• Study Resource
• Explore

Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Student's t-test wikipedia, lookup

Taylor's law wikipedia, lookup

Bootstrapping (statistics) wikipedia, lookup

Confidence interval wikipedia, lookup

Resampling (statistics) wikipedia, lookup

Misuse of statistics wikipedia, lookup

Degrees of freedom (statistics) wikipedia, lookup

Psychometrics wikipedia, lookup

Transcript
```Chapter 9
Inferences from Two Samples
In this chapter we will deal with two samples
from two populations.
The general goal is to compare the
parameters of the two populations.
For the first population we use index 1, for the
second population index 2.
1
Section 9-2
Two Proportions
2
Notation for Two Proportions
For the first population, we let:
p1 = first population proportion
n1 = size of the first sample
x1 = number of successes in the first
sample
x1
^
p1 = n (the first sample proportion)
1
^q = 1 – ^p
1
1
p2, n2 , x2 , p^2, and q^2 are used for the second population.
3
Pooled Sample Proportion
 The pooled sample proportion
is denoted by p and is given by:
x1 + x2
p= n +n
1
2
 We denote
q =1–p
4
Requirements
1. We have two independent random
samples.
2. For each of the two samples, the
number of successes is at least 5 and
the number of failures is at least 5.
5
Tests for Two Proportions
The goal is to compare the two proportions.
H0: p1 = p2
H1: p1  p2 ,
two tails
H1: p1 < p2 , H1: p 1> p2
left tail
right tail
Note: no numerical values for p1 or p2 are
claimed in the hypotheses.
6
Test Statistic for Two Proportions
z=
^ )–(p –p )
( p^1 – p
2
1
2
pq
pq
n 1 + n2
Note: p1 –
p^
1
p=
x1 + x 2
n1 + n2
p2 =0 according to H0
x1
= n
1
and
and
p^
2
x2
=
n2
q=1–p
7
Example:
The table below lists results from a simple
random sample of front-seat occupants
involved in car crashes.
Use a 0.05 significance level to test the claim
that the fatality rate of occupants is lower for
those in cars equipped with airbags.
8
Example:
Requirements are satisfied: two simple random
samples, two samples are independent; Each
has at least 5 successes and 5 failures.
Step 1: Express the claim as p1 < p2.
Step 2: p1 < p2 does not contain equality so it is
the alternative hypothesis. The null
hypothesis is the statement of equality.
9
Example:
H0: p1 = p2
H1: p1 < p2 (original claim)
Step 3: Significance level is 0.05
Step 4: Compute the pooled proportion:
x1  x2
41  52
p

 0.004347
n1  n2 11,541  9,853
With
p  0.004347
it follows
q  0.995653
10
Example:
Step 5: Find the value of the test statistic.
p̂1  p̂2  p1  p2 

z
pq pq

n1 n2

52 
 41
 11, 541  9,853   0


 0.004347  0.995653   0.004347  0.995653
11, 541
9,853
z  1.91
11
Example:
Left-tailed test. Area to left of z = –1.91 is
0.0281 (Table A-2), so the P-value is 0.0281.
12
Example:
Step 6: Because the P-value of 0.0281 is less
than the significance level of  = 0.05, we
reject the null hypothesis of p1 = p2.
Because we reject the null hypothesis, we
conclude that there is sufficient evidence to
support the original claim.
Final conclusion:
the proportion of accident fatalities for
occupants in cars with airbags is less than the
proportion of fatalities for occupants in cars
without airbags.
13
With a
significance
level of  = 0.05
in a left- tailed
test,
the critical value is z = –1.645. The test statistic
of z = –1.91 does fall in the critical region
bounded by the critical value of z = –1.645.
We again reject the null hypothesis.
14
Confidence Interval
Estimate of p1 – p2
( p^1 – p^2 ) – E < ( p1 – p2 ) < ( p^ 1
where E =
z  
^ )+
–p
2
E
p^1 q^1
p^2 q^2
n1 + n2
15
Example:
Use the same sample data to construct a
90% confidence interval estimate of the
difference between the two population
proportions.
Note: 1─ = 0.90, so  = 0.10 and  = 0.05 .
16
Example:
Requirements are satisfied as we saw earlier.
90% confidence interval: z/2 = 1.645
Calculate the margin of error, E
E  z 2
p̂1q̂1 p̂2 q̂2

n1
n2
 41   11, 500   52   9801 

 
 
 

11, 541 11, 541
9, 853 9, 853
 1.645

11, 541
9, 853
 0.001507
17
Example:
Construct the confidence interval
p̂1  p̂2  E  p1  p2   p̂1  p̂2  E
0.003553  0.005278   0.001507
 p1  p2  
0.003553  0.005278   0.001507
0.00323  p1  p2   0.000218
18
Final note:
The confidence interval limits do not contain 0,
implying that there is a significant difference
between the two proportions.
Thus the confidence interval, too, suggests
that the fatality rate is lower for occupants in
cars with air bags than for occupants in cars
without air bags.
19
Two proportions by TI-83/84
•
•
•
•
•
•
•
Press STAT and select TESTS
Scroll down to 2-PropZTest press ENTER
Type in x1: (number of successes in 1st sample)
n1: (number of trials in 1st sample)
x2: (number of successes in 2nd sample)
n2: (number of trials in 2nd sample)
choose H1: p1 ≠p2
<p2
>p2
(two tails) (left tail) (right tail)
• Press on Calculate
• Read test statistic z=… and P-value p=…
20
Two proportions by TI-83/84
• Press STAT and select TESTS
• Scroll down to 2-PropZInt
press ENTER
• Type in x1: (number of successes in 1st sample)
•
n1: (number of trials in 1st sample)
•
x2: (number of successes in 2nd sample)
•
n2: (number of trials in 2nd sample)
C-Level: (confidence level)
• Press on Calculate
21
Section 9-3
Two Means:
Independent Samples
22
Definitions
Two samples are independent if
the sample values selected from
one population are not related to
or somehow paired or matched
with the sample values from the
other population.
23
Notation for the first population:
1 = population mean
σ1 = population standard deviation
n1 = size of the first sample
x1 = sample mean
s1 = sample standard deviation
Corresponding notations for 2, σ2, s2, x
2
and n2 apply to the second population.
24
Requirements
1. σ1 an σ2 are unknown and no assumption is
2. The two samples are independent.
3. Both samples are random samples.
4. Either or both of these conditions are
satisfied: The two sample sizes are both
large (with n1 > 30 and n2 > 30) or both
populations have normal distributions.
25
Tests for Two Means
The goal is to compare the two means.
H 0:  1 =  2
H 1:  1   2 ,
two tails
H 1:  1 <  2 , H 1 :  1 >  2
left tail
right tail
Note: no numerical values for
claimed in the hypotheses.
1 or 2 are
26
Hypothesis Test for Two Means
with Independent Samples:
Test Statistic is
x  x  

t
1
2
1
2
1
 2

2
2
s
s

n1 n2
Note: 1 –
2 =0 according to H0
Degrees of freedom: df = smaller of n1 – 1 and n2 – 1.
27
Example:
A headline in USA Today proclaimed that “Men,
women are equal talkers.” That headline
referred to a study of the numbers of words that
men and women spoke in a day, see below.
Use a 0.05 significance level to test the claim
that men and women speak the same mean
number of words in a day.
28
Example:
Requirements are satisfied: two population
standard deviations are not known and not
assumed to be equal, independent samples,
both samples are large.
Step 1: Express claim as 1 = 2.
Step 2: If original claim is false, then 1 ≠ 2.
Step 3: Alternative hypothesis does not
contain equality, null hypothesis does.
H0 : 1 = 2 (original claim)
H1 : 1 ≠ 2
29
Example:
Step 4: Significance level is 0.05
Step 5: Use a t distribution
Step 6: Calculate the test statistic
x  x  

t
1
2
1
 2

s12 s22

n1 n2
15,668.5  16,215.0  0


 0.676
8632.5 2 7301.22

186
210
30
Example:
Use Table A-3: area in two tails is 0.05, df = 185,
which is not in the table, the closest value is
t = ±1.972
31
Example:
Step 7: Because the test statistic does not fall
within the critical region, fail to reject
the null hypothesis:
1 = 2 (or 1 – 2 = 0).
Final conclusion:
There is sufficient evidence to support the
claim that men and women speak the same
mean number of words in a day.
32
Confidence Interval Estimate of
1 – 2: Independent Samples
(x1 – x2) – E < (µ1 – µ2) < (x1 – x2) + E
where E =
t
s2
s
+
n2
n1
2
1
2
df = smaller n1 – 1 and n2 – 1
33
Example:
Using the given sample data, construct a
95% confidence interval estimate of the
difference between the mean number of
words spoken by men and the mean
number of words spoken by women.
34
Example:
Find the margin of Error, E; use t/2 = 1.972
E  t
2
s12 s22
8632.52 7301.22

 1.972

 1595.4
n1 n2
186
210
Construct the confidence interval use E = 1595.4
and x1  15,668.5 and x2  16,215.0.
x  x  E  
2141.9  
1
2
1
1
 

   1048.9
 2  x1  x2  E
2
35
Tests about two means by TI-83/84
•
•
•
•
•
•
•
•
•
Press STAT and select TESTS
Scroll down to 2-SampTTest press ENTER
Select Input: Data or Stats. For Stats:
Type in x1: (1st sample mean)
sx1: (1st sample st. deviation)
n1: (1st sample size)
x2: (2nd sample mean)
sx2: (2nd sample st. deviation)
n2: (2nd sample size)
choose H1: 1 ≠2
<2
> 2
(two tails) (left tail) (right tail)
36
• choose Pooled: No or Yes (always No)
•
•
•
•
Press on Calculate
and the P-value p=…
Note: the calculator gives a more accurate
P-value than the book does, because it uses
a more accurate formula for degrees of
freedom (see the line df=… in the
calculator). The book adopts a simple but
inaccurate rule df=smaller of n1-1 and n2-1.
37
Intervals for two means by TI-83/84
•
•
•
•
•
•
•
•
•
•
Press STAT and select TESTS
Scroll down to 2-SampTInt press ENTER
Select Input: Data or Stats. For Stats:
Type in x1: (1st sample mean)
sx1: (1st sample st. deviation)
n1: (1st sample size)
x2: (2nd sample mean)
sx2: (2nd sample st. deviation)
n2: (2nd sample size)
C-Level: confidence level
38
Intervals for two means (continued)
•
•
•
•
choose Pooled: No or Yes (always No)
Press on Calculate
Note: the calculator gives a more accurate
confidence interval than the book does,
because it uses a more accurate formula for
degrees of freedom (see the line df=… in
the calculator). The book adopts a simple but
inaccurate rule df=smaller of n1-1 and n2-1.
39
Section 9-4
Two Means:
Matched Pairs
In this section we deal with dependent samples.
In other words, there is some relationship
between the two samples so that each value in
one sample is paired (naturally matched or
coupled) with a corresponding value in the other
sample.
So the two samples can be treated as matched pairs of
values.
40
Examples:
• Blood pressure of patients before they are
given medicine and after they take it.
• Predicted temperature (by Weather
Forecast) and the actual temperature.
• Heights of selected people in the morning
and their heights by night time.
• Test scores of selected students in
Calculus-I and their scores in Calculus-II.
41
Example:
First sample: weights of 5 students in April
Second sample: their weights in September
These weights make 5 matched pairs
Third line: differences between April weights
and September weights (net change in weight
for each student, separately)
In our calculations we only use differences d,
not the values in the two samples.
42
Notation for Dependent Samples
d
=
µd
= mean value of the differences d for the
population of paired data
d
= mean value of the differences d for the
paired sample data (equal to the mean
of the x – y values)
sd
= standard deviation of the differences d
for the paired sample data
n
= number of pairs of data.
individual difference between the two
values of a matched pair
43
Requirements
1. The sample data are dependent (make
matched pairs).
2. Either or both of these conditions is
satisfied: The number of pairs of sample
data is large (n > 30) or the pairs of values
have differences that are from a
population that is approximately normal.
44
Tests for Matched Pairs
The goal is to see whether there is a difference.
H 0:  d = 0
H 1:  d  0 ,
two tails
H1: d < 0 , H1:  d> 0
left tail
right tail
45
Hypothesis Test Statistic for
Matched Pairs:
t=
d – µd
sd
n
Note: d =0 according to H0
degrees of freedom df = n – 1
46
P-values and
Critical Values
Use Table A-3 (t-distribution)
47
Example:
Use a 0.05 significance level to test the
claim that for the population of students,
the mean change in weight from
September to April is 0 kg
(so there is no change, on the average)
48
Example:
Weight gained = April weight – Sept. weight
d denotes the mean of the “April – Sept.”
differences in weight; the claim is d = 0 kg
Step 1: claim is d = 0
Step 2: If original claim is not true, we have
d ≠ 0
Step 3: H0: d = 0 (original claim)
H1: d ≠ 0
Step 4: significance level is  = 0.05
Step 5: use the student t distribution
49
Example:
Step 6: find values of d and sd
differences are: –1, –1, 4, –2, 1
d = 0.2 and sd = 2.4
now compute the test statistic
d  d 0.2  0
t

 0.186
sd
2.4
n
5
Table A-3: df = n – 1, area in two tails is 0.05,
yields a critical value t = ±2.776
50
Example:
Step 7: Because the test statistic does not fall
in the critical region, we fail to reject
the null hypothesis.
51
Example:
We conclude that there is sufficient
evidence to support the claim that for the
population of students, the mean change
in weight from September to April is
equal to 0 kg.
52
Example:
The P-value method:
Using technology, we can find the P-value of
0.8605.
(Using Table A-3 with the test statistic of t =
0.186 and 4 degrees of freedom, we can
determine that the P-value is greater than
0.20.)
We again fail to reject the null hypothesis,
because the P-value is greater than the
significance level of  = 0.05.
53
Confidence Intervals for
Matched Pairs
d – E < µd < d + E
where
E = t/2
sd
n
Critical values of tα/2 : Use Table A-3 with
df = n – 1 degrees of freedom.
54
Example:
Construct a 95% confidence interval estimate
of d , which is the mean of the “April–
September” weight differences of college
students in their freshman year.
d = 0.2, sd = 2.4, n = 5, ta/2 = 2.776
Find the margin of error, E
E  t 2
sd
2.4
 2.776 
 3.0
n
5
55
Example:
Construct the confidence interval:
d  E  d  d  E
0.2  3.0  d  0.2  3.0
2.8  d  3.2
We have 95% confidence that the limits
of ─2.8 kg and 3.2 kg contain the true
value of the mean weight change from
September to April.
56
Dependent samples by TI-83/84
•
•
•
•
•
•
•
Enter 1st sample in list L1 and 2nd sample in L2
Clear screen, type L1─L2→L3 (use STO key)
Press STAT and select TESTS
Scroll down to T-Test for hypotheses testing
or to TInterval for confidence intervals
Select Input: Data (not Stats) and use list L3
Then proceed as if you had just one sample…
57
Section 9-5
Comparing Variation in
Two Samples
58
Requirements
1. The two populations are
independent.
2. The two samples are random
samples.
3. The two populations are each
normally distributed.
The last requirement is strict.
59
Important:
• The first sample must have a larger
sample standard deviation s1 than
the second sample, i.e. we must have
s1 ≥ s2
• If this is not so, i.e. if s1 < s2 , then
we will need to switch the indices 1
and 2, i.e. we need to label the
second sample (and population) as
first, and the first as second.
60
Notation for Hypothesis Tests with Two
Variances or Standard Deviations
s1 = first (larger) sample st. deviation
n1 = size of the first sample
s1 = st. deviation of the first population
s2 n2 s2 are used for the second sample
and population
61
Tests for Two Variances
The goal is to compare the two population
variances (or standard deviations)
H 0: s 1 = s 2
H 1: s 1  s 2 ,
two tails
Note: H1: s < s
1
H 1: s 1 > s 2
right tail
2
is not considered.
Note: no numerical values for
claimed in the hypotheses.
s1 or s2 are
62
Test Statistic for Hypothesis
Tests with Two Variances
F=
s
s
2
1
2
Where s12 is the first (larger) of
the two sample variances
2
Critical Values: Using Table A-5, we obtain
critical F values that are determined by the
following three values:
1. The significance level 
2. Numerator degrees of freedom = n1 – 1
3. Denominator degrees of freedom = n2 – 1
63
Properties of the F Distribution
• The F distribution is not symmetric.
• Values of the F distribution cannot be
negative, i.e. F ≥ 0.
• The exact shape of the F distribution
depends on the two different degrees
of freedom (numerator df and
denominator df)
64
Density curve of F distribution
65
Use of the F Distribution
If the two populations do have equal
s12
variances, then F = 2 will be close to 1
s
2
2 2
because s1 and s2 are close in value.
66
Use of the F Distribution
If the two populations have radically
different variances, then F will be a
large number.
Remember: the larger sample variance is s21 , so
F is either equal to 1 or greater than 1.
67
Conclusions from the F
Distribution
Consequently, a value of F near 1
will be evidence in favor of the
2
conclusion that s1 = s22 .
But a large value of F will be
evidence against the conclusion
of equality of the population
variances.
68
there we reject H0: s1= s2
69
Finding Critical F Values
To find a critical F value corresponding to a
0.05 significance level, refer to Table A-5 and
use the right-tail area of 0.025 or 0.05,
depending on the type of test:
Two-tailed test: use 0.025 in right tail
Right-tailed test: use 0.05 in right tail
70
Example:
Below are sample weights (in g) of quarters
after 1964.
When designing coin vending machines, we
must consider the standard deviations of pre1964 quarters and post-1964 quarters.
Use a 0.05 significance level to test the claim
that the weights of pre-1964 quarters and the
weights of post-1964 quarters are from
populations with the same standard deviation.
71
Example:
Step 1: claim of equal standard deviations is
equivalent to claim of equal variances
s s
2
1
2
2
Step 2: if the original claim is false, then
s s
2
2
H 0 : s 1  s 2 original claim
2
1
Step 3:
H1 : s  s
2
1
2
2
2
2
72
Example:
Step 4: significance level is 0.05
Step 5: involves two population variances, use F
distribution variances
Step 6: calculate the test statistic
2
1
2
2
2
s
0.08700
F 
 1.9729
2
s
0.016194
For the critical values in this two-tailed test,
refer to Table A-5 for the area of 0.025 in the
right tail. The critical value is 1.8752.
73
Example:
Step 7: The test statistic F = 1.9729 does fall
within the critical region, so we reject
the null hypothesis of equal variances.
There is sufficient evidence to warrant
rejection of the claim of equal standard
deviations.
74
Example:
Left tail is not used and
need not be shown !
75
Conclusion:
There is sufficient evidence to warrant rejection
of the claim that the two standard deviations
are equal.
The variation among weights of quarters made
after 1964 is significantly different from the
variation among weights of quarters made
before 1964.
76
Tests about two variances by TI-83/84
•
•
•
•
•
•
•
Press STAT and select TESTS
Scroll down to 2-SampFTest press ENTER
Select Input: Data or Stats. For Stats:
Type in sx1: (1st sample st. deviation)
n1: (1st sample size)
sx2: (2nd sample st. deviation)
n2: (2nd sample size)
choose H1: s1 ≠s2
<s2
>s2
(two tails) (left tail) (right tail)
77
•
•
•
•
Press on Calculate