chapter 7 hypothesis testing with one sample 1 larson/farber 4th ed
TRANSCRIPT
Chapter 7
Hypothesis Testing with One Sample
1Larson/Farber 4th ed.
Hypothesis Tests
Hypothesis test
• A process that uses sample statistics to test a claim about the value of a population parameter.
• For example: An automobile manufacturer advertises that its new hybrid car has a mean mileage of 50 miles per gallon. To test this claim, a sample would be taken. If the sample mean differs enough from the advertised mean, you can decide the advertisement is wrong.
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Stating a Hypothesis
Null hypothesis
• A statistical hypothesis that contains a statement of equality such as ≤, =, or ≥.
• Denoted H0 read “H subzero” or “H naught.”
Alternative hypothesis
• A statement of inequality such as >, ≠, or <.
• Must be true if H0 is false.
• Denoted Ha read “H sub-a.”
complementary statements
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Example: Stating the Null and Alternative Hypotheses
Write the claim as a mathematical sentence. State the null and alternative hypotheses and identify which represents the claim.1.A university publicizes that the proportion of its students who graduate in 4 years is 82%.
Equality condition
Complement of H0
H0:
Ha:
(Claim)p = 0.82
p ≠ 0.82
Solution:
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μ ≥ 2.5 gallons per minute
Example: Stating the Null and Alternative Hypotheses
Write the claim as a mathematical sentence. State the null and alternative hypotheses and identify which represents the claim.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.
Inequality condition
Complement of HaH0:
Ha:(Claim)μ < 2.5 gallons per minute
Solution:
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Types of Errors
• A type I error occurs if the null hypothesis is rejected when it is true.
• A type II error occurs if the null hypothesis is not rejected when it is false.
Actual Truth of H0
Decision H0 is true H0 is false
Do not reject H0 Correct Decision Type II Error
Reject H0 Type I Error Correct Decision
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Example: Identifying Type I and Type II Errors
The USDA limit for salmonella contamination for chicken is 20%. A meat inspector reports that the chicken produced by a company exceeds the USDA limit. You perform a hypothesis test to determine whether the meat inspector’s claim is true. When will a type I or type II error occur? Which is more serious? (Source: United States Department of Agriculture)
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Let p represent the proportion of chicken that is contaminated.
Solution: Identifying Type I and Type II Errors
H0:
Ha:
p ≤ 0.2
p > 0.2
Hypotheses:
(Claim)
0.16 0.18 0.20 0.22 0.24p
H0: p ≤ 0.20 H0: p > 0.20
Chicken meets USDA limits.
Chicken exceeds USDA limits.
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Solution: Identifying Type I and Type II Errors
A type I error is rejecting H0 when it is true.
The actual proportion of contaminated chicken is lessthan or equal to 0.2, but you decide to reject H0.
A type II error is failing to reject H0 when it is false.
The actual proportion of contaminated chicken is greater than 0.2, but you do not reject H0.
H0:Ha:
p ≤ 0.2p > 0.2
Hypotheses:(Claim)
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Solution: Identifying Type I and Type II Errors
H0:Ha:
p ≤ 0.2p > 0.2
Hypotheses:(Claim)
• With a type I error, you might create a health scare and hurt the sales of chicken producers who were actually meeting the USDA limits.
• With a type II error, you could be allowing chicken that exceeded the USDA contamination limit to be sold to consumers.
• A type II error could result in sickness or even death.
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Level of Significance
Level of significance • Your maximum allowable probability of making a
type I error. Denoted by , the lowercase Greek letter alpha.
• By setting the level of significance at a small value, you are saying that you want the probability of rejecting a true null hypothesis to be small.
• Commonly used levels of significance: = 0.10 = 0.05 = 0.01
• P(type II error) = β (beta)
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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.
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Making a Decision
Decision Rule Based on P-value• Compare the P-value with α.
If P ≤ α, then reject H0.
If P > α, then fail to reject H0.
Claim
Decision Claim is H0 Claim is Ha
Reject H0
Fail to reject H0
There is enough evidence to reject the claim
There is not enough evidence to reject the claim
There is enough evidence to support the claim
There is not enough evidence to support the claim
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Example: Interpreting a Decision
You perform a hypothesis test for the following claim. How should you interpret your decision if you reject H0? If you fail to reject H0?
2.Ha (Claim): Consumer Reports states that the mean stopping distance (on a dry surface) for a Honda Civic is less than 136 feet.Solution:
• The claim is represented by Ha.
• H0 is “the mean stopping distance…is greater than or equal to 136 feet.”
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Solution: Interpreting a Decision
• If you reject H0 you should conclude “there is enough evidence to support Consumer Reports’ claim that the stopping distance for a Honda Civic is less than 136 feet.”
• If you fail to reject H0, you should conclude “there is not enough evidence to support Consumer Reports’ claim that the stopping distance for a Honda Civic is less than 136 feet.”
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z0
Steps for Hypothesis Testing
1. State the claim mathematically and verbally. Identify the null and alternative hypotheses.
H0: ? Ha: ?2. Specify the level of significance.
α = ?3. Determine the standardized
sampling distribution and draw its graph.
4. Calculate the test statisticand its standardized value.Add it to your sketch. z
0Test statistic
This sampling distribution is based on the assumption that H0 is true.
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Steps for Hypothesis Testing
5. Find the P-value.
6. Use the following decision rule.
7. Write a statement to interpret the decision in the context of the original claim.
Is the P-value less than or equal to the level of significance?
Fail to reject H0.
Yes
Reject H0.
No
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Section 7.2
Hypothesis Testing for the Mean (Large Samples)
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Using P-values to Make a Decision
Decision Rule Based on P-value
• To use a P-value to make a conclusion in a hypothesis test, compare the P-value with α.
1. If P ≤ α, then reject H0.
2. If P > α, then fail to reject H0.
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Z-Test for a Mean μ
• Can be used when the population is normal and is known, or for any population when the sample size n is at least 30.
• The test statistic is the sample mean
• The standardized test statistic is z
• When n ≥ 30, the sample standard deviation s can be substituted for σ.
xzn
standard error xn
x
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Using P-values for a z-Test for Mean μ
1. State the claim mathematically and verbally. Identify the null and alternative hypotheses.
2. Specify the level of significance.
3. Determine the standardized test statistic.
4. Find the area that corresponds to z.
State H0 and Ha.
Identify α.
Use Table 4 in Appendix B.
xzn
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In Words In Symbols
Using P-values for a z-Test for Mean μ
Reject H0 if P-value is less than or equal to α. Otherwise, fail to reject H0.
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).6. Make a decision to reject or
fail to reject the null hypothesis.
7. Interpret the decision in the context of the original claim.
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In Words In Symbols
Example: Hypothesis Testing Using P-values
In an advertisement, a pizza shop claims that its mean delivery time is less than 30 minutes. A random selection of 36 delivery times has a sample mean of 28.5 minutes and a standard deviation of 3.5 minutes. Is there enough evidence to support the claim at α = 0.01? Use a P-value.
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Solution: Hypothesis Testing Using P-values
• H0:
• Ha:
• = • Test Statistic:
μ ≥ 30 min
μ < 30 min
0.01
28.5 30
3.5 36
2.57
xz
n
• Decision:
At the 1% level of significance, you have sufficient evidence to conclude the mean delivery time is less than 30 minutes.
z0-2.57
0.0051
• P-value
0.0051 < 0.01Reject H0
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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.
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Using Rejection Regions for a z-Test for a Mean μ
1. State the claim mathematically and verbally. Identify the null and alternative hypotheses.
2. Specify the level of significance.
3. Sketch the sampling distribution.
4. Determine the critical value(s).
5. Determine the rejection region(s).
State H0 and Ha.
Identify .
Use Table 4 in Appendix B.
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In Words In Symbols
Using Rejection Regions for a z-Test for a Mean μ
6. Find the standardized test statistic.
7. Make a decision to reject or fail to reject the null hypothesis.
8. Interpret the decision in the context of the original claim.
or if 30
use
xz nn
s
.
If z is in the rejection region, reject H0. Otherwise, fail to reject H0.
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In Words In Symbols
Example: Testing with Rejection Regions
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.
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Solution: Testing with Rejection Regions
• H0:
• Ha:
• = • Rejection Region:
μ ≥ $45,000
μ < $45,000
0.05
43,500 45, 000
5200 30
1.58
xz
n
• Decision: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.
• Test Statistic
z0-1.645
0.05
-1.58
Fail to reject H0
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Section 7.3
Hypothesis Testing for the Mean (Small Samples)
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Finding Critical Values in a t-Distribution
1. Identify the level of significance α.
2. Identify the degrees of freedom d.f. = n – 1.
3. Find the critical value(s) using Table 5 in Appendix B in the row with n – 1 degrees of freedom. If the hypothesis test is
a. left-tailed, use “One Tail, α ” column with a negative sign,
b. right-tailed, use “One Tail, α ” column with a positive sign,
c. two-tailed, use “Two Tails, α ” column with a negative and a positive sign.
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Example: Finding Critical Values for t
Find the critical value t0 for a left-tailed test givenα = 0.05 and n = 21.
Solution:• The degrees of freedom are
d.f. = n – 1 = 21 – 1 = 20.• Look at α = 0.05 in the
“One Tail, α” column. • Because the test is left-
tailed, the critical value is negative.
t0-1.725
0.05
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t-Test for a Mean μ (n < 30, σ Unknown)
t-Test for a Mean
• A statistical test for a population mean.
• The t-test can be used when the population is normal or nearly normal, σ is unknown, and n < 30.
• The test statistic is the sample mean
• The standardized test statistic is t.
• The degrees of freedom are d.f. = n – 1.
xts n
x
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Using the t-Test for a Mean μ(Small Sample)
1. State the claim mathematically and verbally. Identify the null and alternative hypotheses.
2. Specify the level of significance.
3. Identify the degrees of freedom and sketch the sampling distribution.
4. Determine any critical value(s).
State H0 and Ha.
Identify α.
Use Table 5 in Appendix B.
d.f. = n – 1.
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In Words In Symbols
Using the t-Test for a Mean μ(Small Sample)
5. Determine any rejection region(s).
6. Find the standardized test statistic.
7. Make a decision to reject or fail to reject the null hypothesis.
8. Interpret the decision in the context of the original claim.
xts n
If t is in the rejection region, reject H0. Otherwise, fail to reject H0.
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In Words In Symbols
Example: Testing μ with a Small Sample
A used car dealer says that the mean price of a 2005 Honda Pilot LX is at least $23,900. You suspect this claim is incorrect and find that a random sample of 14 similar vehicles has a mean price of $23,000 and a standard deviation of $1113. Is there enough evidence to reject the dealer’s claim at α = 0.05? Assume the population is normally distributed. (Adapted from Kelley Blue Book)
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Solution: Testing μ with a Small Sample
• H0:
• Ha:
• α =
• df = • Rejection Region:
• Test Statistic:
• Decision:
μ ≥ $23,900μ < $23,900
0.0514 – 1 = 13
23, 000 23,9003.026
1113 14
xt
s n
t0-1.771
0.05
-3.026
At the 0.05 level of significance, there is enough evidence to reject the claim that the mean price of a 2005 Honda Pilot LX is at least $23,900
Reject H0
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Section 7.4
Hypothesis Testing for Proportions
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z-Test for a Population Proportion
z-Test for a Population Proportion
• A statistical test for a population proportion.
• Can be used when a binomial distribution is given such that np ≥ 5 and nq ≥ 5.
• The test statistic is the sample proportion .
• The standardized test statistic is z.
ˆ
ˆ
ˆ ˆp
p
p p pzpq n
p̂
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Using a z-Test for a Proportion p
1. State the claim mathematically and verbally. Identify the null and alternative hypotheses.
2. Specify the level of significance.
3. Sketch the sampling distribution.
4. Determine any critical value(s).
State H0 and Ha.
Identify α.
Use Table 5 in Appendix B.
Verify that np ≥ 5 and nq ≥ 5
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In Words In Symbols
Using a z-Test for a Proportion p
5. Determine any rejection region(s).
6. Find the standardized test statistic.
7. Make a decision to reject or fail to reject the null hypothesis.
8. Interpret the decision in the context of the original claim.
If z is in the rejection region, reject H0. Otherwise, fail to reject H0.
p̂ pzpq n
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In Words In Symbols
Example: Hypothesis Test for Proportions
Zogby International claims that 45% of people in the United States support making cigarettes illegal within the next 5 to 10 years. You decide to test this claim and ask a random sample of 200 people in the United States whether they support making cigarettes illegal within the next 5 to 10 years. Of the 200 people, 49% support this law. At α = 0.05 is there enough evidence to reject the claim?
Solution:•Verify that np ≥ 5 and nq ≥ 5.
np = 200(0.45) = 90 and nq = 200(0.55) = 11042Larson/Farber 4th ed.
Solution: Hypothesis Test for Proportions
• H0:
• Ha:
• = • Rejection Region:
p = 0.45
p ≠ 0.45
0.05
ˆ 0.49 0.45
(0.45)(0.55) 200
1.14
p pz
pq n
• Decision:At the 5% level of significance, there is not enough evidence to reject the claim that 45% of people in the U.S. support making cigarettes illegal within the next 5 to 10 years.
• Test Statistic
z0-1.96
0.025
1.96
0.025
1.14
Fail to reject H0
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