Download Biostat lectures_Part 2_Ch 6 to Ch 9

Page 1 of 103
Page 2 of 103
Page 3 of 103
Biostatistics
Dr. Osama Alkam
Distribution of the Difference Between Two Sample Proportions:
No handout is available. Students are referred to the textbook.
Reading Assignment:
Chapter 5 (5.6) in W.W. Daniel.
Chapter 6
Estimation
Introduction:
Recall that statistical inference is the procedure by which a conclusion is reached about the population on the
basis of the information contained in a random sample drawn from the population.
The following chart represents the usual setup of statistical inference
Ask a question about the population → Get Information from a random
sample → Answer the question for the sample → Use the answer for the
sample to estimate (or guess) the answer for the population.
Definition: A descriptive measure for a population is called a parameter.
A descriptive measure for a sample is called a statistic. We will denote a parameter by
corresponding statistic by .
Example: , , are parameters and ̅ , , ̂ are statistics.
Page 4 of 103
and denote its
Biostatistics
Dr. Osama Alkam
Example:
A computer company wants to estimate the average weekly hours adults use computer at home. A random sample of
21 adults was used in the study. The average weekly hours for that sample x was 5.3 hours and the standard
deviation s was 0.9 hours. Based on that information about x and s we may estimate the parameters  and  .
There are two types of estimates:
1) Point Estimate: the value of the statistic is used (directly) to estimate the corresponding population
parameter
2) Interval Estimate (Confidence Interval): the value of the statistic is used to give an interval (L , U) that
contains the corresponding parameter with a specified degree of confidence.
(1 − )100% Confidence Intervals (C.I.):
The general form of a (1 − )100% confidence interval for a parameter
corresponding statistic
such that the distribution of its
is standardized to (0,1) distribution is given by
( statistic value )  ( reliability coefficient )  ( standard error )
The percentage (1 − )100% is called the confidence level. It is the percentage of samples or the level of
confidence the interval contains the unknown population parameter . Typical values of confidence levels are 90%,
95% and 99%.
The reliability coefficient is denoted by
the left of which the area is 1 − .
(
).
It is the value of on the horizontal axis under (0,1) distribution to
Typical values of the reliability coefficient are given in the following table
Page 5 of 103
Confidence Level ( C.L.)
Reliability Coefficient (R.C.)
90%
1.645
95%
1.960
99%
2.575
Biostatistics
Dr. Osama Alkam
Interpretations of Confidence Intervals:
Confidence intervals have the following two types of interpretations
Probabilistic Interpretation: in repeated sampling from a ( normally or approximately normally distributed )
population, (1 − )100% of all intervals of the form
( statistic value )  ( reliability coefficient )  ( standard error )
will ( on the average, in the long run) include the unknown population parameter.
Practical Interpretation: when sampling is done once from a ( normally or approximately normally distributed )
population, we are (1 − )100% confident that the interval
( statistic value )  ( reliability coefficient )  ( standard error )
contains the unknown population parameter.
(1 − )100% Confidence Interval for :
Recall that the sampling distribution of x is normal with mean  and standard deviation 
n
when the population
is (approximately) normally distributed or the sample size is large.
A (1 − )100% confidence interval (C.I.) for the population mean  is given by
̅±
(
)
√
where  is the population standard deviation and n is the sample size.
Example:
The administration of a state hospital wishes to estimate the mean age of all patients (bellow a certain age) admitted
to the hospital with severe cold in December of last year. In a random sample of size 25, the mean age was 13.8
years. From a past thorough study, the population standard deviation is known to be 1.5 years. Given that the ages
are normally distributed , construct a 95% confidence interval for the population mean age and give the practical
interpretation of your answer.
Page 6 of 103
Biostatistics
Dr. Osama Alkam
Solution:
x  13.8 ,   1.5 , n  25 . Thus, 100(1-  )%  95% , so   0.05 and = 0.025. From the standard
normal distribution we find that
13.8  (1.96) 
.
= 1.96. Therefore the required C.I. is
1.5
 13.8  0.588 . Hence, we are 95% confident that the interval
25
(13.8  0.588,13.8  0.588)  (13.212,14.388)
contains the population mean age.
A (1 − )100% Confidence Interval for p:
A (1 − )100% confidence interval (C.I.) for the population proportion
̂±
(
)
is given by
̂ (1 − ̂ )
Where n is the sample size.
Example:
A random sample of 59 table tennis players contains 15 left-handed players. Use this random sample to construct a
95% C.I. for the proportion of left-handed tennis players.
Solution:
The required C.I. is nothing but ̂ ±
(
)
= 1.96 (since
= 0.05,
(
)
(
)
, with ̂ =
= 0.025) .
= 0.2542 and
If we substitute these values we are 95% confident that the proportion of left-handed tennis players belongs to the
interval (0.143, 0.365) .
Reading Assignment:
Chapter 6 ( 6.1, 6.2 ) in W.W. Daniel.
Page 7 of 103
Biostatistics
Dr. Osama Alkam
The t Distribution:
Recall that if ~ ( ,
) then ̅ ~ ( ,
Question: suppose that
~ ( ,
(0,1).
Answer: the distribution of
̅
̅
). Thus, the distribution of
). What is the distribution of
̅
is the standard normal distribution
√
?
√
is the so called “t-distribution” or “ student’s t-distribution” with
√
− 1 degrees of
freedom (d.f.). It was discovered by William S. Gosset in 1908. There is a different t distribution for each sample size,
in other words, it is a class of distributions. When we speak of a specific t distribution, we have to specify the degrees
of freedom.
The t-distribution shape looks like the standard normal distribution but broader than it for small degrees of
freedom. It approaches the standard normal distribution for large sample sizes (practically speaking when n
> 30). Below are the t-distributions for 1, 3, and 10 degrees of freedom.
The t distribution has the following properties:
1) It has a mean of 0.
2) It has standard deviation greater than 1.
3) It is symmetric about the mean ( the y-axis).
4) Its altitude and width depend on the sample size .
5) It approaches the standard normal distribution as n tends to infinity .
Page 8 of 103
Biostatistics
Dr. Osama Alkam
The procedure outlined in the previous section for constructing a confidence interval for the population mean
necessitates a previous knowledge of the population standard deviation  . In most cases, we found ourselves face
to face with a need for constructing a confidence interval for the population mean  without knowing the
standard deviation of the population  .

The percentile

When  is unknown, use the sample standard deviation s as an estimate for  and use (
(
)
of the t-distribution is denoted by
).
(
as reliability coefficient.
For a (1   ) 100% confidence interval, the above mentioned different reliability coefficient is nothing but
It is obtained from a t distribution table.
The reliability coefficient
(
)
).
represents the value on the horizontal axis of the (n  1)  d . f .

t distribution to the left of which the area under the distribution is 1  

(
)
2

.
A (1   ) 100% confidence interval for a population mean, when the population standard
deviation is unknown and the population is normally distributed, is obtained using the formula
Example:
̅±
(
)
×
√
A random sample with size
mean
.
= 10 and variance
= 8 is selected from a normally distributed population with
= 5. Find the probability that the sample mean ̅ is greater than 7.
Solution:
( ̅ > 7) =
̅
√
9 degrees of freedom
Page 9 of 103
>
√
√
= ( > 2.236) = 0.025 because 2.236 ≈
.
for
Biostatistics
Dr. Osama Alkam
Example:
Suppose that a sample of 16 restaurants was randomly selected and the temperature of the coffee sold at each
restaurant in the sample was measured. Suppose that the sample mean temperature is 162 F and the sample
standard deviation is 10 F . Assuming that the temperatures are approximately normally distributed, construct a
95% confidence interval for the mean coffee temperature and give a practical interpretation for the result.
Solution:
95%  (1   )100% , thus   0.05 . Hence, = 0.025. Using the t-distribution table with 15 d.f. we find that
= 2.1315.. Whence the required C.I. for the population mean is given by
.
162  (2.1315 
Therefore, the required 95% C.I. for
10
)  162  2.1315  2.5
16
 162  5.32875
is
(162  5.32875,162  5.32875)  (156.67125,167.32875) .

We infer with 95% confidence that the average coffee temperature in all restaurants is no less than
156.671  F and no greater than 167.329  F .
Reading Assignment:
Chapter 6 ( 6.3 ) in W.W. Daniel .
Confidence Interval for the Difference Between Two Population Means:
I. Populations With Known Standard Deviations :
Recall that the sampling distribution of x1  x2 is normal with mean 1   2 and standard deviation
 12
n1

 22
n2
provided both samples are drawn from a normally distributed population or the sample sizes n1 and n2 are large (
 30 ).
Page 10 of 103
Biostatistics
Dr. Osama Alkam
To construct a (1   )100% confidence interval for 1   2 select two independent random samples of sizes n1
and n2 from two distinct populations having unknown means 1 ,  2 and known standard deviations  1 and  2 .
Then a (1   )100% confidence interval for 1   2 is given by
−
±
(
)
+
where x1 , x2 are the means of the first and second samples respectively.
Remark: If the confidence interval does not include zero then we are (1   )100% confident that 1 and  2 are
not equal.
Example:
The following data summary gives the tensile strengths of plastic bags for two independent random samples,
randomly selected from two different production runs:
Sample 1 :
n1  36,
x1  102.33
Sample 2 :
n2  40,
x2  118.19
Assume that the populations standard deviations are  1  10 and  2  30
(a) construct a 95% confidence interval for 1   2
(b) based on the computed interval in part (a), can we conclude that 1 and  2 are different?
Solution:
(a) (1   )100%  95%    0.05 . Hence
The standard error 
 12
n1

 22
n2

(
)
= 1.96.
102 302

 5.03 .Hence the required 95% C.I. is given by
36 40
(102.33  118.19)  (1.96)  5.03  ( 25.72, 6.00) .
(b) Since the C.I. in part (a) does not contain 0 we conclude that 1   2 .
Page 11 of 103
Biostatistics
Dr. Osama Alkam
II. Populations With Unknown Standard Deviations :

If both samples are large (  30 ) and the standard deviations of the populations are not known
then we can use s1 , s2 instead of  1 ,  2 respectively. Thus a (1   )100% confidence interval
for 1  2 , when  1 and  2 are not known and both samples are large, is given by

±
(
)
+
−
If both samples are not large, then we distinguish between two different cases:
Case 1 : The populations variances are equal
In this case the common variance of both populations is replaced by the pooled samples variance
s 2p 
(n1  1) s12  (n2  1) s22
. The reliability coefficient is obtained using the t  distribution with n1  n2  2
n1  n2  2
degrees of freedom.
Thus a (1   )100% confidence interval for 1   2 , when  1 and  2 are not known but known to be
equal and both samples are not large, is given by
where
−
±
(
)
+
is obtained from the t  distribution with n1  n2  2 degrees of freedom .
Example:
The following information was obtained from two independent samples selected from two normally
distributed populations with unknown but equal populations standard
deviations . Sample 1 :
Sample 2 :
n1  18,
x1  33.75,
s1  5.25
n2  20,
x2  28.50,
s2  4.55
Construct a 95% confidence interval for 1   2 .
Page 12 of 103
Biostatistics
Dr. Osama Alkam
Solution:
  0.05 . Thus
(
)
The pooled variance s 2p 
=
= 2.03 ( . . = 18 + 20 − 2 = 36).
.
17  (5.25) 2  19  (4.55) 2
 23.94 .
36
Hence the required 95% C.I. is given by
(33.75  28.50)  2.03 
23.94 23.94

 (2.02,8.48)
18
20
Case 2 (optional) : The populations variances are not equal
In this case a (1   )100% confidence interval for 1   2 is given by
−
where
) with
(
(
)
=
with
n2  1 degrees of freedom.
±
=
+
(
) with
n1  1 degrees of freedom and
=
Example:
Construct a 95% confidence interval for 1   2 of the previous example assuming that the populations have
unknown and unequal standard deviations
Solution:
=
.
= 2.1098 ( . . = 17) and
=
.
= 2.0930 ( . . = 19) . Hence
.
equals
Biostatistics
Dr. Osama Alkam
Therefore, the required 95% C.I. is given by
(5.25) 2 (4.55) 2

 (1.88,8.62)
18
20
(33.75  28.50)  2.1030 
Confidence Interval for a Population Proportion:
No handout is available. Students are referred to the textbook.
Confidence Interval for the Difference Between Two Population Proportions:
No handout is available. Students are referred to the textbook.
Reading Assignment:
Chapter 6 ( 6.4, 6.5, 6.6 ) in W.W. Daniel .
Determination of Sample Size for Estimating Population Means and Proportions
In the previous lectures we learned how to give an interval that contains the mean μ or the proportion p of the
population with a certain level of confidence. Such an interval is called a confidence interval. The length
interval equals 2
(
)
×
. The quantity
(
)
×
is called the error in estimating
of this
using .
The question is: suppose that we want to construct a (1   )100% confidence interval of length
for the population
mean or proportion, how large should the sample size be?
To answer this question we need to solve the equation
nearest integer. In case
=2
= , we use any previous value of
available, use 0.25 for (1 − ).
(
in
)
×
for n then round it from above to the
=
(
)
. If no previous value of
is
Note: As clear from the above equation, in order to determine the sample size we need to be provided with the
following three items of information: i) the desired length of the confidence interval
iii) the magnitude of the population variance (in case
Example:
ii) the level of confidence
= )
A hospital administrator wishes to estimate the mean weight of babies born in his hospital. How large a sample of
birth records should be taken if he wants a 99% confidence interval that is 1 pound long ? Assume that a reasonable
estimate for  is 1 pound.
Page 14 of 103
Page 15 of 103
Page 16 of 103
Biostatistics
Dr. Osama Alkam
Solution:
(1   )100%  99%    0.01  1  
2
 0.995 .
.
= 2.575.
= 1 and   1 . We need to solve the equation 1 = 2 × 2.575 ×
26.5225. Therefore, the required sample size is 27.
√
for . Clearly,
= (2 × 2.575) =
Example:
Suppose that we want to estimate the proportion of smokers within the population of Jordan with an error less than
0.05 using a 90% confidence interval. How large a sample of Jordanians should be taken in the following cases:
1) no previous information about this proportion is known.
2) a previous value of this proportion is about 0.3.
Solution:
Case1:
Solve the equation 0.05 =
.
×
.
.
×
. × .
sample size should not be less than 269.
= 1.64 ×
.
√
for . Clearly,
=(
.
.
× .
) = 268.69. Hence, the
Case2:
Solve the equation 0.05 =
= 1.64 ×
.
√
for . Clearly,
225.926. Hence in this case, the sample size should not be less than 226
=(
.
× .
.
Confidence Interval for the Variance of a Normally Distributed Population
No handout is available. Students are referred to the textbook.
Confidence Interval for the Ratio of the Variances of Two Normally Distributed Populations
No handout is available. Students are referred to the textbook.
Reading Assignment:
Chapter 6 ( 6.7, 6.8, 6.9, 6.10 ) in W.W. Daniel.
Page 17 of 103
) =
Page 18 of 103
Page 19 of 103
Page 20 of 103
Page 21 of 103
Page 22 of 103
Page 23 of 103
Biostatistics
Dr. Osama Alkam
Chapter 7
Hypothesis Testing
Introduction:
Hypothesis testing is another type of statistical inference. The purpose of hypothesis testing is to aid a researcher in
reaching a conclusion about a population by examining a random sample from that population. Thus the main
objective of hypothesis testing is to assess the evidence provided by a simple random sample of the population, with
a certain level of significance, about a certain claim concerning the population.
General Hypothesis Testing Procedure:
Step1: State the null hypothesis:
:
or
≤
:
or
≥
:
=
The null hypothesis is an established (or conceived) hypothesis about the population parameter. A statement
presumed true in the population unless a strong evidence to the contrary is demonstrated.
Step 2: State the alternative hypothesis:
:
>
or
:
<
or
:
≠
.
The alternative hypothesis is a rival claim to the null hypothesis, suspected by the researcher ; the claim about the
population's parameter that the researcher is trying to find an evidence for.
Step 3 : Select the level of significance
The significance level
 of the test :
 is the probability of rejecting the null hypothesis when it is true. It is also called the
probability of type I error. The other type of error, namely type II error is the error of accepting the null hypothesis
when it is false, it is usually denoted by  .
The significance level should be decided by the researcher on the basis of the significance of the problem itself. It is
usually taken to be 0.1 , 0.05, or 0.01.
Step 4: Determine the rejection region of H 0
Page 24 of 103
Biostatistics
Dr. Osama Alkam
The rejection region of H 0 is located under the standard distribution of . In case
right of

left of −

(
tailed ).
right of

(
)(
, it is to the:
) if H A states that the parameter is greater than that in the null hypothesis (right
(
tailed)
=
) if H A states that the parameter is less than that in the null hypothesis (left
−
(
))
or the left of −
(
)(
equal to that in the null hypothesis (double tailed ) .
−
(
))
if H A states that the parameter is not
Step 5: Compute the test statistic and check whether it is located in the rejection region of H 0 or not. The test
statistic is given by
(assuming that H 0 is true if necessary).
Step 6: State your final conclusion:
If the test statistic is located in the rejection region then conclude that the null hypothesis is false thus the
alternative hypothesis is true otherwise accept the null hypothesis and conclude that the data contained in the
sample does not provide an evidence to conclude that the alternative hypothesis is true.
Example:
The variance of ages in a certain population is 50. A simple random sample with size 100 selected from that
population showed an average age of 31.5 years. Does the sample contain an evidence to conclude that the mean
age of the population is greater than 30 years? Test at
= 0.05.
Solution:
State
,
conclusion..
:
:
≤ 30
> 30
Page 25 of 103
, describe the rejection region of
for
= 0.05, and compute the test statistic. Write down your
Biostatistics
Dr. Osama Alkam
The rejection region is located under the
.
Test statistic=
= 2.12,
√
(0,1) distribution to the right of
.
= 1.64.
which is beyond the rejection region of
. Thus, the sample contains an
evidence to conclude that the mean age of the population is greater than 30 years.
Reading Assignment:
Chapter 7 ( 7.1 ) in W.W. Daniel .
Hypothesis Testing for a Single Population Mean:
Suppose that we want to test a rival claim about  against a null hypothesis H 0 . The possible alternative
hypotheses are

H A :   0 (if we are trying to determine whether the sample provides evidence that the population
mean is greater than 0 )

H A :   0 (if we are trying to determine whether the sample provides evidence that the population
mean is less than 0 )

H A :   0 (if we are trying to determine whether the sample provides evidence to conclude that the
population mean is different from 0 )
The test statistic is given by

Z
x  0

when  is known and (the population is normal or n  30 )
n

Z

t
Page 26 of 103
x  0
when  is unknown and n  30 .
s
n
x  0
( . .=
s
n
− 1) when  is unknown and the population is normal and n  30 .
Biostatistics
Dr. Osama Alkam
Example:
The standard deviation of yield of corn in the United States is about 10 bushels per acre. A random sample of 50
farmers showed an average ̅ = 123.6 bushels per acre. Does this sample provide sufficient evidence that the mean
corn yield of the United States is higher than 120 bushels per acre? Test at the 0.05 level of significance.
Solution:
H 0 :   120
H A :   120
=
= 1.645. Test statistic =
.
123.6  120 3.6  50

 2.5455..
10
10
50
Since the test statistic is located to the right of 1.645, it is in the rejection region of H 0 . Thus we accept H A and
conclude that the mean corn yield of the United States is higher than 120 bushels per acre.
Example:
In a certain state the average SAT math score is believed to be 450. In a sample of 500 students averaged 461 with
a standard deviation 100, is there any reason to believe that the average SAT score in this state is not 450? test at a
0.01 level of significance.
Solution:
H 0 :   450
H A :   450
(
)
=
.
= 2.575. Test statistic =
461  450 11 500

 2.4596..
100
100
500
The test statistic is located in the acceptance region of H 0 . Hence we accept H 0 . There is no reason to believe
that the average math SAT score in that state is different than 450.
Page 27 of 103
Biostatistics
Dr. Osama Alkam
Example:
A chemist measures the haptoglobin concentration (in gm/liter) in blood serum taken from a random sample of eight
healthy adults. The values are:
1.82
3.32
1.07
1.27
0.49
3.79
0.15
1.98
Does this sample contain an evidence that the mean haptoglobin concentration in adults is greater than 1 gm/liter ?
Test at   0.01. Assume that the population is normally distributed.
Solution:
H0 :  1
H A :  1
(
)
=
= 2.998 ( . . = 7)
.
To compute the test statistic we need to compute the mean and the standard deviation of the sample.
8
x
x
i
i 1
8
1.82  3.32  1.07  1.27  0.49  3.79  0.15  1.98
 1.73625
8

8
s
 ( x  x)
i 1
i
7
The test statistic =
2
 1.283031
1.73625  1 0.73625  8

 1.62305..
1.283031
1.283031
8
Since the test statistic is less than 2.998, it is located in the rejection region of H 0 . Hence we accept H 0 . We
conclude that the sample does not contain an evidence that the mean haptoglobin concentration in adults is greater
than 1 gm/liter.
Page 28 of 103
Page 29 of 103
Page 30 of 103
Page 31 of 103
Page 32 of 103
Page 33 of 103
Page 34 of 103
Page 35 of 103
Biostatistics
Dr. Osama Alkam
The p Vlaue for a Test
No handout is available. Students are referred to the textbook.
Reading Assignment:
Chapter 7 ( 7.2 ) in W.W. Daniel .
Testing H0 by Means of a Confidence Interval:
There is a close relationship between confidence intervals and hypothesis testing. When a 95% confidence
interval is constructed, all values in the interval are considered plausible values for the parameter being estimated.
Values outside the interval are rejected as relatively implausible. If the value of the parameter specified by the null
hypothesis is contained in the 95% interval then the null hypothesis cannot be rejected at the 0.05 level. If the value
specified by the null hypothesis is not in the interval then the null hypothesis can be rejected at the 0.05 level. If a
99% confidence interval is constructed, then values outside the interval are rejected at the 0.01 level. Therefore, a
(1   )100% confidence interval for  can be used to test the null hypothesis H 0 :    0 versus the
alternative hypothesis H A :    0 at  level of significance. If the hypothesized mean  0 is contained within the
confidence interval then we can not reject null hypothesis H 0 :    0 otherwise we reject it .
Example:
A survey of 100 similar-sized hospitals revealed a mean daily census in the pediatrics service of 27 with a standard
deviation 6.5. Do these data provide a sufficient evidence to indicate that the population mean is different than 25?
Test the relevant hypotheses using a 95% confidence interval.
Solution:
H 0 :   25
H A :   25
A 95% confidence interval for  is

 

, x  Z (1 0.05 ) 
 x  Z (1 0.05 2 ) 
 
2
n
n

6.5
6.5 

, 27  1.96 
 27  1.96 
   25.726 , 28.274 
100
100 

Since 25 does not belong to the interval
 25.726 , 28.274
we reject the null hypothesis (thus accept the
alternative hypothesis) and deduce from the data that the population mean is different than 25 (with a 0.05 level of
significance).
Page 36 of 103
Biostatistics
Dr. Osama Alkam
Hypothesis Testing for the Difference Between Two Population Means:
Recall the following:

 12  22 
N



,

x

x
1. The sampling distribution for 1
 1
 provided that
2
2 (for independent samples) is
n
n2 

1
both samples are collected from normally distributed populations or both n1 and n2 are large.
2. A (1   )100% confidence interval for 1  2 is

(x 1  x 2 )  Z (1  ) 
2
 12
n1

 22
n2
in case  1 and  2 are known (and the populations are
normal or the samples are large)

(x 1  x 2 )  Z (1  ) 
2

(x 1  x 2 )  t (1  ) 
2
s 12 s 22

in case  1 and  2 are unknown (and the samples are large)
n1 n 2
s p2
n1

s p2
n1
in case  1 and  2 are unknown and equal and the
populations are normal and the samples are small.
Possible Tests About the Difference Between Two Population Means:
The null value of 1  2 is always given in the problem, it is usually 0.
The possible tests are:

H 0 : 1   2  ( 1   2 )0 versus H A : 1   2  ( 1   2 )0

H 0 : 1   2  ( 1   2 )0 versus H A : 1   2  ( 1   2 )0

H 0 : 1   2  ( 1   2 )0 versus H A : 1   2  ( 1   2 )0
Test Statistic:

Z
( x1  x2 )  ( 1   2 )0
 12
n1

 22
n2
the samples are both  30 )
Page 37 of 103
in case  1 and  2 are known and (the populations are normal or
Biostatistics
Dr. Osama Alkam
( x1  x2 )  ( 1   2 )0
in case  1 and  2 are unknown and the samples are both  30 )

Z

in case  1 and  2 are unknown but equal and the populations are normal and the
s12 s22

n1 n2
samples are both  30 , t 
(x 1  x 2 )  ( 1   2 )0
s p2
n1

s p2
s 2p 
( n1  1) s12  (n2  1) s22
, with
n1  n2  2
n2
n1  n 2  2 degrees of freedom.

in case  1 and  2 are unknown but unequal and the populations are normal and the
samples are both  30 , use Cochran's modified t test (optional)
Example:
One particular company was trying to increase the tensile strength of the plastic bags and still hold down
production cost by adjusting temperature and pressure in the production runs. The following is a summary
data from two independent random samples for the tensile strengths of plastic bags from two different
production runs:
Sample 1: n1  36,
x1  102.33,
Sample 2: n2  40,
x2  118.19,
1  10
 2  30
Do these samples provide sufficient evidence to conclude that the mean tensile strength of plastic bags from
the first production run is less than that from the second production run? Conduct a test at the 0.05 level of
significance.
Solution:
H 0 : 1   2  0
H A : 1   2  0
Test statistic 
x1  x 2  0
 12
n1
Page 38 of 103

 22
n2

102.33  118.19
 3.15..,  Z 1   Z 0.95  1.645
100 900

36 40
Biostatistics
Dr. Osama Alkam
Since the test statistic = −3.15. . < −1.645, it is located in the rejection region of H 0 . Thus we accept
H A and deduce that the samples contain an evidence that the mean tensile strength of the first production
run is less than that of the second production run.
Example:
Suppose that in the previous example the standard deviations of the two production runs are not known and
that the standard deviations of the two samples are s1  14.06 and s2  24.44 , respectively. Do the
samples provide sufficient evidence to deduce that the production runs are different in tensile strengths?
Conduct a test with the 0.05 level of significance.
Solution:
H 0 : 1   2  0
H A : 1   2  0
Test statistic 
x1  x 2  0
2
1
2
2

s
s

n1 n 2
102.33  118.19
(14.06) 2 (24.44) 2

36
40
 3.509..
Z (1  )  Z 0.975  1.96
2
Since the test statistic = −3.509 < −1.96, it is located in the rejection region of H 0 . Thus we accept
H A and deduce that the mean tensile strength of the production runs are different.
Example:
The following information was obtained from two independent samples selected from two normally
distributed populations with unknown but equal population standard deviations :
Sample 1: n1  18,
x1  33.75,
s1  5.25
Sample 2: n2  20,
x2  28.5,
s2  4.55
Test whether the difference between the two means 1  2 is greater than 1.25 at the 0.05 significance
level.
Page 39 of 103
Biostatistics
Dr. Osama Alkam
Solution:
H 0 : 1   2  1.25
H A : 1   2  1.25
s 2p 
(n1  1) s12  (n2  1) s22 17  (5.25) 2  19  (4.55) 2

 23.942
n1  n2  2
18  20  2
The test statistic t 
x1  x2  1.25
s 2p
n1

s 2p
n2

33.75  28.5  1.25
 2.516
23.942 23.942

18
20
t 1  t 0.95  1.645 (degrees of freedom is 18+20 - 2 = 36, so we use (0,1)). Since the test statistic =
2.516 > 1.645, it is located in the rejection region of H 0 . Thus we accept H A and deduce that the
difference is greater than 1.25.
Paired Comparisons:
In all previous discussions involving the difference between two means, it was assumed that the samples
are independent. In some situations we need to make use of relating observations resulting from nonindependent samples, for example if the effectiveness of a method of medical treatment is to be assessed
then we don’t consider two independent samples of patients because otherwise there is a possibility for the
presence of extraneous sources of variation that may lead us wrongly to reject of the null hypothesis .
A hypothesis test based on samples with a pair of data for each subject in the sample is known as a paired
comparison test.
The objective in paired comparisons tests is to eliminate a maximum number of sources of extraneous
variation by making the pairs similar with respect to as many variables as possible.
Example:
Suppose that we wish to compare the protection of two sunscreens A and B. If we try the sunscreen A on a
sample of people with a darker complexion than another sample of people on which we try sunscreen B
Page 40 of 103
Biostatistics
Dr. Osama Alkam
then the results will be affected by the complexion darkness rather than the sunscreen type itself. The
sample with a darker skin is less sensitive to sunlight than the other sample. A better way to design the
experiment would be to select just one simple random sample of subjects and let each member of the
sample receive both sunscreens, we could, for example assign sunscreen to the left and sunscreen B to
the right side of each subject's back.
The main idea of paired comparisons: instead of performing the analysis with individual observations, we
use d i , the difference between pairs of observations, as the variable of interest.
The possible tests are:

H 0 : d  ( d )0 versus H A :  d  (  d )0

H 0 : d  ( d )0 versus H A :  d  (  d )0

H 0 :  d  (  d )0 versus H A :  d  (  d )0
n
d  (  d )0
Test Statistic: t 
, degrees of freedom n 1 , d 
sd
n
d
i 1
i
n
,
sd 
(di  d )2
n 1
Example:
The following table provides the weights in kg of 9 obese women before (B) and after (A) a 12 week of
treatment with a very low calorie diet
B
117.3
111.4
98.6
104.3
105.4
100.4
81.7
89.5
78.2
A
83.3
85.9
75.8
82.9
82.3
77.7
62.7
69.0
63.9
Can we deduce from this random sample that the treatment is effective in causing weight reduction in
obese women? test with 0.05 level of significance.
Solution:
Let
=
− . The treatment is considered effective if
against the null hypothesis
Thus
Page 41 of 103
d  0 .
d  0
which will be our alternative hypothesis
Biostatistics
Dr. Osama Alkam
H 0 : d  0
H A : d  0
We need to compute the difference di , i  1, 2,...,9, between each pair of data. The differences are -34, 25.5, -22.8, -21.4, -23.1, -22.7, -19, -20.5, -14.3
d
203.3
22.5889  0
 22.5889 , sd  5.3194 (why?) t 
 12.7395
5.3194
9
9
Since -12.7395 < t 0.95  3.3554 , we accept H A and deduce from the sample that the treatment is
effective.
Reading Assignment:
Chapter 7 ( 7.3, 7.4 ) in W.W. Daniel.
Hypothesis Testing: A Single Population Proportion
Suppose that we want to test a rival claim about p against the null hypothesis H 0 : p  p 0 . The possible
alternative hypotheses are

H A : p  p 0 (if we are trying to determine whether the sample provides evidence that the population
proportion is greater than p 0 )

H A : p  p 0 (if we are trying to determine whether the sample provides evidence that the population
proportion is less than p 0 )

H A : p  p 0 (if we are trying to determine whether the sample provides evidence to conclude that the
population proportion is different from p 0 )
The test statistic is given by Z 
pˆ  p 0
p 0 (1  p 0 )
n
.
The rejection region of HA is located under N(0,1) distribution.
Page 42 of 103
Biostatistics
Dr. Osama Alkam
Example:
In a survey of injection drug users in USA, a researcher found that 18 out of 423 were HIV positive. Does this sample
contain an evidence that less than 5% of injection drug users in USA are HIV positive? Test at
significance.
= 0.1 level of
Solution:
H 0 : p  0.05
H A : p  0.05
−
=−
.
18
 0.05
0.0426  0.05
= −1.28. Test statistic = 423

  0.698321
0.0105968
0.05  0.95
423
Since the test statistic is located to the right of −1.28, it is in the rejection region of H A . Thus we reject H A and
conclude that the sample does not contain an evidence that less than 5% of injection drug users in USA are HIV
positive.
Hypothesis Testing: The Difference Between Two Population Proportions
We want to use two independent random samples, with sizes
, , respectively, selected from two populations, to
test a rival claim about the difference between two proportions p  q against the null hypothesis H 0 : p  q  0 .
The possible alternative hypotheses are

H A : p  q  0 , H A : p  q  0 or H A : p  q  0
p̂ q̂
p* 
Let , be the respective sample proportions and let
the test statistic is given by
Z 
pˆ  qˆ
 1 1
p * (1  p * )   
m n 
The rejection region of HA is located under N(0,1).
Page 43 of 103
.
( mpˆ )  ( nqˆ )
be the pooled estimate p and q, then
m n
Biostatistics
Dr. Osama Alkam
Example:
In a study of nutrition care in nursing homes, a researcher found that among 55 patients with hypertension, 24 were
on sodium-restricted diets. Of 149 patients without hypertension, 36 were on sodium-restricted diets. Can we
conclude from these random samples that the proportion of patients on sodium-restricted diets is higher among all
patients with hypertension? Test at
Solution:
Let ,
= 0.05 level of significance.
be the proportions of patients on sodium-restricted diets among all patients with and without hypertension,
respectively. To test
H 0 : p  q  0 versus H A : p  q  0
∗
24 36

55 149
= 0.2941, test statistic 
 2.71 , which is greater than
1 
 1
0.2941 0.7059  

 55 149 
=
Z 1  Z 0.95  1.645 . So, we accept H A and conclude from these random samples that the proportion of
patients on sodium-restricted diets is higher among patients with hypertension.
Reading Assignment:
Chapter 7 ( 7.5, 7.6 ) in W.W. Daniel.
The Chi-Square Distribution; Inference About Population Variance
Let
be the variance of a random sample with (fixed) size , selected from a normally distributed population with
variance
. Then
distribution with
following
Page 44 of 103
(
)
is a random variable. The distribution of this random variable is called a chi-square
− 1 degrees of freedom (denoted by
( − 1)). A typical chi-square distribution looks like the
Biostatistics
Dr. Osama Alkam
A (1 − )100 percentile
(
)
of the chi-square distribution is denoted by
.
2-tables are similar to t-tables, however, since the 2 –distribution is not symmetric, there is a table that gives 2 for
big and small values of .
( − )
% . .
:


 (n  1)s 2 (n  1)s 2 
A (1 − )100% C. I. for σ is given by 
,
2
2
.




 (1 )


2
2

Example:
A random sample with size 11 is selected from a normally distributed population with variance  2 . Suppose that the
sample variance is s 2  30 . Use this random sample to give a 90% confidence interval for  2 .
Solution:
2
2
  0.1, so  2    0.95
 18.3070 ,  2   0.05
 3.94030
1
2
(d .f .  10) . Thus, the required C.I. is
2
 (11  1)30 (11  1)30 
,

  (16.3872 , 76.1363)
 18.3070 3.94030 
Hypothesis Testing About
:
Suppose that we want to test a rival claim about
hypotheses are:
against a null hypothesis H 0 . The possible alternative
Biostatistics

Dr. Osama Alkam
H A :  2   02 (if we are trying to determine whether the sample provides evidence that the population
variance is greater than  02 )

H A :  2   02 (if we are trying to determine whether the sample provides evidence that the population
variance is less than  02 )

H A :  2   02 (if we are trying to determine whether the sample provides evidence to conclude that the
population variance is different from  02 ).
Test statistic:  2 
(n  1)s 2
 02
. The rejection region of H A is located under the chi-square distribution (with n – 1
d.f) to the left of 12 or to the right of  2 or between  2  and  2 .
1
2
2
Example:
A sample with size n  11 and variance s 2  30 is randomly selected from a normally distributed population with
variance  2 . Does this sample contain an evidence that  2  35 . Test at   0.1 .
Solution:
H 0 :  2  35
H A :  2  35
=
.
= 4.86518 (d.f. = 10). Test statistic =
(11  1)  30
 8.57143
35
Since the test statistic is located to the right of 4.86518 , it is in the rejection region of H A . Thus we reject H A
and conclude that the sample does not contain an evidence that the population variance is less than 35.
Example:
A sample with size n  16 and variance s 2  20 is randomly selected from a normally distributed population with
variance  2 . Does this sample contain an evidence that  2  50 . Test at   0.1 .
Page 46 of 103
Biostatistics
Dr. Osama Alkam
Solution:
H 0 :  2  50
H A :  2  50
=
.
= 7.26094 and
Test statistic =
(16  1)  20
6
50
/
=
.
= 24.9958 (d.f. = 15).
Since the test statistic is not between 7.26094 and 24.9958 , we accept H A (hence reject H 0 ) and conclude
that the sample contains an evidence that the population variance is different than 50.
The F-Distribution; Hypothesis Testing About
Let
,
/
be the variances of two independent random samples with (fixed) sizes
normally distributed populations with variances
and
, respectively. Then
distribution of this random variable is called an
F-distribution with (
looks like the following
Page 47 of 103
− 1, − 1) degrees of freedom (denoted by
(
×
×
and , selected from two
is a random variable. The
− 1, − 1)). A typical F-distribution
Biostatistics
Dr. Osama Alkam
A (1 − )100 percentile
simply,
).
An Important Property:
(
of the F(m – 1 ,n – 1)-distribution is denoted by
)
(
− 1, − 1) =
(
,
)
(
− 1, − 1) (or
Suppose that we want to use two independent samples to test a rival claim against a null hypothesis
. We are
interested in the following alternative hypotheses:

H A :  12 /  22  1 (if we are trying to determine whether the samples provide evidence that the populations
variance ratio is greater than 1.

H A :  12 /  22  1 (if we are trying to determine whether the samples provide evidence that the populations
variance ratio is less than 1)

H A :  12 /  22  1 (if we are trying to determine whether the samples provide evidence that the populations
variance ratio is different from 1).
Test statistic: F 
s12
. The rejection region of H A is located under the F- distribution (with (m-1,n-1) degrees of
s 22
freedom, where m is the size of sample1 (with variance s 12 ) and n is the size of the sample2 (with variance s 22 ) .It is
either to the left of F1 or to the right of F or between F
1

and F  .
2
2
Example: Consider the following information about two independent samples, selected from two normally distributed
populations with variances  12 and  22 .
Sample Size
Sample Variance
Sample1
10
100
Sample2
6
60
Can we conclude from these samples that  12 and  22 are different. Test at   0.1
Solution: To test H 0 :  12 /  22  1
versus
H A :  12 /  22  1
  0.1 . So, F1 /2 (10  1, 6  1)  F0.95 (9,5)  4.77 and
F / 2 (10  1, 6  1)  F0.05 (9,5) 
Page 48 of 103
1
1

 0.275482 .
F0.95 (5,9) 3.63
Biostatistics
Dr. Osama Alkam
s12 100
Test statistics 2 
 1.67 is located between 0.275482 and 4.77. Thus we can not accept H A .So, at this
s2
60
level of significance (and using these samples) we can not conclude that  12 and  22 are different.
Type II Error and the Power of the Test
No handout is available. Students are referred to the textbook.
Determining Sample Size to Control Type II Error
No handout is available. Students are referred to the textbook.
Reading Assignment:
Chapter 6 ( 6.9 ) and Chapter 7(7.7, 7.8, 7.9, 7.10) in W.W. Daniel.
Chapter 8
Analysis of Variance
Introduction:
Page 50 of 103
Page 51 of 103
Page 52 of 103
Page 53 of 103
Page 54 of 103
Biostatistics
Page 55 of 103
Dr. Osama Alkam
Biostatistics
The Completely Randomized Design (CRD)
Dr. Osama Alkam
Biostatistics
Page 57 of 103
Dr. Osama Alkam
Biostatistics
Page 58 of 103
Dr. Osama Alkam
Biostatistics
Page 59 of 103
Dr. Osama Alkam
Biostatistics
Dr. Osama Alkam
Critical value  F1 ( k  1, N  k )  F0.95 (3  1,15  3)  F0.95 (2,12)  3.89 .
Since the value of the test statistic F  1.093  Critical value, we can not accept H A and we deduce that the
mean scores of the three teaching methods are the same.
Reading Assignment:
Chapter 8(8.1, 8.2) in W.W. Daniel.
Biostatistics
Dr. Osama Alkam
Chapter 9
Simple Linear Regression and Correlation
Biostatistics
Reading Assignment:
Cha
Dr. Osama Alkam
Biostatistics
Dr. Osama Alkam
Biostatistics
Dr. Osama Alkam
Biostatistics
Dr. Osama Alkam
Find the regression line and the correlation coefficient r.
Answers:
The regression line is given by
To find r you need the following
=
=
10.7333
√26.5667√4.6667
= 0.963961
Biostatistics
Dr. Osama Alkam
Evaluating the Regression Equation: No handout is available. Students are referred to the textbook.
Reading Assignment: Chapter 9(9.1,9.2,9.3,9.4, 9.7) in W.W. Daniel.
Relative Risk, Odds Ratio and Survival Analysis: This additional topic will be covered if time permits. No handout
is available. Students are referred to Chapter 12 (12,7, 12.8) in W.W. Daniel.
Page 67 of 103
Page 68 of 103
Page 69 of 103
Page 70 of 103
Page 71 of 103
Page 72 of 103
Page 73 of 103
Page 74 of 103
Page 75 of 103
Page 76 of 103
Page 77 of 103
Page 78 of 103
Page 79 of 103
Page 80 of 103
Page 81 of 103
Page 82 of 103
Page 83 of 103
Page 84 of 103
Page 85 of 103
Page 86 of 103
Page 87 of 103
Page 88 of 103
Page 89 of 103
Page 90 of 103
x
0
1
2
3
4
x
0
1
2
3
4
5
x
0
1
2
3
4
5
6
x
0
1
2
3
4
5
6
7
0.01
0.951
0.999
1.000
1.000
1.000
0.01
0.941
0.999
1.000
1.000
1.000
1.000
0.01
0.932
0.998
1.000
1.000
1.000
1.000
1.000
0.01
0.923
0.997
1.000
1.000
1.000
1.000
1.000
1.000
Page 91 of 103
0.05
0.774
0.977
0.999
1.000
1.000
Table II: Cumulative Binomial Probabilities for n = 5
p
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.590 0.328 0.168 0.078 0.031 0.010 0.002 0.000
0.919 0.737 0.528 0.337 0.188 0.087 0.031 0.007
0.991 0.942 0.837 0.683 0.500 0.317 0.163 0.058
1.000 0.993 0.969 0.913 0.812 0.663 0.472 0.263
1.000 1.000 0.998 0.990 0.969 0.922 0.832 0.672
0.9
0.000
0.000
0.009
0.081
0.410
0.95
0.000
0.000
0.001
0.023
0.226
0.99
0.000
0.000
0.000
0.001
0.049
0.05
0.735
0.967
0.998
1.000
1.000
1.000
Table II: Cumulative Binomial Probabilities for n = 6
p
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.531 0.262 0.118 0.047 0.016 0.004 0.001 0.000
0.886 0.655 0.420 0.233 0.109 0.041 0.011 0.002
0.984 0.901 0.744 0.544 0.344 0.179 0.070 0.017
0.999 0.983 0.930 0.821 0.656 0.456 0.256 0.099
1.000 0.998 0.989 0.959 0.891 0.767 0.580 0.345
1.000 1.000 0.999 0.996 0.984 0.953 0.882 0.738
0.9
0.000
0.000
0.001
0.016
0.114
0.469
0.95
0.000
0.000
0.000
0.002
0.033
0.265
0.99
0.000
0.000
0.000
0.000
0.001
0.059
0.05
0.698
0.956
0.996
1.000
1.000
1.000
1.000
Table II: Cumulative Binomial Probabilities for n = 7
p
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.478 0.210 0.082 0.028 0.008 0.002 0.000 0.000
0.850 0.577 0.329 0.159 0.063 0.019 0.004 0.000
0.974 0.852 0.647 0.420 0.227 0.096 0.029 0.005
0.997 0.967 0.874 0.710 0.500 0.290 0.126 0.033
1.000 0.995 0.971 0.904 0.773 0.580 0.353 0.148
1.000 1.000 0.996 0.981 0.938 0.841 0.671 0.423
1.000 1.000 1.000 0.998 0.992 0.972 0.918 0.790
0.9
0.000
0.000
0.000
0.003
0.026
0.150
0.522
0.95
0.000
0.000
0.000
0.000
0.004
0.044
0.302
0.99
0.000
0.000
0.000
0.000
0.000
0.002
0.068
0.05
0.663
0.943
0.994
1.000
1.000
1.000
1.000
1.000
Table II: Cumulative Binomial Probabilities for n = 8
p
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.430 0.168 0.058 0.017 0.004 0.001 0.000 0.000
0.813 0.503 0.255 0.106 0.035 0.009 0.001 0.000
0.962 0.797 0.552 0.315 0.145 0.050 0.011 0.001
0.995 0.944 0.806 0.594 0.363 0.174 0.058 0.010
1.000 0.990 0.942 0.826 0.637 0.406 0.194 0.056
1.000 0.999 0.989 0.950 0.855 0.685 0.448 0.203
1.000 1.000 0.999 0.991 0.965 0.894 0.745 0.497
1.000 1.000 1.000 0.999 0.996 0.983 0.942 0.832
0.9
0.000
0.000
0.000
0.000
0.005
0.038
0.187
0.570
0.95
0.000
0.000
0.000
0.000
0.000
0.006
0.057
0.337
0.99
0.000
0.000
0.000
0.000
0.000
0.000
0.003
0.077
x
0
1
2
3
4
5
6
7
8
x
0
1
2
3
4
5
6
7
8
9
x
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
0.01
0.914
0.997
1.000
1.000
1.000
1.000
1.000
1.000
1.000
0.01
0.904
0.996
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
0.01
0.860
0.990
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
Page 92 of 103
0.05
0.630
0.929
0.992
0.999
1.000
1.000
1.000
1.000
1.000
Table II: Cumulative Binomial Probabilities for n = 9
p
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.387 0.134 0.040 0.010 0.002 0.000 0.000 0.000
0.775 0.436 0.196 0.071 0.020 0.004 0.000 0.000
0.947 0.738 0.463 0.232 0.090 0.025 0.004 0.000
0.992 0.914 0.730 0.483 0.254 0.099 0.025 0.003
0.999 0.980 0.901 0.733 0.500 0.267 0.099 0.020
1.000 0.997 0.975 0.901 0.746 0.517 0.270 0.086
1.000 1.000 0.996 0.975 0.910 0.768 0.537 0.262
1.000 1.000 1.000 0.996 0.980 0.929 0.804 0.564
1.000 1.000 1.000 1.000 0.998 0.990 0.960 0.866
0.9
0.000
0.000
0.000
0.000
0.001
0.008
0.053
0.225
0.613
0.95
0.000
0.000
0.000
0.000
0.000
0.001
0.008
0.071
0.370
0.99
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.003
0.086
0.05
0.599
0.914
0.988
0.999
1.000
1.000
1.000
1.000
1.000
1.000
Table II: Cumulative Binomial Probabilities for n = 10
p
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.349 0.107 0.028 0.006 0.001 0.000 0.000 0.000
0.736 0.376 0.149 0.046 0.011 0.002 0.000 0.000
0.930 0.678 0.383 0.167 0.055 0.012 0.002 0.000
0.987 0.879 0.650 0.382 0.172 0.055 0.011 0.001
0.998 0.967 0.850 0.633 0.377 0.166 0.047 0.006
1.000 0.994 0.953 0.834 0.623 0.367 0.150 0.033
1.000 0.999 0.989 0.945 0.828 0.618 0.350 0.121
1.000 1.000 0.998 0.988 0.945 0.833 0.617 0.322
1.000 1.000 1.000 0.998 0.989 0.954 0.851 0.624
1.000 1.000 1.000 1.000 0.999 0.994 0.972 0.893
0.9
0.000
0.000
0.000
0.000
0.000
0.002
0.013
0.070
0.264
0.651
0.95
0.000
0.000
0.000
0.000
0.000
0.000
0.001
0.012
0.086
0.401
0.99
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.004
0.096
0.05
0.463
0.829
0.964
0.995
0.999
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
Table II: Cumulative Binomial Probabilities for n = 15
p
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.206 0.035 0.005 0.000 0.000 0.000 0.000 0.000
0.549 0.167 0.035 0.005 0.000 0.000 0.000 0.000
0.816 0.398 0.127 0.027 0.004 0.000 0.000 0.000
0.944 0.648 0.297 0.091 0.018 0.002 0.000 0.000
0.987 0.836 0.515 0.217 0.059 0.009 0.001 0.000
0.998 0.939 0.722 0.403 0.151 0.034 0.004 0.000
1.000 0.982 0.869 0.610 0.304 0.095 0.015 0.001
1.000 0.996 0.950 0.787 0.500 0.213 0.050 0.004
1.000 0.999 0.985 0.905 0.696 0.390 0.131 0.018
1.000 1.000 0.996 0.966 0.849 0.597 0.278 0.061
1.000 1.000 0.999 0.991 0.941 0.783 0.485 0.164
1.000 1.000 1.000 0.998 0.982 0.909 0.703 0.352
1.000 1.000 1.000 1.000 0.996 0.973 0.873 0.602
1.000 1.000 1.000 1.000 1.000 0.995 0.965 0.833
1.000 1.000 1.000 1.000 1.000 1.000 0.995 0.965
0.9
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.002
0.013
0.056
0.184
0.451
0.794
0.95
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.001
0.005
0.036
0.171
0.537
0.99
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.010
0.140
x
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
x
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
0.01
0.818
0.983
0.999
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
0.01
0.778
0.974
0.998
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
Page 93 of 103
0.05
0.358
0.736
0.925
0.984
0.997
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
Table II: Cumulative Binomial Probabilities for n = 20
p
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.122 0.012 0.001 0.000 0.000 0.000 0.000 0.000
0.392 0.069 0.008 0.001 0.000 0.000 0.000 0.000
0.677 0.206 0.035 0.004 0.000 0.000 0.000 0.000
0.867 0.411 0.107 0.016 0.001 0.000 0.000 0.000
0.957 0.630 0.238 0.051 0.006 0.000 0.000 0.000
0.989 0.804 0.416 0.126 0.021 0.002 0.000 0.000
0.998 0.913 0.608 0.250 0.058 0.006 0.000 0.000
1.000 0.968 0.772 0.416 0.132 0.021 0.001 0.000
1.000 0.990 0.887 0.596 0.252 0.057 0.005 0.000
1.000 0.997 0.952 0.755 0.412 0.128 0.017 0.001
1.000 0.999 0.983 0.872 0.588 0.245 0.048 0.003
1.000 1.000 0.995 0.943 0.748 0.404 0.113 0.010
1.000 1.000 0.999 0.979 0.868 0.584 0.228 0.032
1.000 1.000 1.000 0.994 0.942 0.750 0.392 0.087
1.000 1.000 1.000 0.998 0.979 0.874 0.584 0.196
1.000 1.000 1.000 1.000 0.994 0.949 0.762 0.370
1.000 1.000 1.000 1.000 0.999 0.984 0.893 0.589
1.000 1.000 1.000 1.000 1.000 0.996 0.965 0.794
1.000 1.000 1.000 1.000 1.000 0.999 0.992 0.931
1.000 1.000 1.000 1.000 1.000 1.000 0.999 0.988
0.9
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.002
0.011
0.043
0.133
0.323
0.608
0.878
0.95
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.003
0.016
0.075
0.264
0.642
0.99
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.001
0.017
0.182
0.05
0.277
0.642
0.873
0.966
0.993
0.999
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
Table II: Cumulative Binomial Probabilities for n = 25
p
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.072 0.004 0.000 0.000 0.000 0.000 0.000 0.000
0.271 0.027 0.002 0.000 0.000 0.000 0.000 0.000
0.537 0.098 0.009 0.000 0.000 0.000 0.000 0.000
0.764 0.234 0.033 0.002 0.000 0.000 0.000 0.000
0.902 0.421 0.090 0.009 0.000 0.000 0.000 0.000
0.967 0.617 0.193 0.029 0.002 0.000 0.000 0.000
0.991 0.780 0.341 0.074 0.007 0.000 0.000 0.000
0.998 0.891 0.512 0.154 0.022 0.001 0.000 0.000
1.000 0.953 0.677 0.274 0.054 0.004 0.000 0.000
1.000 0.983 0.811 0.425 0.115 0.013 0.000 0.000
1.000 0.994 0.902 0.586 0.212 0.034 0.002 0.000
1.000 0.998 0.956 0.732 0.345 0.078 0.006 0.000
1.000 1.000 0.983 0.846 0.500 0.154 0.017 0.000
1.000 1.000 0.994 0.922 0.655 0.268 0.044 0.002
1.000 1.000 0.998 0.966 0.788 0.414 0.098 0.006
1.000 1.000 1.000 0.987 0.885 0.575 0.189 0.017
1.000 1.000 1.000 0.996 0.946 0.726 0.323 0.047
1.000 1.000 1.000 0.999 0.978 0.846 0.488 0.109
1.000 1.000 1.000 1.000 0.993 0.926 0.659 0.220
1.000 1.000 1.000 1.000 0.998 0.971 0.807 0.383
1.000 1.000 1.000 1.000 1.000 0.991 0.910 0.579
1.000 1.000 1.000 1.000 1.000 0.998 0.967 0.766
1.000 1.000 1.000 1.000 1.000 1.000 0.991 0.902
1.000 1.000 1.000 1.000 1.000 1.000 0.998 0.973
1.000 1.000 1.000 1.000 1.000 1.000 1.000 0.996
0.9
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.002
0.009
0.033
0.098
0.236
0.463
0.729
0.928
0.95
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.001
0.007
0.034
0.127
0.358
0.723
0.99
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.002
0.026
0.222
Page 94 of 103
Page 95 of 103
Page 96 of 103
Page 97 of 103
Page 98 of 103
Page 99 of 103
Page 100 of 103
Page 101 of 103
Page 102 of 103
Page 103 of 103