BFC 34303

CIVIL ENGINEERING STATISTICS

Chapter 6
Significance Testing
Faculty of Civil and Environmental Engineering
Universiti Tun Hussein Onn Malaysia

Hypothesis Testing
A statistical hypothesis is a statement about a population parameter
developed for the purpose of testing.
In statistical analysis we make a claim, that is, state a hypothesis, then
follow up with tests to verify the assertion or to determine that it is untrue.
Since populations are mostly so large that it is not feasible to study all the
items or people in the population, we take a sample from the population.
We can, therefore, test a statement to determine whether the sample
does or does not support the statement concerning the population.
Hypothesis testing (or significance testing) is a procedure based on
sample evidence and probability theory to determine whether the
hypothesis is a reasonable statement.

1
 5-Step Procedure for Testing a Hypothesis

1 State the null 2 3

hypothesis, 𝐻 Select a level of Identify the test
and alternative significance, 𝛼 statistic
hypothesis, 𝐻

Do not reject 𝐻 5 Take a sample 4

Formulate a
and arrive at a
decision rule
Reject 𝐻 decision

1 State the null hypothesis, 𝐻 and alternative hypothesis, 𝐻

The null hypothesis, 𝑯𝒐 is a statement about the value of the population

parameter.
The alternative hypothesis, 𝑯𝒂 is a statement that is accepted if the
sample data provide enough evidence that the null hypothesis is false.
This is also called the research hypothesis.

For example: A recent article indicated the mean life of bituminous roads
is 6 years. We want to investigate if this is different in tropical regions. The
null and alternative hypotheses are written as follows:
𝐻 :𝜇=6
𝐻 :𝜇≠6
4

2
2 Select a level of significance, 𝛼

The level of significance, 𝜶 is the probability of rejecting the null

hypothesis when it is true.
There is no single level of significance that is applied to all tests. A
decision is made to use either the 0.05 level (often stated as the 5%
level), the 0.01 level, the 0.10 level, or any other level between 0 and 1.

3 Identify the test statistic

A test statistic is a value, determined from sample information, used to

determine whether to reject the null hypothesis.
There are many test statistics, such as the 𝑧, 𝑡, 𝐹 and 𝜒 test statistics.
5

4 Formulate a decision rule

A decision rule is a statement of the specific conditions under which the

null hypothesis is rejected and the conditions under which it is not
rejected.
For example:
Region of rejection for a
right-tailed test for sampling
Do not reject 𝐻 Reject 𝐻 distribution of the 𝑧-statistic
with 0.05 level of
|
significance
0 Z
1.65
0.95 probability 0.05 probability
Critical value

3
5 Make a decision

The final step is computing the test statistic, comparing it to the critical
value and making a decision to reject or not reject the null hypothesis,
𝑯𝒐 .
It should be noted that only one of two decisions is possible in hypothesis
testing – either “do not reject” or “reject” the null hypothesis.
There is always a possibility that the null hypothesis is rejected when it
should not be rejected (Type I error). Also, there is a definable chance that
the null hypothesis is not rejected when it should be rejected (Type II
error).

Type I and Type II Errors

When we make decisions or conclusions from a statistical test, there are
two types of errors we may commit:

Type I Error Type II Error

• Reject 𝐻 when we know 𝐻 • Do not reject 𝐻 when we

is true know 𝐻 is false
• Denoted by 𝛼 • Denoted by 𝛽
• Also called false positive • Also called false negative

4
 Decision 𝐻 is true 𝐻 is false
Type I Error Correct decision
Reject 𝐻
𝜶 1−𝛼

Correct decision Type II Error

Do not reject 𝐻
1−𝛽 𝜷

Note:
Either type of error is undesirable and we would like the value of both 𝛼
and 𝛽 to be small.

One-Tailed and Two-Tailed Tests of Significance

One-Tailed Test Two-Tailed Test
• Allows you to determine if one • Allows you to determine if two
mean is greater or less than means are different from one
another mean, but not both. another.
• A direction must be chosen • A direction does not have to be
prior to testing. specified prior to testing.
• In other words, a one-tailed • In other words, a two-tailed test
test tells you the effect of a will taken into account the
change in one direction and possibility of both a positive
not the other. and a negative effect.
• 𝐻 : 𝜇=𝜇 • 𝐻 :𝜇 = 𝜇
• 𝐻 : 𝜇 ≤ 𝜇 𝑜𝑟 𝐻 : 𝜇 ≥ 𝜇 • 𝐻 :𝜇 ≠ 𝜇
• 𝐻 : 𝜇 > 𝜇 𝑜𝑟 𝐻 : 𝜇 < 𝜇

10

5
 Rejection regions for one-tailed and two-tailed tests:

One-tailed Test Two-tailed Test

Do not reject 𝐻 Reject 𝑯𝒐 Reject 𝑯𝒐 Do not reject 𝐻 Reject 𝑯𝒐

| |
0 0
Critical Critical Critical
Value Value Value

Reject 𝑯𝒐
Do not reject 𝐻

|
0
Critical
Value 11

p-Value in Hypothesis Testing

In testing a hypothesis, we compare the test statistic to a critical value. A
decision is made to either reject 𝐻 or not to reject 𝐻 .
In recent years, spurred by the availability of computer software,
additional information is often reported on the strength of the rejection or
acceptance, i.e. how confident are we in rejecting the null hypothesis.
This approach reports the probability of getting a value of the test statistic
at least as extreme as the value actually obtained.
This process compares the probability that is called the 𝒑-value with the
significance level, 𝛼.
The 𝑝-value is defined as the probability of observing a sample value as
extreme as, or more extreme than, the value observed, given that the null
hypothesis is true.
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6
Decision rules using the 𝑝-value:
• Reject 𝐻 if the 𝑝-value is equal to or less than the significance level,
i.e. 𝑝 ≤ 𝛼.
• Do not reject 𝐻 if the 𝑝-value is greater than the significance level, i.e.
𝑝 > 𝛼.

If the 𝑝-value is equal to or less than:

• 0.10, we have some evidence that 𝐻 is not true.
• 0.05, we have strong evidence that 𝐻 is not true.
• 0.01, we have very strong evidence that 𝐻 is not true.
• 0.001, we have extremely strong evidence that 𝐻 is not true.

13

One-tailed Test

𝒑-value

𝛼
𝒑-value

| Z | Z
0 0

Calculated Calculated
𝑧-value 𝑧-value

𝑝 > 𝛼  Do not reject 𝐻 𝑝 ≤ 𝛼  Reject 𝐻

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7
 Two-tailed Test

𝒑-value 𝒑-value

𝛼 𝛼
𝛼 𝛼
2 2
2 2
| Z | Z
0 0

Calculated Calculated Calculated Calculated

𝑧-value 𝑧-value 𝑧-value 𝑧-value

𝑝 > 𝛼  Do not reject 𝐻 𝑝 ≤ 𝛼  Reject 𝐻

15

One-Sample Hypothesis Testing

A hypothesis about a population mean (using one sample) can be tested
when sampling is from any of the following:
𝑋−𝜇
• A normally distributed population with known 𝑧= 𝜎
standard deviation. 𝑛

• A normally distributed population with 𝑋−𝜇

unknown standard deviation and 𝒏 < 𝟑𝟎. 𝑡= 𝑠
𝑛
• A normally distributed population with
unkown standard deviation and 𝒏 ≥ 𝟑𝟎.
𝑋−𝜇
𝑧= 𝑠
• A population that is not normally distributed 𝑛
but, assuming 𝒏 ≥ 𝟑𝟎 (the central limit
theorem applies)
16

8
Example 6.1
A dean of an engineering school claims that his students have above
average intelligence. The mean population IQ is 100 with a standard
deviation of 15. A random sample of 30 students was selected and the
mean IQ was found to be 112. Is there sufficient evidence to support the
dean’s claim?

𝐻 : 𝜇 ≤ 100
𝐻 : 𝜇 > 100

Assume level of significance, 𝛼 = 0.05

𝑋−𝜇
Assume the data is normally distributed. 𝑛 ≥ 30 and the 𝑧= 𝜎
standard deviation is known. So, use the 𝑧-statistic. 𝑛

17

The rejection region

(one-tailed test):
0.45 Reject 𝐻
𝛼 = 0.05

Z
0

1.65 Critical 𝑧-value taken

from 𝑧-table,
𝑋−𝜇 112 − 100 corresponding to an
𝑧= 𝜎 = = 4.38 area under the table
15 of 0.45
𝑛 30

Since the calculated 𝑧 (4.38) is greater than the critical 𝑧 (1.65) and falls in
the rejection region, we can reject 𝑯𝒐 . Thus, we accept 𝐻 , which says the
mean is above the average of 100. The dean’s claim is therefore
supported.
18

9
Areas under the Standard Normal Curve (z-Table) showing values for P(0 ≤ Z ≤ z)
0 z

Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621
1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830
1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015
1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177
1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319
1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441
1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545
1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633
1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706
1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767
2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817
2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857
2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890
2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916
2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936
2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952

19

Alternative method using 𝑝-value:

𝑝-value = 0.000032
𝑋−𝜇 112 − 100
𝑧= 𝜎 = = 4.38 (For z-values more
15 than 3.4, use online
𝑛 30 calculator)

Z
0

4.38

Since the 𝑝-value (0.000032) is less than the significance level 𝛼 (0.05),
we can reject 𝑯𝒐 . Thus, we accept 𝐻 , which says the mean is above the
average of 100. The dean’s claim is therefore supported.

https://www.zscorecalculator.com
20

10
Example 6.2
A manufacturing company wants to see if the job satisfaction of its
assembly line workers differ when their work is machine-paced rather
than self-paced. A matched pairs study was performed on a sample of
workers. Workers’ satisfaction was assessed in each setting. The
response variable is the difference in satisfaction scores, i.e. self-paced
minus machine-paced.
If the job satisfaction scores follow a normal distribution with a standard
deviation of 60, and data from a sample of 18 workers gives a sample
mean difference score of 17, can it be concluded that there is a difference
in job satisfaction between machine-paced and self-paced assembly line
workers?

21

𝐻 : 𝜇=0 (There is no average difference in satisfaction scores)

𝐻 : 𝜇≠0 (There is an average difference in satisfaction scores)

Assume a level of significance, 𝛼 = 0.05.

𝑋−𝜇
Given that the data is normally distributed and the 𝑧= 𝜎
standard deviation is known, use the 𝑧-statistic. 𝑛

The rejection region

(two-tailed test): Reject 𝑯𝒐 Reject 𝑯𝒐
𝛼 = 0.025 0.4750 𝛼 = 0.025

-1.96 Critical 𝑧-value, corresponding

0 1.96
to an area under the table of
0.4750
22

11
Areas under the Standard Normal Curve (z-Table) showing values for P(0 ≤ Z ≤ z)
0 z

Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621
1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830
1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015
1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177
1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319
1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441
1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545
1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633
1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706
1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767
2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817
2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857
2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890
2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916
2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936
2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952

23

𝑋−𝜇 17 − 0
𝑧= 𝜎 = = 1.20
60
𝑛 18

Since the calculated 𝑧 (1.20) is smaller than the critical 𝑧 (1.96) and does
not fall in the rejection region, we do not reject 𝑯𝒐 . Thus, we accept 𝐻 ,
which says there is no average difference in satisfaction scores between
machine-paced and self-paced assembly line workers.
In other words, machine-paced and self-paced assembly line workers are
equally satisfied with their jobs.

12
 𝑝-value = 0.1151 + 0.1151
Alternative method using 𝑝-value: = 0.2302

0.5 – 0.3849 =
𝑋−𝜇 17 − 0 0.1151
𝑧= 𝜎 = = 1.20
60
𝑛 18 Z
0

-1.20 1.20

Since the 𝑝-value (0.2302) is greater than the significance level 𝛼 (0.05),
we do not reject 𝑯𝒐 . Thus, we accept 𝐻 , which says there is no average
difference in satisfaction scores between machine-paced and self-paced
assembly line workers.
In other words, machine-paced and self-paced assembly line workers are
equally satisfied with their jobs.

25

Areas under the Standard Normal Curve (z-Table) showing values for P(0 ≤ Z ≤ z)
0 z

Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621
1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830
1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015
1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177
1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319
1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441
1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545
1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633
1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706
1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767
2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817
2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857
2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890
2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916
2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936
2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952

26

13
Example 6.3
The mean speed of vehicles on highways in a country was found to be
115 km/h. The national highway authority wants to test a new Automated
Speed Enforcement (ASE) system to see if it has either a positive or
negative effect on speed of vehicles, or no effect at all.
A sample of 20 highways with the new ASE system has a mean of 100
km/h with a standard devation of 30 km/h. Did the new ASE system affect
speed of vehicles?

𝐻 : 𝜇 = 115
𝐻 : 𝜇 ≠ 115

Assume a level of significance, 𝛼 = 0.05.

27

Given 𝑛 < 30 and the standard deviation is 𝑋−𝜇

𝑡= 𝑠
unknown, we use the 𝑡-statistic and apply the
𝑛
central limit theorem.

The rejection region

(two-tailed test): Reject 𝑯𝒐 Reject 𝑯𝒐

Critical 0 Critical
𝒕-value 𝒕-value

Reject 𝐻 if the calculated 𝑡-value falls in the rejection region.

28

14
Degree of freedom = 20 – 1 = 19

Critical 𝑡-value = 2.093 (From the 𝑡 distribution table, for a two-tailed test
with 𝛼 = 5% and degree of freedom = 19)

𝑋−𝜇 100 − 115

𝑡= 𝑠 = = −2.236
30
𝑛 20

Since the calculated 𝑡-value (-2.236) is less than the critical 𝑡-value (-2.093),
we reject 𝑯𝒐 . Thus, we accept 𝐻 , which states that there is a difference in
mean speeds with the new ASE system. So, the new ASE system affects
speed of vehicles.

29

Level of significance for One-Tailed Test, 𝛼/2 Student’s t-

Degree of 0.10 0.05 0.025 0.01 0.005 0.001 0.0005 distribution
freedom,
Level of significance for Two-Tailed Test, 𝛼 table
df
0.20 0.10 0.05 0.02 0.01 0.002 0.001
The table gives the
16 1.337 1.746 2.120 2.583 2.921 3.686 4.015 values of t , df
17 1.333 1.740 2.110 2.567 2.898 3.646 3.965
18 1.330 1.734 2.101 2.552 2.878 3.610 3.922 where
19 1.328 1.729 2.093 2.539 2.861 3.579 3.883
P(Tdf > t , df ) = 
20 1.325 1.725 2.086 2.528 2.845 3.552 3.850
21 1.323 1.721 2.080 2.518 2.831 3.527 3.819
22 1.321 1.717 2.074 2.508 2.819 3.505 3.792
and t /2, df
23 1.319 1.714 2.069 2.500 2.807 3.485 3.767
24 1.318 1.711 2.064 2.492 2.797 3.467 3.745 where
25 1.316 1.708 2.060 2.485 2.787 3.450 3.725
26 1.315 1.706 2.056 2.479 2.779 3.435 3.707 P(Tdf > t /2, df ) = /2
27 1.314 1.703 2.052 2.473 2.771 3.421 3.690
28 1.313 1.701 2.048 2.467 2.763 3.408 3.674
29 1.311 1.699 2.045 2.462 2.756 3.396 3.659
30 1.310 1.697 2.042 2.457 2.750 3.385 3.646
30

15
Two-Sample Hypothesis Testing
Two-sample hypothesis testing is designed to test if there is a difference
between two means from two different populations.
Examples:
• Is there any difference in the mean amount of residential real estate
sold by male and female agents in Johor?
• Is there a difference in the average number of accidents along Jalan
Batu Pahat – Kluang during day time and night time?
We have learned previously that the distributions of sample means follows
the normal distribution, assuming that 𝑛 ≥ 30. Thus, the distribution of
their differences will also follow the normal distribution (𝑧-distribution). If
𝑛 < 30, we use the 𝑡-distribution.

31

The test statistics for the differences between two sample means:

𝑋 −𝑋 𝑋 −𝑋
𝑧= 𝑡=
𝑠 𝑠 𝑠 𝑠
+ +
𝑛 𝑚 𝑛 𝑚

where
𝑋 = mean of first sample 𝑋 𝑠 = pooled estimate of the population variance
𝑋 = mean of second sample 𝑌 𝑛−1 𝑠 + 𝑚−1 𝑠
=
𝑛+𝑚−2
𝑠 = variance of first sample 𝑋
𝑠 = variance of second sample 𝑌 Note: Degree of freedom = 𝑛 + 𝑚– 2

𝑛 = size of first sample 𝑋

𝑚 = size of second sample 𝑌
32

16
Example 6.4
There have been complaints that airport staff respond too slowly to locals
at the airport. In fact, it is claimed that foreigners receive faster service. A
study was conducted to investigate this issue. The following sample
information was collected:

Sample Mean Sample Standard

Passenger Sample
Response Time Deviation
Type Size
(minutes) (minutes)
Local 5.5 0.4 50
Foreigner 5.3 0.3 100

At the 0.01 significance level, is it reasonable to conclude the mean

response time is longer for local passengers?
33

Let 𝜇 be the mean response time for locals, and 𝜇 be the mean
response time for foreigners.

𝐻 : 𝜇 ≤𝜇
𝐻 : 𝜇 >𝜇

Level of significance, 𝛼 = 0.01.

Given 𝑛 and 𝑚 are large enough that the 𝑋 −𝑋

𝑧=
distribution of the sample means follows the 𝑠 𝑠
normal distribution, we use the 𝑧-statistic 𝑛 + 𝑚

34

17
 The rejection region
(one-tailed test): 0.49 Reject 𝐻
𝛼 = 0.01

Z
0

𝑋 −𝑋 5.5 − 5.3 2.33 Critical 𝑧-value taken from 𝑧-

𝑧= = = 3.13 table, corresponding to an
𝑠 𝑠 0.4 0.3 area under the table of 0.49

𝑛 + 𝑚 50
+ 100

Since the calculated 𝑧-value (3.13) is greater than the critical 𝑧-value (2.33)
and falls in the rejection region, we reject 𝑯𝒐 . Thus, we accept 𝐻 and it is
reasonable to conclude that the mean response time for local passengers is
greater than the mean response time for foreigners.

35

Areas under the Standard Normal Curve (z-Table) showing values for P(0 ≤ Z ≤ z)
0 z

Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621
1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830
1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015
1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177
1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319
1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441
1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545
1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633
1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706
1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767
2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817
2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857
2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890
2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916
2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936
2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952

36

18