# notes

#### Document technical information

Format ppt
Size 10.7 MB
First found May 22, 2018

#### Document content analysis

Category Also themed
Language
Type
not defined
Concepts
no text concepts found

#### Transcript

```Statistics for Business
(Env)
Chapter 10
Statistical Inferences Based on
Two Samples
1
Statistical Inferences Based on Two Samples
10.1 Comparing Two Population Means by Using
Independent Samples: Variances Known
10.2 Comparing Two Population Means by Using
Independent Samples: Variances Unknown
10.3 Paired Difference Experiments
10.4 Comparing Two Population Proportions by Using
Large, Independent Samples
2
Comparing Two Population Means by Using
Independent Samples: Variances Known
• Suppose a random sample has been taken from each
of two different populations
• Suppose that the populations are independent of
each other
– Then the random samples are independent of
each other
• Then the sampling distribution of the difference in
sample means is normally distributed
3
Do the achievement scores for children taught by method A differ
from the scores for children taught by method B?
4
A research design that uses a separate sample for
each treatment condition (or for each population) is
called an independent-measures research design or
a between-subjects design.
The goal of an independent-measures research study
is to evaluate the mean difference between two
populations (or between two treatment conditions).
5
Sampling Distribution of the
Difference of Two Sample Means #1
• Suppose population 1 has mean µ1 and variance σ12
– From population 1, a random sample of size n1 is selected
which has mean
and variance s12
• Suppose population 2 has mean µ2 and variance σ22
– From population 2, a random sample of size n2 is selected
which has mean
and variance s22
• Then the sample distribution of the difference of two
sample means…
6
Sampling Distribution of the
Difference of Two Sample Means #2
• Is normal, if each of the sampled populations
is normal
– Approximately normal if the sample sizes n1 and
n2 are large
• Has mean
= µ1 – µ2
• Has standard deviation
 x x 
1
2

2
1
n1


2
2
n2
7
If you select one score from each of these two
populations, the closest two values are X1 =50 and X2
=30. The two values that are farthest apart are X1 =70
and X2 =20.
µ2
µ1
8
Sampling Distribution of the
Difference of Two Sample Means #3
9
z-Based Confidence Interval for the Difference in
Means (Variances Known)
• Let
be the mean of a sample of size n1 that has
been randomly selected from a population with
mean m1 and standard deviation 1
• Let
be the mean of a sample of size n2 that has
been randomly selected from a population with m2
and 2
• Suppose each sampled population is normally
distributed or that the samples sizes n1 and n2 are
large
• Suppose the samples are independent of each other,
then …
10
z-Based Confidence Interval for the Difference in
Means
Continued
• A 100(1 – ) percent confidence interval for
the difference in populations µ1–µ2 is

12  22 
x1  x2   z  2


n1 n2 



11
Example 10.1 The Bank Customer Waiting Time
Case #1
• A random sample of size 100 waiting times observed under
the current system of serving customers has a sample mean
of 8.79
– Call this population 1
– Assume population 1 is normal or sample size is large
– The variance is 4.7
• A random sample of size 100 waiting times observed under
the new system of time of 5.14
– Call this population 2
– Assume population 2 is normal or sample size is large
– The variance is 1.9
• Then if the samples are independent …
12
Example 10.1 The Bank Customer Waiting Time
Case #2
• At 95% confidence, z/2 = z0.025 = 1.96, and
x1  x2   z 2

12  22
4.7 1.9 

 8.79  5.14   1.96


n1
n2
100
100


 3.65  0.5035 
 3.15 , 4.15 
• According to the calculated interval, the bank
manager can be 95% confident that the new system
reduces the mean waiting time by between 3.15 and
4.15 minutes
13
z-Based Test About the Difference in Means
(Variances Known)
• Test the null hypothesis about
H0: µ1 – µ2 = D0
– D0 = µ1 – µ2 is the claimed difference between the
population means
– D0 is a number whose value varies depending on
the situation
– Often D0 = 0, and the null means that there is no
difference between the population means
14
z-Based Test About the Difference in Means
(Variances Known)
• Use the notation from the confidence interval
statement on a prior slide
• Assume that each sampled population is
normal or that the samples sizes n1 and n2 are
large
15
Test Statistic (Variances Known)
• The test statistic is

x1  x2   D0
z
2
1
n1

2
2
n2
• The sampling distribution of this statistic is a
standard normal distribution
• If the populations are normal and the samples
are independent ...
16
z-Based Test About the Difference in
Means (Variances Known)
• Reject H0: µ1 – µ2 = D0 in favor of a particular
alternative hypothesis at a level of significance
if the appropriate rejection point rule holds
(i.e. calculated z is in the rejection region).
• Rules are on the next slide…
17
Hypothesis Tests for
Two Population Means
Two Population Means, Known Population Variances
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: μ1  μ2
H1: μ1 < μ2
H0: μ1 ≤ μ2
H1: μ1 > μ2
H0: μ1 = μ2
H1: μ1 ≠ μ2
i.e.,
i.e.,
i.e.,
H0: μ1 – μ2  0
H1: μ1 – μ2 < 0
H0: μ1 – μ2 ≤ 0
H1: μ1 – μ2 > 0
H0: μ1 – μ2 = 0
H1: μ1 – μ2 ≠ 0
18
Hypothesis tests for μ1 – μ2
Two Population Means, Known Population Variances
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: μ1 – μ2  0
H1: μ1 – μ2 < 0
H0: μ1 – μ2 ≤ 0
H1: μ1 – μ2 > 0
H0: μ1 – μ2 = 0
H1: μ1 – μ2 ≠ 0


-z
Reject H0 if Z < -Z
z
Reject H0 if Z > Z
/2
-z/2
/2
z/2
Reject H0 if Z < -Z/2
or Z > Z/2
19
EXAMPLE
Two cities, Boston and
Kingston are both in
Massachusetts.
The mean household income
in Boston is \$38,000. The
population s.d. is known to
be \$6,000 for a sample of 40
households.
The mean income in Kingston
is \$35,000 for a sample of 35
households. The population s.d.
is known to be \$7,000.
At the .01 significance level can
we conclude the mean income
in Boston is more?
EXAMPLE
Step 1
State the null and alternate
hypotheses.
H0: µB < µK
H1: µB > µK
Step 4
State the decision rule.
The null hypothesis is
rejected if t is greater than
2.326 or p < .01.
Step 2
Select the level of significance.
The .01 significance level is stated
in the problem.
Step 3
Find the appropriate test
statistic. Since both samples
are more than 30, we can use
z as the test statistic.
21
EXAMPLE
Step 5: Compute the value of z and make a decision.
t
\$38,000  \$35,000
 1.98
(\$6,000) 2 (\$7,000) 2

40
35
Because the computed Z of 1.98
< critical Z of 2.26, the p-value of .0239 > .01 (), the
decision is not to reject the H0. We cannot conclude
that the mean household income in Boston is larger.
22
Comparing Two Population Means by Using
Independent Samples: Variances Unknown
• In general, the true values of the population
variances σ12 and σ22 are not known
• They have to be estimated from the sample
variances s12 and s22, respectively
23
Comparing Two Population Means by Using
Independent Samples: Variances Unknown #2
•
•
Also need to estimate the standard deviation
of the sampling distribution of the difference
between sample means
Two approaches:
1. If it can be assumed that σ12 = σ22 = σ2, then
calculate the “pooled estimate” of σ2
2. If σ12 ≠ σ22, then use approximate methods
24
Pooled Estimate of σ2
• Assume that σ12 = σ22 = σ2
• The pooled estimate of σ2 is the weighted averages
of the two sample variances, s12 and s22
• The pooled estimate of σ2 is denoted by sp2
s 2p

n1  1s12  n2  1s22

n1 n 2 2
• The estimate of the population standard deviation of
the sampling distribution is
 x1  x2 
2
s p 
1
1 
 
 n1 n2 
25
One sample compared with 2 samples statistics
Assume that σ12 = σ22 = σ2
Mean
SS2
df2
26
t-Based Confidence Interval for the Difference in
Means (Variances Unknown)
• Select independent random samples from two normal
populations with equal variances
• A 100(1 – ) percent confidence interval for the difference in
populations µ1 – µ2 is




1
1
x1  x2   t  2 s 2p    
 n n 

2 
 1


• where
s 2p

n1  1s12  n2  1s22

n1 n 2 2
• and t/2 is based on (n1+n2-2) degrees of freedom (df)
27
Finding the value of the test statistic requires two
steps:
Step One: Pool the sample
standard deviations.
s 2p 
2
(n1  1) s1
2
 (n2  1) s2
n1  n2  2
Step Two: Determine the value of t from the
following formula.
t
X1  X 2
 1
1 

s p  

 n1 n 2 
28
Hypothesis tests for μ1 – μ2
Two Population Means, Unknown Population Variances
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: μ1 – μ2  0
H1: μ1 – μ2 < 0
H0: μ1 – μ2 ≤ 0
H1: μ1 – μ2 > 0
H0: μ1 – μ2 = 0
H1: μ1 – μ2 ≠ 0


-t
Reject H0 if t < -t
t
Reject H0 if t > t
/2
-t/2
/2
t/2
Reject H0 if t < -t/2
or t > t/2
29
Example:
A recent EPA study
compared the highway fuel
economy of domestic and
imported passenger cars. A
sample of 15 domestic cars
revealed a mean of 33.7 mpg
with a sample standard
deviation of 2.4 mpg.
A sample of 12 imported cars
revealed a mean of 35.7 mpg
with a sample standard
deviation of 3.9. At the .05
significance level can the EPA
conclude that the mpg is
higher on the imported cars?
30
Example:
(continued)
Step 1
State the null and alternate
hypotheses.
H0: µD > µI
H1: µD < µI
Step 2
State the level of significance.
The .05 significance level is
stated in the problem.
Step 3
Find the appropriate test statistic. Both
samples are less than 30, so we use the t
distribution.
31
Example:
(continued)
Step 4
The decision rule is to reject H0 if
t<-1.708. There are n1 + n2 – 2 or
25 degrees of freedom.
Step 5
We compute the pooled
variance.
2
2
(
n

1
)(
s
)

(
n

1
)(
s
2
1
1
2
2)
sp 
n1  n 2  2
(15  1)( 2.4) 2  (12  1)(3.9) 2

 9.918
15  12  2
32
Example:
(continued)
We compute the value of t as follows.
t 
X1  X 2
 1
1 

s 

n

n
2 
 1
33.7  35.7
2
p

1 
 1
9.918


 15 12 
 1.640
33
Example:
(continued)
Since a computed z of –1.64 > critical z
of –1.71, H0 can not be rejected.
There is insufficient sample
evidence to claim a higher mpg on
the imported cars.
-1.71
-1.64
34
Example: Comparing Mean weights
To show if boys are heavier than girls of the same age, a
survey is conducted in which a sample of 15 boys shows a
mean weight of 41Kg and a standard deviation of 3Kg.
A group of 10 girls of the same age shows a mean weight
of 38Kg and a standard deviation of 2Kg.
Assuming both the weights of boys and girls follow the
normal distribution. At the level of significant 0.05, test if the
average weight of boys is greater than the average weight
of girls of the same age.
Step 1
State the null and alternate
hypotheses.
H0: µg > µb
H1: µg < µb
35
Example:
(continued)
Step 2
State the level of significance.
The .05 significance level is
stated in the problem.
Step 3
Find the appropriate test statistic. Both
samples are less than 30, so we use the t
distribution.
Step 4
The decision rule is to reject H0 if t > t0.05 =1.714.
There are n1 + n2 – 2 or 23 degrees of freedom.
36
Example:
(continued)
Step 5
Compute the pooled variance and t.
S2p = [(15-1)*32 + (10-1)*22]/ (15+10-2)
= 7.04
Sp = 2.65
t = (41-38) / sqrt(7.04*(1/15 + 1/10))
= 2.77
=1.714
Since t =2.77 > t0.05 =1.714, we reject H0 .
So the mean weight of boys is larger than
the mean weight of girls of the same age.
37
Example: Directed reading activities in the classroom
A class of 21 third-graders participates in these activities for 8 weeks
while a control classroom of 23 third-graders follows the same
curriculum without the activities. After the 8 weeks, all children take a
reading test (scores in table). At a level of significance 0.05, can we
conclude directed reading activities help improve reading ability?
Step 1
State the null and
alternate hypotheses.
H0: µ1 = µ2
H1: µ1 = µ2
38
Example: Directed reading activities
(continued)
Step 2
State the level of significance.
The .05 significance level is
stated in the problem.
Step 3
Find the appropriate test statistic. Both
samples are less than 30, so we use the t
distribution.
Step 4
The decision rule is to reject H0 if t > t0.025 =1.97 or
t < -t0.025 . There are n1 + n2 – 2 or 42 degrees of
freedom.
39
Example: Directed reading activities
(continued)
Step 5
Compute the pooled variance and t.
S2p = [(21-1)*11.012 + (23-1)*17.152]/ (21+23-2)
= 211.79
t = (51.48-41.52) / sqrt(211.79*(1/21 + 1/23))
= 9.96/4.39=2.27
Since t =2.27 > t0.025 =1.97, we reject H0 .
So there are significant difference between
the 2 group.
40
Example: Directed reading activities
(continued)
Step 1
State the null and alternate
hypotheses.
H0: µ2 > µ1
H1: µ2 < µ1
Step 5
Compute the pooled variance and t.
S2p = [(21-1)*11.012 + (23-1)*17.152]/ (21+23-2)
= 211.79
t = (51.48-41.52) / sqrt(211.79*(1/21 + 1/23))
= 9.96/4.39=2.27
There are n1 + n2 – 2 or 42 degrees of
freedom. The rule is to reject H0 if t > t0.05
=1.65.
41
Pooled Variance t Test: Example
You are a financial analyst for a brokerage firm. Is there a
difference in dividend yield between stocks listed on the NYSE
& NASDAQ? You collect the following data:
NYSE NASDAQ
Number
21
25
Sample mean
3.27
2.53
Sample std dev 1.30
1.16
Assuming both populations are
approximately normal with
equal variances, is
there a difference in average
yield ( = 0.05)?
Chap 9-42
Calculating the Test Statistic
The test statistic is:

X  X   μ  μ 
t

1
2
1
1 1
S   
 n1 n2 
2
p
2
3.27  2.53   0
1 
 1
1.5021 

 21 25 
2
2





n

1
S

n

1
S
21  11.30 2  25  11.16 2
2
1
1
2
2
S 

p
(n1  1)  (n2  1)
(21 - 1)  (25  1)
 2.040
 1.5021
43
Solution
H0: μ1 - μ2 = 0 i.e. (μ1 = μ2)
H1: μ1 - μ2 ≠ 0 i.e. (μ1 ≠ μ2)
 = 0.05
df = 21 + 25 - 2 = 44
Critical Values: t = ± 1.96
Reject H0
Reject H0
.025
-1.96
.025
0
1.96
t
2.040
Test Statistic:
Decision:
3.27  2.53
t
 2.040 Reject H0 at  = 0.05
1 
 1
Conclusion:
1.5021  

 21 25 
There is evidence of a
difference in means.
44
Two kinds of studies
So far, we have studied :
Independent samples
two sets of sample data that come from two independent
populations (e.g. women and men, or students from program
A and from program B).
However, sometimes we want to study Paired samples
two sets of sample data that come from related populations
(e.g. “before treatment” and “after treatment”).
45
Paired/Dependent Samples
Independent samples are samples that are not related in any way.
Dependent samples are samples that are
paired or related in some fashion.
*The same subjects measured at two different
points in time (repeated-measures).
*Matched or paired observations
*Hypothesis test proceeds just as in the one
sample case.
46
Paired-Sample t Test: Example
Assume you work in the finance department. Is the new financial
package faster (=0.05 level)? You collect the following
processing times for same set of jobs:
Existing System (1)
9.98 Seconds
9.88
9.84
9.99
9.94
9.84
9.86
10.12
9.90
9.91
New Software (2) Difference Di
9.88 Seconds
.10
9.86
.02
9.75
.09
9.80
.19
9.87
.07
9.84
.00
9.87
- .01
9.98
.14
9.83
.07
9.86
.05
D

D
i
n
SD 
 .072
 D  D
i
n 1
 .06215
47
2
Paired-Sample t Test: Example
Is the new financial package faster (  0.05 level)?
H0: mD  0
H1: mD > 0
Reject
 .05
 .05
D = .072
Critical Value=1.8331
df = n - 1 = 9
Test Statistic
t
D  mD
.072  0

 3.66
SD / n .06215/ 10
t
1.8331
Decision: Reject H0
t Stat. in the rejection zone.
3.66
Conclusion: The new software package
is faster.
48
Paired-Sample: Example-twins
Suppose we collect 8 pairs of twins. The first twin in the pair is healthy; the
second is not. For each twin, we measure grey matter density (gmd).
Is grey matter density in the populations significantly different ?
Processed data from the 8 pairs is shown below (units not given).
Consider the population differences, D = X1 - X2,
Hypothesis Testing Involving
Paired Observations
(continued)
If σD is unknown, we can estimate
the unknown population standard
deviation with a sample standard S D 
deviation:
where D is the mean of the differences
sd is the (sample) s.d. of the differences
n is the number of pairs (differences)
The test statistic for D is now
a t statistic, with n-1 d.f.
n
2
(D

D
)
 i
i 1
n 1
D  μD
t
SD
n
50
Confidence Interval of Paired
Observations, σD Unknown
(continued)
The confidence interval for μD is
SD
D  t n1
n
n
where SD is:
SD 
2
(D

D
)
 i
i 1
n 1
51
Hypothesis Testing for
Mean Difference, σD Unknown
Paired Samples
Lower-tail test:
Upper-tail test:
Two-tail test:
H0: μD  0
H1: μD < 0
H0: μD ≤ 0
H1: μD > 0
H0: μD = 0
H1: μD ≠ 0


-t
Reject H0 if t < -t
t
Reject H0 if t > t
Where t has n - 1 d.f.
Chap 9-52
/2
-t/2
/2
t/2
Reject H0 if t < -t/2
or t > t/2
Paired Samples Example
• Assume you send your salespeople to a “customer
service” training workshop. Has the training made a
difference in the number of complaints? You collect the
following data:
Salesperson
Chen
Li
Zhang
Wang
Wan
Number of Complaints:
Before (1)
After (2)
6
20
3
0
4
4
6
2
0
0
(2) - (1)
Difference, Di
- 2
-14
- 1
0
- 4
-21
D =
D
i
n
= -4.2
SD 
 (D
2

D
)
i
n 1
 5.67
Chap 9-53
Paired Samples: Solution
• Has the training made a difference in the number of complaints
(at the 0.01 level)?
H0: μD = 0
H1: μD  0
 = .01 d.f. = n - 1 = 4
Critical Value = ± 4.604
D = -4.2
Test Statistic:
D  μ D  4 .2  0
t

  1 .6 6
S D / n 5 .6 7 / 5
Reject
Reject
/2
/2
- 4.604
4.604
- 1.66
Decision: Do not reject H0
(t stat is not in the reject region)
Conclusion: There is not a significant
change in the number of complaints.
Chap 9-54
An independent testing
agency is comparing the daily
rental cost for renting a
compact car from Hertz and
Avis.
A random sample of eight
cities revealed the following
information. At the .05
significance level can the
testing agency conclude that
there is a difference in the
rental charged?
City
Atlanta
Chicago
Cleveland
Denver
Honolulu
Kansas City
Miami
Seattle
Hertz
(\$)
42
56
45
48
37
45
41
46
Avis (\$)
40
52
43
48
32
48
39
50
EXAMPLE 4
Step 1
Ho: md = 0
H1: md  0
Step 4
H0 is rejected if
t < -2.365 or t > 2.365;
or if p-value < .05.
We use the t distribution with n-1 or 7
degrees of freedom.
Step 2
The stated
significance level
is .05.
Step 3
The appropriate test
statistic is the paired ttest.
Step 5
Perform the calculations and
make a decision.
56
City
Hertz Avis
d
d2
Atlanta
42
40
2
4
Chicago
56
52
4
16
Cleveland
45
43
2
4
Denver
48
48
0
0
Honolulu
37
32
5
25
Kansas City
45
48
-3
9
Miami
41
39
2
4
Seattle
46
50
-4
16
57
d 8.0
d

 1.00
n
8
2



d
d 2 
n
sd 
n 1
d
t
sd

n
82
78 
8  3.1623

8 1
1.00
3.1623
 0.894
8
58
P(t>.894) = .20 for a one-tailed t-test at 7
degrees of freedom.
Because 0.894 is less than the critical value, the p-value of .20 > a of .05, do not
reject the null hypothesis. There is no difference in the mean amount charged by
Hertz and Avis.
59
Comparing Two Population
Proportions
Goal: test a hypothesis or form a confidence interval
for the difference between two population proportions
(p1 – p2).
Assumptions:
two independent samples
from two populations
n1p1  5 , n1(1-p1)  5
n2p2  5 , n2(1-p2)  5
The point estimate for
the difference is
p s1  p s2
60
Two Population Proportions
Since we begin by assuming the null hypothesis is true,
we assume p1 = p2 and pool the two ps estimates
The pooled estimate for the
overall proportion is:
X1  X2
p
n1  n2
where X1 and X2 are the numbers from
samples 1 and 2 with the characteristic of
interest
61
Two Population Proportions
(continued)
The test statistic for p1 – p2 is a Z statistic:

p
Z
where
p
s1

 p s2   p1  p 2 
1 1
p (1  p)   
 n1 n2 
X1  X2
X
X
, p s1  1 , p s2  2
n1  n2
n1
n2
62
Confidence Interval for
Two Population Proportions
Population
proportions
p
s1
The confidence interval for
p1 – p2 is:

 p s2  Z
p s1 (1  p s1 )
n1

p s2 (1  p s2 )
n2
63
Example
Are unmarried workers more likely to be absent from work than
married workers?
A sample of 250 married workers showed 22 missed more than 5
days last year, while a sample of 300 unmarried workers showed 35
missed more than five days. Use a .05 significance level.
64
The null and the alternate hypotheses
H0: pU < pM
H1: pU > pM
The null hypothesis is rejected if the computed value of z is greater than 1.65
or the p-value < .05.
The pooled proportion
35  22
pc 
=
.1036
300  250
65
z
35
22

300 250
 1.10
.1036 (1  .1036 ) .1036 (1  .1036 )

300
250
Because the calculated z of 1.10 < a critical z of 1.65 (
of .05), the null hypothesis is not rejected. We cannot
conclude that a higher proportion of unmarried
workers miss more days in a year than the married
workers.
66
Chapter Ten
Two-Sample Tests of Hypothesis
ONE- Conduct a test of hypothesis about the difference
between two independent population means with Known/
Unknown Variances
TWO- Conduct a test of hypothesis regarding the
difference in two population proportions with Known/
Unknown Variances
THREE- Understand the difference between dependent
and independent samples.
FOUR- Conduct a test of hypothesis about the mean
difference between paired or dependent observations.
67
Table of the Standard Normal Distribution
z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
0.00
0.0000
0.0398
0.0793
0.1179
0.1554
0.1915
0.2257
0.2580
0.2881
0.3159
0.3413
0.3643
0.3849
0.4032
0.4192
0.4332
0.4452
0.4554
0.4641
0.4713
0.4772
0.4821
0.4861
0.4893
0.4918
0.4938
0.4953
0.4965
0.4974
0.4981
0.4987
0.01
0.0040
0.0438
0.0832
0.1217
0.1591
0.1950
0.2291
0.2611
0.2910
0.3186
0.3438
0.3665
0.3869
0.4049
0.4207
0.4345
0.4463
0.4564
0.4649
0.4719
0.4778
0.4826
0.4864
0.4896
0.4920
0.4940
0.4955
0.4966
0.4975
0.4982
0.4987
0.02
0.0080
0.0478
0.0871
0.1255
0.1628
0.1985
0.2324
0.2642
0.2939
0.3212
0.3461
0.3686
0.3888
0.4066
0.4222
0.4357
0.4474
0.4573
0.4656
0.4726
0.4783
0.4830
0.4868
0.4898
0.4922
0.4941
0.4956
0.4967
0.4976
0.4982
0.4987
0.03
0.0120
0.0517
0.0910
0.1293
0.1664
0.2019
0.2357
0.2673
0.2967
0.3238
0.3485
0.3708
0.3907
0.4082
0.4236
0.4370
0.4484
0.4582
0.4664
0.4732
0.4788
0.4834
0.4871
0.4901
0.4925
0.4943
0.4957
0.4968
0.4977
0.4983
0.4988
0.04
0.0160
0.0557
0.0948
0.1331
0.1700
0.2054
0.2389
0.2704
0.2995
0.3264
0.3508
0.3729
0.3925
0.4099
0.4251
0.4382
0.4495
0.4591
0.4671
0.4738
0.4793
0.4838
0.4875
0.4904
0.4927
0.4945
0.4959
0.4969
0.4977
0.4984
0.4988
0.05
0.0199
0.0596
0.0987
0.1368
0.1736
0.2088
0.2422
0.2734
0.3023
0.3289
0.3531
0.3749
0.3944
0.4115
0.4265
0.4394
0.4505
0.4599
0.4678
0.4744
0.4798
0.4842
0.4878
0.4906
0.4929
0.4946
0.4960
0.4970
0.4978
0.4984
0.4989
0.06
0.0239
0.0636
0.1026
0.1406
0.1772
0.2123
0.2454
0.2764
0.3051
0.3315
0.3554
0.3770
0.3962
0.4131
0.4279
0.4406
0.4515
0.4608
0.4686
0.4750
0.4803
0.4846
0.4881
0.4909
0.4931
0.4948
0.4961
0.4971
0.4979
0.4985
0.4989
0.07
0.0279
0.0675
0.1064
0.1443
0.1808
0.2157
0.2486
0.2794
0.3078
0.3340
0.3577
0.3790
0.3980
0.4147
0.4292
0.4418
0.4525
0.4616
0.4693
0.4756
0.4808
0.4850
0.4884
0.4911
0.4932
0.4949
0.4962
0.4972
0.4979
0.4985
0.4989
0.08
0.0319
0.0714
0.1103
0.1480
0.1844
0.2190
0.2517
0.2823
0.3106
0.3365
0.3599
0.3810
0.3997
0.4162
0.4306
0.4429
0.4535
0.4625
0.4699
0.4761
0.4812
0.4854
0.4887
0.4913
0.4934
0.4951
0.4963
0.4973
0.4980
0.4986
0.4990
0.09
0.0359
0.0753
0.1141
0.1517
0.1879
0.2224
0.2549
0.2852
0.3133
0.3389
0.3621
0.3830
0.4015
0.4177
0.4319
0.4441
0.4545
0.4633
0.4706
0.4767
0.4817
0.4857
0.4890
0.4916
0.4936
0.4952
0.4964
0.4974
0.4981
68
0.4986
0.4990
69
```