimation
One-tail tests
T test for the mean (σ unknown)
Potential hypothesis-testing pitfalls and ethical
considerations
K. H. Hamed, Cairo University
Lecture 9 – Slide 2
What is a Hypothesis?
• A hypothesis is a
claim (assumption)
about the population
parameter
– Examples of parameters:
• population mean
• Population variance
– The parameter must be identified before
analysis
K. H. Hamed, Cairo University
Lecture 9 – Slide 3
The Null Hypothesis, H0
• States the assumption (numerical) to be
tested
– e.g.: The average number of children in Cairo
Homes is at least three (H 0 : μ ≥ 3 )
• Is always
– about a population parameter (e.g., H 0 : μ ≥ 3 ),
– not about a sample statistic (e.g., H 0 : X ≥ 3 )
K. H. Hamed, Cairo University
Lecture 9 – Slide 4
The Null Hypothesis, H0
(continued)
• Begins with the assumption that the null
hypothesis is true
– Similar to the notion of innocent until
proven guilty
• Refers to the status quo (current state)
• Always contains the “=” sign
• May or may not be rejected
K. H. Hamed, Cairo University
Lecture 9 – Slide 5
The Alternative Hypothesis, H1
• Is the opposite of the null hypothesis
– e.g.: The average number of children in
Cairo homes is less than 3 ( H1 : μ < 3 )
•
•
•
•
Challenges the status quo
Never contains the “=” sign
May or may not be accepted
Is generally the hypothesis that is
believed (or needed to be proven) to be
true by the researcher
K. H. Hamed, Cairo University
Lecture 9 – Slide 6
Hypothesis Testing Process
• Mean number of accidents has been equal to 200
accidents per year.
• It is claimed that a new traffic law has decreased the
mean rate to less than 200 accidents per year.
• A sample is taken and the sample mean
as 180 accidents.
x
is calculated
• The null hypothesis is rejected if there is a reason to
believe that the calculated x = 180 is significantly
different than the assumed 200 accidents per year.
K. H. Hamed, Cairo University
Lecture 9 – Slide 7
Reason for Rejecting H0
Sampling Distribution of X
... Therefore,
we reject the
null hypothesis
that μ = 200.
It is unlikely that
we would get a
sample mean of
this value ...
If H0 is true
180
μ = 200
K. H. Hamed, Cairo University
X
Lecture 9 – Slide 8
Level of Significance, α
• Defines the region of unlikely values of the
sample statistic if null hypothesis is true
– Called rejection region of the sampling
distribution
• Is designated by α , (level of significance)
– Typical values are 0.01, 0.05, 0.10
• Is selected by the researcher at the
beginning
• Provides the critical value(s) of the test
K. H. Hamed, Cairo University
Lecture 9 – Slide 9
Level of Significance
and the Rejection Region
α
H0: μ ≥ 3
H1: μ < 3
H0: μ ≤ 3
H1: μ > 3
Rejection
Regions
0
0
H0: μ = 3
H1: μ ≠ 3
Critical
Value(s)
α
α/2
0
K. H. Hamed, Cairo University
Lecture 9 – Slide 10
Errors in Making Decisions
• Type I Error
– Rejecting a true null hypothesis
– Has serious consequences
The probability of Type I Error is α
• Called level of significance
• Set by researcher
• Type II Error
– Failure to reject a false null hypothesis
– The probability of Type II Error is β
– The power of the test is (1 − β )
K. H. Hamed, Cairo University
Lecture 9 – Slide 11
Result Probabilities
The Truth
Decision
H 0 True H 0 False
Do Not
Reject
H0
1- α
Type II
Error ( β )
Reject
H0
Type I
Error
(α )
Power
(1 - β )
K. H. Hamed, Cairo University
Lecture 9 – Slide 12
Relationship Between
Type I & II Errors
If one reduces the probability of one error, the other
increases (everything else is unchanged).
K. H. Hamed, Cairo University
Lecture 9 – Slide 13
How to Choose between
Type I and Type II Errors
• Choice depends on the cost of the errors
• Choose smaller Type I Error when the cost of
rejecting the maintained hypothesis is high
– A criminal trial: convicting an innocent person
• Choose larger Type I Error when you have an
interest in changing the status quo
– A decision in a startup company about a new
piece of software
K. H. Hamed, Cairo University
Lecture 9 – Slide 14
Critical Values
Approach to Testing
• Convert sample statistic (e.g.: X ) to test
statistic (e.g.: Z, t or F –statistic)
• Obtain critical value(s) for a specified α
from a table or computer
– If the test statistic falls in the critical region,
reject H0
– Otherwise do not reject H0
K. H. Hamed, Cairo University
Lecture 9 – Slide 15
p-Value Approach
• Convert Sample Statistic (e.g. X ) to Test Statistic (e.g.
Z, t or F –statistic)
• Obtain the p-value from a table or computer
– p-value: Probability of obtaining a test statistic
more extreme ( ≤ or ≥ ) than the observed sample
value given H0 is true
– Called observed level of significance
• Compare the p-value with α
– If p-value ≥ α , do not reject H0
– If p-value ≤ α , reject H0
K. H. Hamed, Cairo University
Lecture 9 – Slide 16
General Steps in
Hypothesis Testing
e.g.: Test the assumption that the true mean number of of
children in Cairo homes is at least three (σ Known)
H0 :μ ≥ 3
1. State the H0
2. State the H1
H1 : μ < 3
3. Choose
α =0.05
α
n = 100
Z test
4. Choose n
5. Choose Test
K. H. Hamed, Cairo University
General Steps
6. Set up critical value(s)
Lecture 9 – Slide 17
(continued)
Reject H0
α
-1.645
100 households surveyed
7. Collect data
8. Compute test statistic
and p-value
9. Make statistical decision
10. Express conclusion
Z
Computed test stat =-2,
p-value = .0228
Reject null hypothesis
The true mean number of children
in Cairo homes is less than 3
K. H. Hamed, Cairo University
Lecture 9 – Slide 18
One-tail Z-Test for Mean
(σ Known)
• Assumptions
– Population is normally distributed
– If not normal, requires large samples
– Null hypothesis has ≤ or ≥ sign only
• Z test statistic
X − μX
X −μ
=
Z=
σX
σ/ n
K. H. Hamed, Cairo University
Lecture 9 – Slide 19
Rejection Region
H0: μ ≤ μ0
H1: μ > μ0
H0: μ ≥ μ0
H1: μ < μ0
Reject H0
Reject H0
α
α
0
Z Must Be Significantly
Below 0 to reject H0
Z
Z
0
Small values of Z don’t
contradict H0
Don’t Reject H0 !
K. H. Hamed, Cairo University
Lecture 9 – Slide 20
Example: One Tail Test
Q. Does an average
sample of irrigation
water contain less than
368 ppm (mg/L) of TDS?
A random sample of 25
water samples showed
that X = 372.5. The
water quality division
has specified σ to be 15
mg/L. Test at the α = 0.05
level.
H0: μ ≤ 368
H1: μ > 368
K. H. Hamed, Cairo University
Lecture 9 – Slide 21
Finding Critical Value: One Tail
What is Z given α = 0.05?
Standardized Cumulative
Normal Distribution Table
(Portion)
σZ =1
Z
.95
α = .05
0 1.645 Z
Critical Value
= 1.645
.04
.05
.06
1.6 .9495 .9505 .9515
1.7 .9591 .9599 .9608
1.8 .9671 .9678 .9686
1.9 .9738 .9744 .9750
K. H. Hamed, Cairo University
Lecture 9 – Slide 22
H0: μ ≤ 368
H1: μ > 368
Example Solution:
One Tail Test
Test Statistic:
α = 0.5
Z=
n = 25
Critical Value: 1.645
X −μ
σ
=1.50
n
Decision:
Reject
Do Not Reject at α = .05
.05
Conclusion:
0 1.645 Z
No evidence that true
mean is more than 368
1.50
K. H. Hamed, Cairo University
Lecture 9 – Slide 23
p -Value Solution
p-Value is P(Z ≥ 1.50) = 0.0668
Use the
alternative
hypothesis
to find the
direction
of the
rejection
region.
P-Value =.0668
1.0000
- .9332
.0668
0
From Z Table:
Lookup 1.50 to
Obtain .9332
Z
1.50
Z Value of Sample
Statistic
K. H. Hamed, Cairo University
Lecture 9 – Slide 24
p -Value Solution
(continued)
(p-Value = 0.0668) ≥ (α = 0.05)
Do Not Reject.
p Value = 0.0668
Reject
α = 0.05
0
1.50
1.645
Z
Test Statistic 1.50 is in the Do Not Reject Region
K. H. Hamed, Cairo University
Lecture 9 – Slide 25
Example: Two-Tail Test
Q. Does an average
irrigation water
sample contain 368
mg/L of TDS? A
random sample of 25
water samples
showed X = 372.5.
The company has
specified σ to be 15
mg/L. Test at the α =
0.05 level.
H0: μ = 368
H1: μ ≠ 368
K. H. Hamed, Cairo University
Lecture 9 – Slide 26
Example Solution: Two-Tail Test
H0: μ = 368
H1: μ ≠ 368
Test Statistic:
Z=
α = 0.05
n = 25
Critical Value: ±1.96
-1.96
372.5 − 368
= 1.50
15
25
Decision:
Do Not Reject at α = .05
.025
0 1.96
1.50
σ
=
n
Reject
.025
X −μ
Conclusion:
No Evidence that True
Mean is Not 368
Z
K. H. Hamed, Cairo University
Lecture 9 – Slide 27
p-Value Solution
(p Value = 0.1336) ≥ (α = 0.05)
Do Not Reject.
p Value = 2 x 0.0668
Reject
Reject
α = 0.05
0
1.50
1.96
Z
Test Statistic 1.50 is in the Do Not Reject Region
K. H. Hamed, Cairo University
Lecture 9 – Slide 28
t Test: σ Unknown
• Assumption
– Population is normally distributed
– If not normal, requires a large sample
• t test statistic with n-1 degrees of freedom
–
t=
X −μ
S/ n
K. H. Hamed, Cairo University
Lecture 9 – Slide 29
Example: One-Tail t Test
Does an average
irrigation water sample
contain less than 368
mg/L of TDS? A random
sample of 36 water
samples boxes showed
X = 372.5, and s = 15.
Test at the α = 0.01
level.
σ is not given
H0: μ ≤ 368
H1: μ > 368
K. H. Hamed, Cairo University
Lecture 9 – Slide 30
Example Solution: One-Tail
H0: μ ≤ 368
H1: μ > 368
Test Statistic:
t=
α = 0.01
n = 36, df = 35
Critical Value: 2.4377
Decision:
Do Not Reject at α = .01
Reject
.01
0 2.4377 t35
1.80
X − μ 372.5 − 368
=
= 1.80
S
15
n
36
Conclusion:
No evidence that true
mean is more than 368
K. H. Hamed, Cairo University
Lecture 9 – Slide 31
p -Value Solution
(p Value is between .025 and .05) ≥ (α = 0.01).
Do Not Reject.
p Value = [.025, .05]
Reject
α = 0.01
0
t
2.4377 35
Test Statistic 1.80 is in the Do Not Reject Region
1.80
K. H. Hamed, Cairo University
Lecture 9 – Slide 32