Transcript Chapter 8

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.
1
Main Objectives
We will study hypothesis testing for
1. population proportion p
2. population mean 
3. population standard deviation 
2
Example
Claim: the XSORT method of gender
selection increases the likelihood of having a
baby girl.
This is a claim about proportion (of girls)
To test this claim 14 couples (volunteers) were
subject to XSORT treatment.
If 6 or 7 or 8 have girls, the method probably
does not increase the likelihood of a girl.
If 13 or 14 couples have girls, the method is
probably increases the likelihood of a girl.
3
Rare Event Rule for
Inferential Statistics
If, under a given assumption, the
probability of a particular observed
event is exceptionally small, we
conclude that the assumption is
probably not correct.
4
Components of a
Formal
Hypothesis Test
5
Null Hypothesis: H0
• The null hypothesis (denoted by H0)
is a statement that the value of a
population parameter (such as
proportion, mean, or standard
deviation) is equal to some claimed
value.
•
We test the null hypothesis directly.
•
Either reject H0 or fail to reject H0
(in other words, accept H0 ).
6
Alternative Hypothesis: H1
• The alternative hypothesis (denoted
by H1) is the statement that the
parameter has a value that somehow
differs from the null hypothesis.
• The symbolic form of the alternative
hypothesis must use one of these
symbols: , <, >.
(not equal, less than, greater than)
7
Example 1
Claim: the XSORT method of gender selection
increases the likelihood of having a baby girl.
We express this claim in symbolic form: p>0.5
(here p denotes the proportion of baby girls)
Null hypothesis must say “equal to”, so
H0 : p=0.5
Alternative hypothesis must express difference:
H1 : p>0.5
Original claim is now the alternative hypothesis
8
Example 1 (continued)
We always test the null hypothesis.
If we reject the null hypothesis, then the original
clam is accepted.
Final conclusion would be: XSORT method
increases the likelihood of having a baby girl.
If we fail to reject the null hypothesis, then the
original clam is rejected.
Final conclusion would be: XSORT method
does not increase the likelihood of having a
baby girl.
9
Example 2
Claim: for couples using the XSORT method
the likelihood of having a baby girl is 50%
Express this claim in symbolic form: p=0.5
(again p denotes the proportion of baby girls)
Null hypothesis must say “equal to”, so
H0 : p=0.5
Alternative hypothesis must express difference:
H1 : p0.5
Original claim is now the null hypothesis
10
Example 2 (continued)
If we reject the null hypothesis, then the
original clam is rejected.
Final conclusion would be: for couples using
the XSORT, the likelihood of having a baby
girl is not 0.5
If we fail to reject the null hypothesis, then the
original clam is accepted.
Final conclusion would be: for couples using
the XSORT the likelihood of having a baby
girl is indeed equal to 0.5
11
Example 3
Claim: for couples using the XSORT method the
likelihood of having a baby girl is at least 0.5
Express this claim in symbolic form: p≥0.5
(again p denotes the proportion of baby girls)
Null hypothesis must say “equal to”, so
H0 : p=0.5 (this agrees with the claim!)
Alternative hypothesis must express difference:
H1 : p<0.5
Original claim is now the null hypothesis
12
Example 3 (continued)
If we reject the null hypothesis, then the
original clam is rejected.
Final conclusion would be: for couples using
the XSORT, the likelihood of having a baby
girl is less 0.5
If we fail to reject the null hypothesis, then the
original clam is accepted.
Final conclusion would be: for couples using
the XSORT the likelihood of having a baby
girl is indeed at least 0.5
13
General rules:
• If the null hypothesis is rejected, the
alternative hypothesis is accepted.
• If the null hypothesis is accepted, the
alternative hypothesis is rejected.
• Acceptance or rejection of the null
hypothesis is an initial conclusion.
• Always state the final conclusion
expressed in terms of the original claim,
not in terms of the null hypothesis or the
alternative hypothesis.
14
Type I Error
• A Type I error is the mistake of
rejecting the null hypothesis when it
is actually true.
• The symbol  (alpha) is used to
represent the probability of a type I
error.
15
Type II Error
• A Type II error is the mistake of
accepting the null hypothesis when it
is actually false.
• The symbol  (beta) is used to
represent the probability of a type II
error.
16
Type I and Type II Errors
17
Example
Claim:
a new medicine has a greater success
rate, p>p0, than the old (existing) one.
Null hypothesis:
H0 : p=p0
Alternative hypothesis: H1 : p>p0
(agrees with the original claim)
18
Example (continued)
Type I error: the null hypothesis is true, but
we reject it => we accept the claim, hence
we adopt the new (inefficient, potentially
harmful) medicine.
This is a critical error, should be avoided!
Type II error: the alternative hypothesis is
true, but we reject it => we reject the claim,
hence we decline the new medicine and
continue using the old one (no harm…).
19
Significance Level
The probability of the type I error
(denoted by ) is also called the
significance level of the test.
It characterizes the chances that the test
fails (i.e., type I error occurs)
It must be a small number. Typical values
used in practice:  = 0.1, 0.05, or 0.01
(in percents, 10%, 5%, or 1%).
20
Testing hypothesis
Step 1: compute Test Statistic
The test statistic is a value used in
making a decision about the null
hypothesis.
The test statistic is computed by a
specific formula depending on the
type of the test.
21
Section 8-3
Testing a Claim About a
Proportion
22
Notation
n = number of trials

x
p = n (sample proportion)
p = population proportion (must be
specified in the null hypothesis)
q=1–p
23
Requirements for Testing Claims
About a Population Proportion p
1) The sample observations are a simple
random sample.
2) The conditions for a binomial distribution
are satisfied.
3) The conditions np  5 and nq  5 are both
satisfied, so the binomial distribution of
sample proportions can be approximated
by a normal distribution with µ = np and
 = npq . Note: p is the assumed
proportion not the sample proportion.
24
Test Statistic for Testing
a Claim About a Proportion

z=
p–p
pq
n
Note: p is the value specified
in the null hypothesis; q = 1-p
25
Example 1 again:
Claim: the XSORT method of gender
selection increases the likelihood of
having a baby girl.
Null hypothesis:
Alternative hypothesis:
H0 : p=0.5
H1 : p>0.5
Suppose 14 couples treated by XSORT
gave birth to 13 girls and 1 boy.
Test the claim at a 5% significance level
26
Compute the test statistic:
pˆ  13 14  0.929
pˆ  p 0.929  0.5
z

 3.21
pq
0.5 0.5 
n
14
27
Draw the diagram (the normal curve)
On the diagram, mark a region of extreme
values that agree with the alternative
hypothesis:
Sample proportion of: pˆ  0.929
or
Test Statistic z = 3.21
28
Critical Region
The critical region (or rejection region)
is the set of all values of the test
statistic that cause us to reject the null
hypothesis.
For example, see the red-shaded region
in the previous figure.
29
Critical Value
A critical value is a value that separates the
critical region (where we reject the null
hypothesis) from the values of the test
statistic that do not lead to rejection of the null
hypothesis.
See the previous figure where the critical
value is z = 1.645. It corresponds to a
significance level of  = 0.05.
30
Significance Level
The significance level (denoted by ) is
the probability that the test statistic will
fall in the critical region (when the null
hypothesis is actually true).
31
Conclusion of the test
Since the test statistic (z=3.21) falls in the
critical region (z>1.645), we reject the
null hypothesis.
Final conclusion: the original claim is
accepted, the XSORT method of gender
selection indeed increases the
likelihood of having a baby girl.
32
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
33
Right-tailed Test
H0: p=0.5
 is in the right tail
H1: p>0.5
Points Right
34
Critical value for a
right-tailed test
A right-tailed test requires
one (positive) critical value:
z
35
Left-tailed Test
H0: p=0.5
 is in the left tail
H1: p<0.5
Points Left
36
Critical value for a
left-tailed test
A left-tailed test requires
one (negative) critical value:
─z
37
Two-tailed Test
H0: p=0.5  is divided equally between
H1: p0.5
the two tails of the critical
region
Means less than or greater than
38
Critical values for a
two-tailed test
A two-tailed test requires
two critical values:
z/2
and
─z/2
39
P-Value
The P-value (or p-value or probability
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.
40
Example 1 (continued)
P-value is the area to the right of the test
statistic z = 3.21.
We refer to Table A-2 (or use calculator) to find
that the area to the right of z = 3.21 is 0.0007.
P-value = 0.0007
41
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.
42
Example 1 (continued)
P-value = 0.0007
It is smaller than  = 0.05.
Hence the null hypothesis must be rejected
43
P-Value
Critical region P-value = area to the right of
in the right tail: 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
(see the following diagram)
44
Procedure for Finding P-Values
Figure 8-5
45
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
46
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).
47
P-Value method:
If P-value is small (  ), reject H0.
If P-value is not small (> ), accept H0.
If the P is low, the null must go.
If the P is high, the null will fly.
48
Testing hypothesis by TI-83/84
•
•
•
•
•
•
Press STAT and select TESTS
Scroll down to 1-PropZTest press ENTER
Type in p0: (claimed proportion, from H0)
x: (number of successes)
n: (number of trials)
choose H1: p ≠p0
<p0
>p0
(two tails) (left tail) (right tail)
• Press on Calculate
• Read the test statistic z=…
• and the P-value p=…
49
Do we prove a claim?
• A statistical test cannot 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
(such as not enough evidence to
convict a suspect).
50
Section 8-4
Testing a Claim About a
Mean:  Known
51
Notation
n = sample size
x = sample mean
 = claimed population mean (from H0)
 = known value of the population
standard deviation
52
Requirements for Testing Claims About
a Population Mean (with  Known)
1) The value of the population standard
deviation  is known.
2) Either or both of these conditions is
satisfied: The population is normally
distributed or n > 30.
53
Test Statistic for Testing a Claim
About a Mean (with  Known)
x–µ
z= 
n
54
Example:
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.
Test the claim that men have a mean weight
greater than 166.3 lb.
55
Example:
Requirements are satisfied:  is known
(26 lb), sample size is 40 (n > 30)
We can express claim as  > 166.3 lb
It does not contain equality, so it is the
alternative hypothesis.
H0:  = 166.3 lb null hypothesis
H1:  > 166.3 lb alternative hypothesis
(and original claim)
56
Example:
Let us set significance level to  = 0.05
Next we calculate z
z
x  x

n
172.55  166.3

 1.52
26
40
It is a right-tailed test, so P-value is
the area is to the right of z = 1.52;
57
Example:
Table A-2: area to the left of z = 1.52
is 0.9357, so the area to the right is
1 – 0.9357 = 0.0643.
The P-value is 0.0643
The P-value of 0.0643 is greater than
the significance level of  = 0.05, we
fail to reject the null hypothesis.
P-value = 0.0643
 = 166.3
or
z=0
x  172.55
or
z = 1.52
58
Example:
The traditional method:
Use critical value z = 1.645 instead of finding
the P-value. Since z = 1.52 does not fall in
the critical region, again fail to reject the null
hypothesis.
59
Testing hypothesis by TI-83/84
• Press STAT and select TESTS
• Scroll down to Z-Test press ENTER
• Choose Data or Stats. For Stats:
• Type in 0: (claimed mean, from H0)
•
: (known st. deviation)
•
x: (sample mean)
•
n: (sample size)
choose H1:  ≠0
<0
>0
(two tails) (left tail) (right tail)
60
• (continued)
• Press on Calculate
• Read the test statistic z=…
• and the P-value p=…
61
Section 8-5
Testing a Claim About a
Mean:  Not Known
62
Notation
n = sample size
x = sample mean
 = claimed population mean (from H0)
s = sample standard deviation
63
Requirements for Testing Claims
About a Population
Mean (with  Not Known)
1) The value of the population standard
deviation  is not known.
2) Either or both of these conditions is
satisfied: The population is normally
distributed or n > 30.
64
Test Statistic for Testing a
Claim About a Mean
(with  Not Known)
x–µ
t= s
n
P-values and Critical Values
Found in Table A-3 or by calculator
Degrees of freedom (df) = n – 1
65
Example:
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 and standard deviation s = 26.33 lb.
Do not assume that the population standard
deviation  is known.
Test the claim that men have a mean weight
greater than 166.3 lb.
66
Example:
Requirements are satisfied:  is not
known, sample size is 40 (n > 30)
We can express claim as  > 166.3 lb
It does not contain equality, so it is the
alternative hypothesis.
H0:  = 166.3 lb null hypothesis
H1:  > 166.3 lb alternative hypothesis
(and original claim)
67
Example:
Let us set significance level to  = 0.05
Next we calculate t
x   x 172.55  166.3
t

 1.501
s
26.33
n
40
df = n – 1 = 39
area of 0.05, one-tail yields
critical value t = 1.685;
68
Example:
t = 1.501 does not fall in the critical
region bounded by t = 1.685, we fail
to reject the null hypothesis.
 = 166.3
or
z=0
x  172.55
Critical value
t = 1.685
or
t = 1.52
69
Example:
Final conclusion:
Because we fail to reject the null hypothesis, we
conclude that there is not sufficient evidence to
support a conclusion that the population mean
is greater than 166.3 lb.
70
Normal Distribution Versus
Student t Distribution
The critical value in the preceding example
was t = 1.782, but if the normal distribution
were being used, the critical value would have
been z = 1.645.
The Student t critical value is larger (farther to
the right), showing that with the Student t
distribution, the sample evidence must be
more extreme before we can consider it to be
significant.
71
P-Value Method
 Use software or a TI-83/84 Plus
calculator.
 If technology is not available, use Table
A-3 to identify a range of P-values
(this will be explained in Section 8.6)
72
Testing hypothesis by TI-83/84
•
•
•
•
•
•
•
•
Press STAT and select TESTS
Scroll down to T-Test press ENTER
Choose Data or Stats. For Stats:
Type in 0: (claimed mean, from H0)
x: (sample mean)
sx: (sample st. deviation)
n: (sample size)
choose H1:  ≠0
<0
>0
(two tails) (left tail) (right tail)
73
• (continued)
• Press on Calculate
• Read the test statistic t=…
• and the P-value p=…
74
Section 8-6
Testing a Claim About a
Standard Deviation or
Variance
75
Notation
n = sample size
s = sample standard deviation
s2 = sample variance
 = claimed value of the population standard
deviation (from H0 )
2 = claimed value of the population
variance (from H0 )
76
Requirements for Testing
Claims About  or  2
1. The sample is a simple random
sample.
2. The population has a normal
distribution. (This is a much stricter
requirement than the requirement of a
normal distribution when testing
claims about means.)
77
Chi-Square Distribution
Test Statistic

2
n  1s



2
2
78
Critical Values for
Chi-Square Distribution
• Use Table A-4.
• The degrees of freedom df = n –1.
79
Table A-4
Table A-4 is based on cumulative areas
from the right.
Critical values are found in Table A-4 by
first locating the row corresponding to the
appropriate number of degrees of freedom
(where df = n –1).
Next, the significance level  is used to
determine the correct column.
The following examples are based on a
significance level of  = 0.05.
80
Critical values
Right-tailed test: needs one critical value
Because the area to the right of the
critical value is 0.05, locate 0.05 at the
top of Table A-4.
Area 
Area 1-

c rit ic al v alue
81
Critical values
Left-tailed test: needs one critical value
With a left-tailed area of 0.05, the area to
the right of the critical value is 0.95, so
locate 0.95 at the top of Table A-4.
Area 
Area 1-
1 -
c rit ic al v alue
82
Critical values
Two-tailed test: needs two critical values
Critical values are two different positive
numbers, both taken from Table A-4
Divide a significance level of 0.05 between
the left and right tails, so the areas to the
right of the two critical values are 0.975
and 0.025, respectively.
Locate 0.975 and 0.025 at top of Table A-4
83
Critical values for a two-tailed test
Area 
Area 
Area 1-
 
lef t critical v alue

 
right crit ical v alue
84
Finding a range for P-value
• A useful interpretation of the P-value: it is
observed level of significance.
• Compare your test statistic 2 with critical
values shown in Table A-4 on the line with
df=n-1 degrees of freedom.
• Find the two critical values that enclose your
test statistic. Determine the significance
levels 1 and 2 for those two critical values.
• Your P-value is between 1 and 2
(see examples below)
85
Examples:
Right-tailed test: If the test statistic 2 is
between critical values corresponding to
the areas 1 and 2 , then your P-value
is between 1 and 2 .
Left-tailed test: If the test statistic 2 is
between critical values corresponding to
the areas 1-1 and 1-2 , then your
P-value is between 1 and 2 .
86
Examples:
Two-tailed test: If the test statistic 2 is
between critical values corresponding to
the areas 1 and 2 , then your P-value
is between 21 and 22 .
Two-tailed test: If the test statistic 2 is
between critical values corresponding to
the areas 1-1 and 1-2 , then your
P-value is between 21 and 22 .
(Note: for two-tailed tests, multiply the
areas by two)
87
Finding the exact P-value by TI-83/84
• Use the test statistic 2 and the calculator function
2 cdf to compute the area of the tail:
2 cdf(teststat,999,df) gives the area of the right tail
(to the right from the test statistic)
2 cdf(-999,teststat,df) gives the area of the left tail
(to the left from the test statistic)
Multiply the area of the tail by 2
if you have a two-tailed test
88