Download Chapter 8 Hypothesis Testing

Chapter 8
Section 8.2
Basics of Hypothesis Testing
Hypothesis Testing
8.2 Basics of Hypothesis Testing
Objective
8.3 Testing about a Proportion p
For a population parameter (p, µ, σ) we wish
to test whether a predicted value is close to
the actual value (based on sample values).
8.4 Testing about a Mean µ (σ known)
8.5 Testing about a Mean µ (σ unknown)
8.6 Testing about a Standard Deviation σ
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Example 1
Definitions
Claim: The XSORT method of gender selection
increases the likelihood of birthing a girl.
(i.e. increases the proportion of girls born)
In statistics, a Hypothesis is a claim or
statement about a property of a population.
A Hypothesis Test is a standard procedure
for testing a claim about a property of a
population.
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To test the claim, use a hypothesis test (about
a proportion) on a sample of 14 couples:
If 6 or 7 have girls, the method probably doesn’t
increase the probability of birthing a girl.
Ch. 8 will cover hypothesis tests about a
• Proportion p
• Mean µ (σ known or σ unknown)
• Standard Deviation σ
If 13 or 14 couples have girls, this method probably
does increase the probability of birthing a girl.
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Rare Event Rule for
Inferential Statistics
This will be explained in Section 8.3
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Components of a
Hypothesis Test
If, under a given assumption, the
probability of a particular event is
exceptionally small, we conclude the
assumption is probably not correct.
Null Hypothesis: H0
Example:
Suppose we assume the probability of pigs flying is 10-10
If we find a farm with 100 flying pigs, we conclude
our assumption probably wasn’t correct
Alternative Hypothesis: H1
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Null Hypothesis: H0
Alternative Hypothesis: H1
The null hypothesis (denoted H0)
is a statement that the value of a
population parameter (p, µ, σ) is
equal to some claimed value.
The alternative hypothesis (denoted H1)
is a statement that the parameter has a
value that somehow differs from the null
hypothesis.
We test the null hypothesis directly. It
will either reject H0 or fail to reject H0
The difference will be one of <, >, ≠
(less than, greater than, doesn’t equal)
(i.e. accept H0)
Example
H0: p = 0.6
Example
H1: p < 0.6
H0: p = 0.6
H1: p < 0.6
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Example 1
Example 1
Claim: The XSORT method of gender selection
increases the likelihood of birthing a girl.
Claim: The XSORT method of gender selection
increases the likelihood of birthing a girl.
• Let p denote the proportion of girls born.
Continued
If we reject the null hypothesis, then the
original clam is accepted.
• The claim is equivilent to “p>0.5”
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The null hypothesis must say “equal to”:
H0 : p = 0.5
Conclusion: The XSORT method increases the
likelihood of having a baby girl.
If we fail to reject the null hypothesis, then the
original clam is rejected.
The alternative hypothesis states the difference:
H1 : p > 0.5
Conclusion: The XSORT method does not increase
the likelihood of having a baby girl.
Here, the original claim is the alternative hypothesis
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Note: We always test the null hypothesis
Example 2
Example 2
Claim: For couples using the XSORT method, the
likelihood of having a girl is 50%
Claim: For couples using the XSORT method, the
likelihood of having a girl is 50%
• Again, let p denote the proportion of girls born.
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Continued
If we reject the null hypothesis, then the
original clam is rejected.
• The claim is equivalent to “p=0.5”
Conclusion: For couples using the XSORT method, the
likelihood of having a girl is not 0.5
The null hypothesis must say “equal to”:
H0 : p = 0.5
If we fail to reject the null hypothesis, then the
original clam is accepted.
The alternative hypothesis states the difference:
H1 : p ≠ 0.5
Conclusion: For couples using the XSORT method, the
likelihood of having a girl is indeed 0.5
Here, the original claim is the null hypothesis
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Note: We always test the null hypothesis
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Example 3
Example 3
Claim: For couples using the XSORT method, the
likelihood of having a girl is at least 50%
Claim: For couples using the XSORT method, the
likelihood of having a girl is at least 50%
• Again, let p denote the proportion of girls born.
If we reject the null hypothesis, then the
original clam is rejected.
• The claim is equivalent to “p ≥ 0.5”
Conclusion: For couples using the XSORT method, the
likelihood of having a girl is less than 0.5
The null hypothesis must say “equal to”:
H0 : p = 0.5
If we fail to reject the null hypothesis, then the
original clam is accepted.
The alternative hypothesis states the difference:
H1 : p < 0.5
Continued
we can’t use ≥ or ≤ in the alternative
hypothesis, so we test the negation
Conclusion: For couples using the XSORT method, the
likelihood of having a girl is at least 0.5
Here, the original claim is the null hypothesis
Note: We always test the null hypothesis
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Type I Error
General rules
• If the null hypothesis is rejected,
the alternative hypothesis is accepted.
• A Type I error is the mistake of rejecting
the null hypothesis when it is actually true.
H0 rejected → H1 accepted
• If the null hypothesis is accepted,
the alternative hypothesis is rejected.
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H0 accepted → H1 rejected
• Acceptance or rejection of the null hypothesis
is called an initial conclusion.
• The final conclusion is always expressed in
terms of the original claim. Not in terms of the
null hypothesis or alternative hypothesis.
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• Also called a “True Negative”
True: means the actual hypothesis is true
Negative: means the test rejected the hypothesis
• The symbol  (alpha) is used to represent
the probability of a type I error.
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Type II Error
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Type I and Type II Errors
• A Type II error is the mistake of accepting
the null hypothesis when it is actually false.
• Also called a “False Positive”
False: means the actual hypothesis is false
Positive: means the test failed to reject the hypothesis
• The symbol  (beta) is used to represent
the probability of a type II error.
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Example 4
Example 4
Claim: A new medication has greater success rate (p)
than that of the old (existing) machine (p0)
Continued
Claim: A new medication has greater success rate (p)
than that of the old (existing) machine (p0)
H0 : p = p0
• p: Proportion of success for the new medication
• p0: Proportion of success for the old medication
H1 : p > p0
Type I error
H0 is true, but we reject it → We accept the claim
• The claim is equivalent to “p > p0”
So we adopt the new (inefficient, potentially harmful) medicine.
Null hypothesis:
H 0 : p = p0
Alternative hypothesis:
H 1 : p > p0
(This is called a critical error, must be avoided)
Type II error
H1 is true, but we reject it → We reject the claim
So we decline the new medicine and continue with the old one.
Here, the original claim is the null hypothesis
(no direct harm…)
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Critical Region
Significance Level
Consider a parameter (p, µ, σ, etc.)
• The probability of a type I error (denoted ) is
also called the significance level of the test.
The “guess” for the parameter will have a probability
that follows a certain distribution (z, t, χ2,etc.)
• Characterizes the chance the test will fail.
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(i.e. the chance of a type I error)
• Used to set the “significance” of a hypothesis test.
(i.e. how reliable the test is in avoiding type I errors)
• Lower significance → Lower chance of type I error
Note: This is just like what we used to calculate CIs.
Using the significance level α, we determine the
region where the guessed value becomes unusual.
This is known as the critical region.
The region is described using critical value(s).
(Like those used for finding confidence intervals)
• Values used most:  = 0.1, 0.05, 0.01
(i.e. 10%, 5%, 1%, just like with CIs)
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Example
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Testing a Claim Using a Hypothesis Test
p follows a z-distribution
1. State the H0 and H1
If we guess p > p0 the critical region is defined by the
right tail whose area is α
2. Compute the test statistic
Depends on the value being tested
tα
3. Compute the critical region for the test statistic
If we guess p < p0 the critical region is defined by the
left tail whose area is α
Depends on the distribution of the test statistic (z, t, χ2)
Depends on the significance level α
Found using the critical values
-tα
4. Make an initial conclusion from the test
If we guess p ≠ p0 the critical region is defined by the
two tails whose areas are α/2
-tα/2
Reject H0 (accept H1) if the test statistic is within the critical region
Accept H0 if test statistic is not within the critical region
5. Make a final conclusion about the claim
tα/2
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State it in terms of the original claim
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Example 5
Claim: The XSORT method of gender selection
increases the likelihood of birthing a girl.
Suppose 14 couples using XSORT had 13 girls and 1 boy.
Test the claim at a 5% significance level
1. State H0 and H1
H0 : p = 0.5
H1 : p > 0.5
2. Find the test statistic
3. Find the critical region
4. Initial conclusion
5. Final conclusion
We accept the claim
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Notation
Section 8.3
Testing a claim about a Proportion
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Objective
For a population with proportion p, use a
sample (with a sample proportion) to test
a claim about the proportion.
Testing a proportion uses the standard
normal distribution (z-distribution)
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Requirements
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Test Statistic
(1) The sample used is a a simple random sample
Denoted z (as in z-score) since
the test uses the z-distribution.
(i.e. selected at random, no biases)
(2) Satisfies conditions for a Binomial distribution
(3) n p0 ≥ 5 and n q0 ≥ 5
Note: 2 and 3 satisfy conditions for the normal
approximation to the binomial distribution
Note: p0 is the assumed proportion, not the sample proportion
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Types of Hypothesis Tests:
Two-tailed, Left-tailed, Right-tailed
Traditional method:
If the test statistic falls within the
critical region, reject H0.
The tails in a distribution are the extreme
regions where values of the test statistic
agree with the alternative hypothesis
If the test statistic does not fall within
the critical region, fail to reject H0
(i.e. accept H0).
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Left-tailed Test “<”
H0: p = 0.5
Right-tailed Test “>”
H0: p = 0.5
 significance level
H1: p < 0.5
 significance level
H1: p > 0.5
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Area = 
Area = 
-z
z
(Negative)
(Positive)
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Example 1
Two-tailed Test “≠”
H0: p = 0.5
H1: p ≠ 0.5
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The XSORT method of gender selection is believed to
increases the likelihood of birthing a girl.
 significance level
14 couples used the XSORT method and resulted in the
birth of 13 girls and 1 boy.
Using a 0.05 significance level, test the claim that the
XSORT method increases the birth rate of girls.
Area = /2
-z/2
(Assume the normal birthrate of girls is 0.5)
Area = /2
z/2
What we know:
p0 = 0.5
n = 14
Claim: p > 0.5
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x = 13
using
p = 0.9286
α = 0.05
n p0 = 14*0.5 = 7
n q0 = 14*0.5 = 7
Since n p0 > 5 and n q0 > 5, we can perform a hypothesis test.
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Example 1
What we know:
p0 = 0.5
n = 14
Claim: p > 0.5
x = 13
P-Value
p = 0.9286
α = 0.01
using
The P-value is the probability of getting a
value of the test statistic that is at least as
extreme as the one representing the
sample data, assuming that the null
hypothesis is true.
H0 : p = 0.5
H1 : p > 0.5
Right-tailed
Test statistic:
zα = 1.645
Critical value:
z = 3.207
Example
z in critical region
z Test statistic
zα Critical value
Initial Conclusion: Since z is in the critical region, reject H0
P-value = P(Z > z)
Final Conclusion: We Accept the claim that the XSORT
method increases the birth rate of girls
P-value = area to the right of
the test statistic
Critical region
in the left tail:
P-value = area to the left of the
test statistic
Critical region
in two tails:
P-value = twice the area in the tail
beyond the test statistic
z zα
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P-Value
Critical region
in the right tail:
p-value
(area)
P-Value method:
If P-value   , reject H0.
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If P-value >  , fail to reject H0.
If the P is low, the null must go.
If the P is high, the null will fly.
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Calculating P-value for a Proportion
Caution
Stat → Proportions → One sample → with summary
Don’t confuse a P-value with a proportion p.
Know this distinction:
P-value = probability of getting a test statistic
at least as extreme as the one
representing sample data
p = population proportion
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Calculating P-value for a Proportion
Calculating P-value for a Proportion
Enter the number of successes (x) and the
number of observations (n)
Enter the Null proportion (p0) and select the
alternative hypothesis (≠, <, or >)
Then hit Calculate
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Example 1 Using P-value
Calculating P-value for a Proportion
What we know:
p0 = 0.5
n = 14
Claim: p > 0.5
The resulting table shows both the
test statistic (z) and the P-value
H0 : p = 0.5
x = 13
using
p = 0.9286
α = 0.01
Stat → Proportions→ One sample → With summary
H1 : p > 0.5
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Number of successes:
13
Number of observations:
14
● Hypothesis Test
Null: proportion=
0.5
Alternative
>
P-value = 0.0007
P-value
Test statistic
P-value = 0.0007
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Initial Conclusion: Since p-value < α (α = 0.05), reject H0
Final Conclusion: We Accept the claim that the XSORT
method increases the birth rate of girls
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Example 2
Do we prove a claim?
Problem 32, pg 424
Mendel’s Genetics Experiments
When Gregor Mendel conducted his famous
hybridization experiments with peas, one such
experiment resulted in 580 offspring peas, with 26.2%
of them having yellow pods. According to Mendel’s
theory, ¼ of the offspring peas should have yellow
pods. Use a 0.05 significance level to test the claim that
the proportion of peas with yellow pods is equal to ¼.
A statistical test cannot definitely prove
a hypothesis or a claim.
Our conclusion can be only stated like this:
The available evidence is not strong enough
to warrant rejection of a hypothesis or a
claim
What we know:
We can say we are 95% confident it holds.
p0 = 0.25
n = 580
Claim: p = 0.25
“The only definite is that there are no definites” -Unknown
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p = 0.262
using
α = 0.05
n p0 = 580*0.25 = 145
n q0 = 580*0.75 = 435
Since n p0 > 5 and n q0 > 5, we can perform a hypothesis test.
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Example 2 Using P-value
Example 2
What we know:
p0 = 0.25
n = 580
Claim: p = 0.25
p = 0.262
using
What we know:
α = 0.05
p0 = 0.25
n = 580
Claim: p = 0.25
H0 : p = 0.25
H0 : p = 0.25
H1 : p ≠ 0.25 Two-tailed
H1 : p ≠ 0.25
p = 0.262
using
α = 0.05
Stat → Proportions→ One sample → With summary
Number of successes: 152
Number of observations: 580
● Hypothesis Test
Null: proportion=
Test statistic:
≠
x = np = 580*0.262 ≈ 152
zα = 1.960
zα = -1.960
0.25
Alternative
P-value = 0.5021
z = 0.667
Critical value:
z not in critical region
Initial Conclusion: Since z is not in the critical region, accept H0
Initial Conclusion: Since P-value > α, accept H0
Final Conclusion: We Accept the claim that the proportion
of peas with yellow pods is equal to ¼
Final Conclusion: We Accept the claim that the proportion
of peas with yellow pods is equal to ¼
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Section 8.4
Testing a claim about a mean
(σ known)
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Objective
For a population with mean µ (with σ known),
use a sample (with a sample mean) to test a
claim about the mean.
Testing a mean (when σ known) uses the
standard normal distribution (z-distribution)
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Notation
Requirements
(1) The population standard deviation σ is known
(2) One or both of the following:
The population is normally distributed
or
n > 30
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Example 1
Test Statistic
People have died in boat accidents because an obsolete
estimate of the mean weight of men (166.3 lb.) was used.
Denoted z (as in z-score) since
the test uses the z-distribution.
A random sample of n = 40 men yielded the mean
x = 172.55 lb. Research from other sources suggests that
the population of weights of men has a standard deviation
given by  = 26 lb.
Use a 0.1 significance level to test the claim that men have
a mean weight greater than 166.3 lb.
What we know:
µ0 = 166.3
n = 40
Claim: µ > 166.3
x = 172.55 σ = 26
using
α = 0.1
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Example 1 Using Critical Region
What we know:
µ0 = 166.3
n = 40
Claim: µ > 166.3
Calculating P-value for a Mean
x = 172.55 σ = 26
using
(σ known)
α = 0.05
Stat → Z statistics → One sample → with summary
H0 : µ = 166.3
H1 : µ > 166.3 Right-tailed
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Test statistic:
zα = 1.282
z = 1.520
Critical value:
z in critical region
Initial Conclusion: Since z is in the critical region, reject H0
Final Conclusion: We Accept the claim that the actual mean
weight of men is greater than 166.3 lb.
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Calculating P-value for a Mean
Calculating P-value for a Mean
(σ known)
(σ known)
Then hit Calculate
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Example 1 Using P-value
Calculating P-value for a Mean
(σ known)
What we know:
µ0 = 166.3
The resulting table shows both the
test statistic (z) and the P-value
x = 172.55 σ = 26
n = 40
Claim: µ > 166.3
using
α = 0.05
H0 : µ = 166.3
Stat → Z statistics→ One sample → With summary
H1 : µ > 166.3
Standard deviation:
26
Null: proportion=
Sample size:
40
Alternative
Sample mean: 172.55
● Hypothesis Test
166.3
>
P-value = 0.0642
P-value
Test statistic
Initial Conclusion: Since P-value < α, reject H0
P-value = 0.0642
Final Conclusion: We Accept the claim that the actual mean
weight of men is greater than 166.3 lb.
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Example 2 Using Critical Region
Example 2
What we know:
Weight of Bears
A sample of 54 bears has a mean weight of 237.9 lb.
µ0 = 250
Claim: µ < 250
H1 : µ < 250
using
x = 237.9 σ = 37.8
using
α = 0.05
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Left-tailed
Test statistic:
–zα = –1.645
x = 237.9 σ = 37.8
n = 54
n = 54
H0 : µ = 250
Assuming that σ is known to be 37.8 lb. use a 0.05
significance level to test the claim that the population mean
of all such bear weights is less than 250 lb.
What we know:
µ0 = 250
Claim: µ < 250
z = –2.352
Critical value:
α = 0.05
z in critical region
Initial Conclusion: Since z is in the critical region, reject H0
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Example 2 Using P-value
What we know:
µ0 = 250
n = 54
Claim: µ < 250
x = 237.9 σ = 37.8
using
α = 0.05
H0 : µ = 250
Stat → Z statistics→ One sample → With summary
H1 : µ < 250
Standard deviation:
Sample mean: 237.9
Sample size:
37.8
● Hypothesis Test
Null: proportion=
Alternative
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250
<
P-value = 0.0093
Initial Conclusion: Since P-value < α, reject H0
Final Conclusion: We Accept the claim that the mean weight
of bears is less than 250 lb.
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Final Conclusion: We Accept the claim that the mean weight
of bears is less than 250 lb.
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