Download Ch8a

Chapter 8
Hypothesis Testing
8.2 Basics of Hypothesis Testing
8.3 Testing about a Proportion p
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
1
Section 8.2
Basics of Hypothesis Testing
Objective
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).
2
Definitions
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.
Ch. 8 will cover hypothesis tests about a
• Proportion p
• Mean µ (σ known or σ unknown)
• Standard Deviation σ
3
Example 1
Claim: The XSORT method of gender selection
increases the likelihood of birthing a girl.
(i.e. increases the proportion of girls born)
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.
If 13 or 14 couples have girls, this method probably
does increase the probability of birthing a girl.
This will be explained in Section 8.3
4
Rare Event Rule for
Inferential Statistics
If, under a given assumption, the
probability of a particular event is
exceptionally small, we conclude the
assumption is probably not correct.
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
5
Components of a
Hypothesis Test
Null Hypothesis: H0
Alternative Hypothesis: H1
6
Null Hypothesis: H0
The null hypothesis (denoted H0)
is a statement that the value of a
population parameter (p, µ, σ) is
equal to some claimed value.
We test the null hypothesis directly. It
will either reject H0 or fail to reject H0
(i.e. accept H0)
Example
H0: p = 0.6
H1: p < 0.6
7
Alternative Hypothesis: H1
The alternative hypothesis (denoted H1)
is a statement that the parameter has a
value that somehow differs from the null
hypothesis.
The difference will be one of <, >, ≠
(less than, greater than, doesn’t equal)
Example
H0: p = 0.6
H1: p < 0.6
8
Example 1
Claim: The XSORT method of gender selection
increases the likelihood of birthing a girl.
• Let p denote the proportion of girls born.
• The claim is equivilent to “p>0.5”
The null hypothesis must say “equal to”:
H0 : p = 0.5
The alternative hypothesis states the difference:
H1 : p > 0.5
Here, the original claim is the alternative hypothesis
9
Example 1
Continued
Claim: The XSORT method of gender selection
increases the likelihood of birthing a girl.
If we reject the null hypothesis, then the
original clam is accepted.
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.
Conclusion: The XSORT method does not increase
the likelihood of having a baby girl.
Note: We always test the null hypothesis
10
Example 2
Claim: For couples using the XSORT method, the
likelihood of having a girl is 50%
• Again, let p denote the proportion of girls born.
• The claim is equivalent to “p=0.5”
The null hypothesis must say “equal to”:
H0 : p = 0.5
The alternative hypothesis states the difference:
H1 : p ≠ 0.5
Here, the original claim is the null hypothesis
11
Example 2
Continued
Claim: For couples using the XSORT method, the
likelihood of having a girl is 50%
If we reject the null hypothesis, then the
original clam is rejected.
Conclusion: For couples using the XSORT method, the
likelihood of having a girl is not 0.5
If we fail to reject the null hypothesis, then the
original clam is accepted.
Conclusion: For couples using the XSORT method, the
likelihood of having a girl is indeed 0.5
Note: We always test the null hypothesis
12
Example 3
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.
• The claim is equivalent to “p ≥ 0.5”
The null hypothesis must say “equal to”:
H0 : p = 0.5
The alternative hypothesis states the difference:
H1 : p < 0.5
we can’t use ≥ or ≤ in the alternative
hypothesis, so we test the negation
Here, the original claim is the null hypothesis
13
Example 3
Continued
Claim: For couples using the XSORT method, the
likelihood of having a girl is at least 50%
If we reject the null hypothesis, then the
original clam is rejected.
Conclusion: For couples using the XSORT method, the
likelihood of having a girl is less than 0.5
If we fail to reject the null hypothesis, then the
original clam is accepted.
Conclusion: For couples using the XSORT method, the
likelihood of having a girl is at least 0.5
Note: We always test the null hypothesis
14
General rules
• If the null hypothesis is rejected,
the alternative hypothesis is accepted.
H0 rejected → H1 accepted
• If the null hypothesis is accepted,
the alternative hypothesis is rejected.
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.
15
Type I Error
• A Type I error is the mistake of rejecting
the null hypothesis when it is actually true.
• 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.
16
Type II Error
• 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.
17
Type I and Type II Errors
18
Example 4
Claim: A new medication has greater success rate (p)
than that of the old (existing) machine (p0)
• p: Proportion of success for the new medication
• p0: Proportion of success for the old medication
• The claim is equivalent to “p > p0”
Null hypothesis:
H0 : p = p0
Alternative hypothesis:
H1 : p > p0
Here, the original claim is the null hypothesis
19
Example 4
Continued
Claim: A new medication has greater success rate (p)
than that of the old (existing) machine (p0)
H0 : p = p0
Type I error
H1 : p > p0
H0 is true, but we reject it → We accept the claim
So we adopt the new (inefficient, potentially harmful) medicine.
(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.
(no direct harm…)
20
Significance Level
• The probability of a type I error (denoted ) is
also called the significance level of the test.
• Characterizes the chance the test will fail.
(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
• Values used most:  = 0.1, 0.05, 0.01
(i.e. 10%, 5%, 1%, just like with CIs)
21
Critical Region
Consider a parameter (p, µ, σ, etc.)
The “guess” for the parameter will have a probability
that follows a certain distribution (z, t, χ2,etc.)
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)
22
Example
p follows a z-distribution
If we guess p > p0 the critical region is defined by the
right tail whose area is α
tα
If we guess p < p0 the critical region is defined by the
left tail whose area is α
-tα
If we guess p ≠ p0 the critical region is defined by the
two tails whose areas are α/2
-tα/2
tα/2
23
Testing a Claim Using a Hypothesis Test
1. State the H0 and H1
2. Compute the test statistic
Depends on the value being tested
3. Compute the critical region for the test statistic
Depends on the distribution of the test statistic (z, t, χ2)
Depends on the significance level α
Found using the critical values
4. Make an initial conclusion from the test
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
State it in terms of the original claim
24
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
25
26
Section 8.3
Testing a claim about a Proportion
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)
27
Notation
28
Requirements
(1) The sample used is a a simple random sample
(i.e. selected at random, no biases)
(2) Satisfies conditions for a Binomial distribution
(3) np0 ≥ 5 and nq0 ≥ 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
29
Test Statistic
Denoted z (as in z-score) since
the test uses the z-distribution.
30
Traditional method:
If the test statistic falls within the
critical region, reject H0.
If the test statistic does not fall within
the critical region, fail to reject H0
(i.e. accept H0).
31
Types of Hypothesis Tests:
Two-tailed, Left-tailed, Right-tailed
The tails in a distribution are the extreme
regions where values of the test statistic
agree with the alternative hypothesis
32
Left-tailed Test “<”
H0: p = 0.5
 significance level
H1: p < 0.5
Area = 
-z
(Negative)
33
Right-tailed Test “>”
H0: p = 0.5
 significance level
H1: p > 0.5
Area = 
z
(Positive)
34
Two-tailed Test “≠”
H0: p = 0.5
H1: p ≠ 0.5
Area = /2
-z/2
 significance level
Area = /2
z/2
35
Example 1
The XSORT method of gender selection is believed to
increases the likelihood of birthing a girl.
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.
(Assume the normal birthrate of girls is 0.5)
What we know:
p0 = 0.5
n = 14
Claim: p > 0.5
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.
36
Example 1
What we know:
p0 = 0.5
n = 14
Claim: p > 0.5
x = 13
using
p = 0.9286
α = 0.01
H0 : p = 0.5
H1 : p > 0.5
Right-tailed
Test statistic:
Critical value:
zα = 1.645
z = 3.207
z in critical region
Initial Conclusion: Since z is in the critical region, reject H0
Final Conclusion: We Accept the claim that the XSORT
method increases the birth rate of girls
37
P-Value
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.
Example
z Test statistic
zα Critical value
P-value = P(Z > z)
p-value
(area)
z zα
38
P-Value
Critical region
in the right tail:
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
39
P-Value method:
If P-value   , reject H0.
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.
40
Caution
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
41
Calculating P-value for a Proportion
Stat → Proportions → One sample → with summary
42
Calculating P-value for a Proportion
Enter the number of successes (x) and the
number of observations (n)
43
Calculating P-value for a Proportion
Enter the Null proportion (p0) and select the
alternative hypothesis (≠, <, or >)
Then hit Calculate
44
Calculating P-value for a Proportion
The resulting table shows both the
test statistic (z) and the P-value
P-value
Test statistic
P-value = 0.0007
45
Example 1 Using P-value
What we know:
p0 = 0.5
n = 14
Claim: p > 0.5
H0 : p = 0.5
H1 : p > 0.5
x = 13
using
p = 0.9286
α = 0.01
Stat → Proportions→ One sample → With summary
Number of successes:
13
Number of observations:
14
● Hypothesis Test
Null: proportion=
Alternative
0.5
>
P-value = 0.0007
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
46
Do we prove a claim?
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
We can say we are 95% confident it holds.
“The only definite is that there are no definites” -Unknown
47
Example 2
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.
What we know:
p0 = 0.25
n = 580
Claim: p = 0.25
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.
48
Example 2
What we know:
p0 = 0.25
n = 580
Claim: p = 0.25
p = 0.262
using
α = 0.05
H0 : p = 0.25
H1 : p ≠ 0.25 Two-tailed
Test statistic:
zα = 1.960
zα = -1.960
z = 0.667
Critical value:
z not in critical region
Initial Conclusion: Since z is not in the critical region, accept H0
Final Conclusion: We Accept the claim that the proportion
of peas with yellow pods is equal to ¼
49
Example 2 Using P-value
What we know:
p0 = 0.25
n = 580
Claim: p = 0.25
H0 : p = 0.25
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=
0.25
Alternative
≠
x = np = 580*0.262 ≈ 152
P-value = 0.5021
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
50
51
Section 8.4
Testing a claim about a mean
(σ known)
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)
52
Notation
53
Requirements
(1) The population standard deviation σ is known
(2) One or both of the following:
The population is normally distributed
or
n > 30
54
Test Statistic
Denoted z (as in z-score) since
the test uses the z-distribution.
55
Example 1
People have died in boat accidents because an obsolete
estimate of the mean weight of men (166.3 lb.) was used.
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
56
Example 1 Using Critical Region
What we know:
µ0 = 166.3
n = 40
Claim: µ > 166.3
x = 172.55 σ = 26
using
α = 0.05
H0 : µ = 166.3
H1 : µ > 166.3 Right-tailed
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.
57
Calculating P-value for a Mean
(σ known)
Stat → Z statistics → One sample → with summary
58
Calculating P-value for a Mean
(σ known)
59
Calculating P-value for a Mean
(σ known)
Then hit Calculate
60
Calculating P-value for a Mean
(σ known)
The resulting table shows both the
test statistic (z) and the P-value
P-value
Test statistic
P-value = 0.0642
61
Example 1 Using P-value
What we know:
µ0 = 166.3
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
Sample mean: 172.55
Standard deviation: 26
Sample size:
● Hypothesis Test
Null: proportion=
Alternative
40
166.3
>
P-value = 0.0642
Initial Conclusion: Since P-value < α, reject H0
Final Conclusion: We Accept the claim that the actual mean
weight of men is greater than 166.3 lb.
62
Example 2
Weight of Bears
A sample of 54 bears has a mean weight of 237.9 lb.
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
n = 54
Claim: µ < 250
x = 237.9 σ = 37.8
using
α = 0.05
63
Example 2 Using Critical Region
What we know:
µ0 = 250
n = 54
Claim: µ < 250
x = 237.9 σ = 37.8
using
α = 0.05
H0 : µ = 250
H1 : µ < 250
Left-tailed
Test statistic:
–zα = –1.645
z = –2.352
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 mean weight
of bears is less than 250 lb.
64
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
Sample mean: 237.9
Standard deviation: 37.8
Sample size:
● Hypothesis Test
Null: proportion=
Alternative
54
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.
65