Week 7 - Seminar

Report
7
Hypothesis Testing
Useful videos/websites:
Video on writing hypothesis statements:
http://screencast.com/t/ZO8kMt6v
Overview Video on hypothesis testing:
http://screencast.com/t/qdz8F0yp
Difference between Type I and II Errors:
http://screencast.com/t/fhWKC3ypM8
Section 7.1
Introduction to
Hypothesis Testing
Definition
A hypothesis is a statement or claim
regarding a characteristic of one or more
populations.
What do you mean by “claim”?
What do you mean by “claim”?
Let’s measure the
raisins in each box with
a random variable, x.
H0: mu = 2 scoops
H1: mu not = 2 scoops
Which one is the claim?
Null hypothesis H0
contains a statement of equality such as , = or .
Alternative hypothesis Ha or H1 contains a
statement of inequality such as < ,  or >
For example:
H0: Mr. Smith is innocent of the crime.
H1: Mr. Smith is guilty of the crime. `
Writing Hypotheses
Write the claim about the population.
Then, write its complement.
Note: Either hypothesis, the null or
the alternative, can represent the
claim.
Writing Hypotheses
Write the claim about the population. Then, write its complement.
Either hypothesis, the null or the alternative, can represent the
claim.
A hospital claims its ambulance response time is less
than 10 minutes.
claim
A consumer magazine claims the proportion of cell
phone calls made during evenings and weekends is at
most 60%.
claim
Four Outcomes
Correct
Type II
Error
Type I
Error
Correct
H0: Mr. Smith is innocent of the crime.
H1: Mr. Smith is guilty of the crime.
Hypothesis Testing Errors
• Type I error: Reject a true Null hypothesis
(i.e. innocent person found guilty)
α = alpha = probability of Type I error
• Type II error: Do not reject a false Null
hypothesis
(i.e. guilty man goes free)
β = beta = probability of Type II error
Two-tailed vs. One-tailed tests
Ho: parameter = some value
1. two-tailed test:
Equal versus not equal hypothesis H1: parameter
some value
2. left-tailed test:
Equal versus less than
3. right-tailed test:
Equal versus greater than
Ho: parameter = some value
(or greater)
H1: parameter
some value
Ho: parameter = some value
(or less)
H1: parameter
some value
Types of Hypothesis Tests
Ha is more probable
Ha is more probable
Ha is more probable
Right-tail test
Left-tail test
Two-tail test
One-tailed or two?
1. A university publicizes that the proportion of
its students who graduate in 4 years is 82%.
Solution:
H0: p = 0.82
Ha: p ≠ 0.82
Two-tailed test
½ P-value
area
-z
½ P-value
area
0
z
z
14
One-tailed or two?
2. A water faucet manufacturer announces that
the mean flow rate of a certain type of faucet
is less than 2.5 gallons per minute.
Solution:
H0: μ ≥ 2.5 gpm
Ha: μ < 2.5 gpm
Left-tailed test
P-value area
-z
0
z
One-tailed or two?
3. A cereal company advertises that the mean
weight of the contents of its 20-ounce size
cereal boxes is more than 20 ounces.
Solution:
H0: μ ≤ 20 oz
Ha: μ > 20 oz
Right-tailed test
P-value ar
ea
0
z
z
Hypothesis Test Strategy
1. Begin by assuming the equality condition in the null hypothesis is
true. This is regardless of whether the claim is represented by the null
hypothesis or by the alternative hypothesis.
2. Collect data from a random sample taken from the
population and calculate the necessary sample statistics.
3. If the sample statistic has a low probability of being drawn
from a population in which the null hypothesis is true, you will
reject H0. (As a consequence, you will support the alternative
hypothesis.)
4. If the probability is not low enough, fail to reject H0.
Two Methods:
 P-value Method – uses the probability of obtaining a sample
statistics with a value as extreme (or more) than the one
determined by sample data.
 Critical Value Method – define a ‘rejection region’ using a
critical value (similar to the critical z).
P-values
P-value (or probability value)
 The probability, if the null hypothesis is true, of obtaining a
sample statistic with a value as extreme or more extreme
than the one determined from the sample data.
 Depends on the nature of the test.
19
Larson/Farber 4th ed.
P-values
The P-value is the probability of obtaining a sample
statistic with a value as extreme or more extreme than
the one determined by the sample data.
P-value = indicated area
Area in
left tail
Area in
right tail
z
z
For a right tail test
For a left tail test
If z is negative,
twice the area
in the left tail
If z is positive,
twice the area
in the right tail
z
z
For a two-tail test
Finding P-values: 1-tail Test
The test statistic for a right-tail test is z = 1.56. Find the
P-value.
Area in right tail
z = 1.56
Answer:
The area to the right of z = 1.56 is 1 – .9406 = 0.0594.
The P-value is 0.0594.
Finding P-values: 2-tail Test
The test statistic for a two-tail test is z = –2.63. Find the
corresponding P-value.
z = –2.63
Answer:
The area to the left of z = –2.63 is 0.0043.
The P-value is 2(0.0043) = 0.0086.
P-value Method
In Words
In Symbols
1. State the claim mathematically and
verbally. Identify the null and
alternative hypotheses.
State H0 and Ha.
2. Specify the level of significance.
Identify .
3. Determine the standardized test
statistic.
z
4. Find the area that corresponds
to z.
Use Table 4 in
Appendix B.
x 
 n
23
P-value Method – Part II
In Words
In Symbols
5. Find the P-value.
a. For a left-tailed test, P = (Area in left tail).
b. For a right-tailed test, P = (Area in right tail).
c. For a two-tailed test, P = 2(Area in tail of test
statistic).
Reject H0 if P-value
6. Make a decision to reject or
is less than or equal
fail to reject the null hypothesis.
to . Otherwise, fai
l to reject H0.
7. Interpret the decision in the
context of the original claim.
24
1. Identify Hypothesis and indicate
which one is the claim.
H0 :
Ha :
2. Identify level of significance
=
3. Compute the test statistic.
[Depends on what parameter is
being tested.]
z
x 
 n
4. Find the area corresponding to z. Use preferred method for finding area in tail.
5. Find the P-value
a.
b.
6. Make a Decision
Reject H0 if p-value <= , otherwise fail to
reject.
7. Interpret the decision in the
context of the original claim.
For a one-tailed test, P = (Area in left tail).
For a two-tailed test, P = 2(Area in tail).
Example: Hypothesis Testing Using P-values
You think that the average
franchise investment
information shown in the graph
is incorrect, so you randomly
select 30 franchises and determine the necessary investment
for each. The sample mean investment is $135,000 with a
standard deviation of $30,000. Is there enough evidence to
support your claim at  = 0.05? Use the P-value method.
26
1. Identify Hypothesis and indicate
which one is the claim.
2. Identify level of significance
3. Compute the test statistic.
[Depends on what parameter is
being tested.]
4. Find the area corresponding to z.
5. Find the P-value
6. Make a Decision
7. Interpret the decision in the
context of the original claim.
H0 is claim
= 0.05
x  $135,000
z
x 
 n
Solution: Hypothesis Testing Using P-values
z

x 

• P-value
n
135, 000  143, 260
30, 000
 1.51
30
-1.51
0
z
1. Identify Hypothesis and indicate
which one is the claim.
2. Identify level of significance
3. Compute the test statistic.
[Depends on what parameter is
being tested.]
4. Find the area corresponding to z.
5. Find the P-value
6. Make a Decision
7. Interpret the decision in the
context of the original claim.
H0 is claim
= 0.05
x   135,000  143, 260
z

 1.51
 n
30,000 30
Solution: Hypothesis Testing Using P-values
z

x 

n
135, 000  143, 260
30, 000
 1.51
30
• P-value
P = 2(0.0655)
= 0.1310
0.0655
-1.51
0
z
30
1. Identify Hypothesis and indicate
which one is the claim.
2. Identify level of significance
3. Compute the test statistic.
[Depends on what parameter is
being tested.]
H0 is claim
= 0.05
x   135,000  143, 260
z

 1.51
 n
30,000 30
4. Find the area corresponding to z. Area to the left of -1.51 is equal to 0.0655
5. Find the P-value
P = 2(.0655) because this is a two-tailed test.
P = .1310
6. Make a Decision
Since .1310 > , we fail to reject H0.
7. Interpret the decision in the
context of the original claim.
Since the claim was that the average (mu)
was equal to $143,260, we say “there is not
enough evidence to reject the claim”.
Example: Testing with P-values
Employees in a large accounting firm claim that
the mean salary of the firm’s accountants is less
than that of its competitor’s, which is $45,000. A
random sample of 30 of the firm’s accountants
has a mean salary of $43,500 with a standard
deviation of $5200. At α = 0.05, test the
employees’ claim.
32
1.
H0: Mu = $45,000 (or more)
H1: Mu < $45,000
(left-tailed test)
2.
Alpha = .05
3.
Xbar = $43,500, Sx = $5200
(Since n >= 30, we can use a
z-statistic)
4.
Compute z-statistic (See pg 387)
z = -1.58
5.
Find the P-value:
P(z <= -1.58) = .0571
6.
Since the P-value of .0571 is NOT less
than or equal to alpha (.05), we DO NOT
REJECT H0
7.
There is not sufficient evidence to
support the claim that the mean salary
Test Decisions with P-values
The decision about whether there is enough
evidence to reject the null hypothesis can be
made by comparing the P-value to the value of
the level of significance of the test.
If
If
,
reject the null hypothesis.
fail to reject the null hypothesis.
Interpreting the Decision
Claim
Decision
Claim is H0
Reject H0
Fail to
reject H0
Claim is Ha
There is enough
evidence to
reject the claim.
There is enough
evidence to
support the
claim.
There is not
enough
evidence to
reject the claim.
There is not
enough
evidence to
support the
claim.
Critical Value method
In Words
In Symbols
1. State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
2. Specify the level of significance.
3. Find the standardized test statistic
4. Determine the critical value(s) &
rejection region(s).
z
x 
 n
or if n  30 use   s.
Use Table 4 in
Appendix B.
37
Critical Value Z method – Part II
In Words
5. Make a decision to reject or fail
to reject the null hypothesis.
In Symbols
If z is in the rejection
region, reject H0.
Otherwise, fail to
reject H0.
6. Interpret the decision in the
context of the original claim.
Larson/Farber 4th ed.
38
Rejection Regions and Critical Values
Rejection region (or critical region)
 The range of values for which the null hypothesis is not probable.
 If a test statistic falls in this region, the null hypothesis is rejected.
 A critical value z0 separates the rejection region from the
nonrejection region.
39
Rejection Regions and Critical Values
Finding Critical Values in a Normal Distribution
1. Specify the level of significance .
2. Decide whether the test is left-, right-, or two-tailed.
3. Find the critical value(s) z0. If the hypothesis test is
a. left-tailed, find the z-score that corresponds to an area o
f ,
b. right-tailed, find the z-score that corresponds to an area
of 1 – ,
c. two-tailed, find the z-score that corresponds to ½ and
1 – ½.
4. Sketch the standard normal distribution. Draw a vertical li
ne at each critical value and shade the rejection region(s).
40
Example: Finding Critical Values
Find the critical value and rejection region for a two-tailed test
with  = 0.05.
Solution:
1 – α = 0.95
½α = 0.025
z0
-z0 = -1.96
½α = 0.025
0 z0 =z01.96
z
The rejection regions are to the left of -z0 = -1.96 and
to the right of z0 = 1.96.
41
Decision Rule Based on Rejection Region
To use a rejection region to conduct a hypothesis test,
calculate the standardized test statistic, z. If the
standardized test statistic
1. is in the rejection region, then reject H0.
2. is not in the rejection region, then fail to reject H0.
Fail to reject Ho.
Fail to reject H0.
Reject H0.
z < z0 z
Reject Ho.
0
0
Left-Tailed Test
z Fail to reject H
0
0
Reject H0
z0
z > zz0
Right-Tailed Test
Reject H0
z0
z0
z < -zTwo-Tailed
0
Test
0
z >zz0
42
Example: Testing with Critical Value
Employees in a large accounting firm claim that
the mean salary of the firm’s accountants is less
than that of its competitor’s, which is $45,000. A
random sample of 30 of the firm’s accountants
has a mean salary of $43,500 with a standard
deviation of $5200. At
α = 0.05, test the employees’ claim.
43
Example – Critical Z
•
•
•
•
• Test Statistic
x   43,500  45, 000
z

 n
5200 30
H0: μ ≥ $45,000
Ha: μ < $45,000
 = 0.05
Rejection Region:
0.05
0
-1.58
z
 1.58
• Decision: Fail to reject H0
At the 5% level of significance,
there is not sufficient evidence
to support the employees’
claim that the mean salary is
less than $45,000.
Zc=-1.645
44
Section 7.2
Hypothesis Testing for the Mean
Large Samples (n 30)
The z-Test for a Mean
The z-test is a statistical test for a population mean. The z-test
can be used:
(1) if the population is normal and s is known or
(2) when the sample size, n, is at least 30.
The test statistic is the sample mean and the standardized
test statistic is z.
When n
30, use s in place of
.
The z-Test for a Mean (P-value)
A cereal company claims the mean sodium content in one
serving of its cereal is no more than 230 mg. You work for a
national health service and are asked to test this claim. You find
that a random sample of 52 servings has a mean sodium
content of 232 mg and a standard deviation of 10 mg. At
= 0.05, do you have enough evidence to reject the company’s
claim?
1. Write the null and alternative hypothesis.
2. State the level of significance.
= 0.05
3. Determine the sampling distribution.
Since the sample size is at least 30, the sampling distribution is normal.
4. Find the test statistic and standardize it.
n = 52
s = 10
Test statistic
5. Calculate the P-value for the test statistic.
Since this is a right-tail test, the P-value
is the area found to the right
of z = 1.44 in the normal distribution.
From the table P = 1 – 0.9251
P = 0.0749.
Area in right tail
z = 1.44
6. Make your decision.
Compare the P-value to .
Since 0.0749 > 0.05, fail to reject H0.
7. Interpret your decision.
There is not enough evidence to reject the claim
that the mean sodium content of one serving of its cereal is
no more than 230 mg.
The z-Test for a Mean (Critical Value)
A cereal company claims the mean sodium content in one
serving of its cereal is no more than 230 mg. You work for a
national health service and are asked to test this claim. You
find that a random sample of 52 servings has a mean
sodium content of 232 mg and a standard deviation of 10
mg. At
= 0.05, do you have enough evidence to reject
the company’s claim?
1. Write the null and alternative hypothesis.
2. State the level of significance.
= 0.05
3. Determine the sampling distribution.
Since the sample size is at least 30, the sampling distribution is normal.
Since Ha contains the > symbol, this is a right-tail test.
Rejection
region
z0
1.645
4. Find the critical value.
5. Find the rejection region.
6. Find the test statistic and standardize it.
n = 52
= 232 s = 10
7. Make your decision.
z = 1.44 does not fall in the rejection region, so fail to reject H 0
8. Interpret your decision.
There is not enough evidence to reject the company’s claim that
there is at most 230 mg of sodium in one serving of its cereal.
Using the P-value of a Test to Compare Areas
= 0.05
z0 = –1.645
Rejection area
0.05
z0
z = –1.23
P = 0.1093
z
For a P-value decision, compare areas.
If
reject H0.
If
fail to reject H0.
For a critical value decision, decide if z is in the rejection region
If z is in the rejection region, reject H0. If z is not in the rejection
region, fail to reject H0.
Section 7.3
Hypothesis Testing for the Mean
Small Samples (n < 30)
The t Sampling Distribution
Find the critical value t0 for a left-tailed test given
and n = 18.
= 0.01
d.f. = 18 – 1 = 17
t0 = –2.567
Area in
left tail
t0
Find the critical values –t0 and t0 for a two-tailed test given
= 0.05 and n = 11.
–t0 = –2.228 and t0 = 2.228
d.f. = 11 – 1 = 10
t0
t0
Testing
–Small Sample
A university says the mean number of classroom hours per
week for full-time faculty is 11.0. A random sample of the
number of classroom hours for full-time faculty for one week is
listed below. You work for a student organization and are asked
to test this claim. At
= 0.01, do you have enough evidence
to reject the university’s claim?
11.8 8.6 12.6 7.9 6.4 10.4 13.6 9.1
1. Write the null and alternative hypothesis
2. State the level of significance
= 0.01
3. Determine the sampling distribution
Since the sample size is 8, the sampling distribution
is a t-distribution with 8 – 1 = 7 d.f.
Since Ha contains the ≠ symbol, this is a two-tail test.
4. Find the critical values.
5. Find the rejection region.
–t0
–3.499
t0
3.499
6. Find the test statistic and standardize it
n=8
= 10.050 s = 2.485
7. Make your decision.
t = –1.08 does not fall in the rejection region, so fail to reject H0 at
8. Interpret your decision.
There is not enough evidence to reject the university’s claim that
faculty spend a mean of 11 classroom hours.
= 0.01
Section 7.4
Hypothesis Testing for Proportions
Test for Proportions
p is the population proportion of successes. The
test statistic is
.
(the proportion of sample successes)
If
and
the sampling distribution for
The standardized test statistic is:
is normal.
Test for Proportions - Example
A communications industry spokesperson claims that
over 40% of Americans either own a cellular phone
or have a family member who does. In a random
survey of 1036 Americans, 456 said they or a family
member owned a cellular phone. Test the
spokesperson’s claim at
= 0.05. What can you
conclude?
1. Write the null and alternative hypothesis.
2. State the level of significance.
= 0.05
3. Determine the sampling distribution.
1036(.40) > 5 and 1036(.60) > 5. The sampling distribution is
normal.
Rejection
region
4. Find the critical value.
5. Find the rejection region.
1.645
6. Find the test statistic and standardize it.
n = 1036
x = 456
7. Make your decision.
z = 2.63 falls in the rejection region, so reject H0
8. Interpret your decision.
There is enough evidence to support the claim that over 40% of
Americans own a cell phone or have a family member who does.
1. State the null and alternative hypotheses.
A company claims the mean lifetime of its AA
batteries is more than 16 hours.
A. H0: μ > 16 Ha: μ ≤ 16
B. H0: μ < 16 Ha: μ ≥ 16
C. H0: μ ≤ 16 Ha: μ > 16
D. H0: μ ≥ 16 Ha: μ < 16
Copyright © 2007 Pearson Education, I
nc. Publishing as Pearson Addison-We
sley
Slide 7- 62
2. State the null and alternative hypotheses.
A student claims the mean cost of a
textbook is at least $125.
A. H0: μ > 125 Ha: μ ≤ 125
B. H0: μ < 125 Ha: μ ≥ 125
C. H0: μ ≤ 125 Ha: μ > 125
D. H0: μ ≥ 125 Ha: μ < 125
Copyright © 2007 Pearson Education, I
nc. Publishing as Pearson Addison-We
sley
Slide 7- 63
3. You are testing the claim that the mean cost
of a new car is more than $25,200. How should
you interpret a decision that rejects the null
hypothesis?
A. There is enough evidence to reject the claim.
B. There is enough evidence to support the
claim.
C. There is not enough evidence to reject the
claim.
D. There is not enough evidence to support the
claim.
Copyright © 2007 Pearson Education, I
nc. Publishing as Pearson Addison-We
sley
Slide 7- 64
True or false:
Given H0: μ = 40 Ha: μ ≠ 40 and P = 0.0436.
You would reject the null hypothesis at the
0.05 level of significance.
A. True
B. False
Copyright © 2007 Pearson Education, I
nc. Publishing as Pearson Addison-We
sley
Slide 7- 65
Answers
Answers:
1. (C)
2. (D)
3. (B)
4. (A)

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