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Statistical Inference
Making decisions regarding the
population base on a sample
Decision Types
• Estimation
– Deciding on the value of an unknown parameter
• Hypothesis Testing
– Deciding a statement regarding an unknown parameter
is true of false
• All decisions will be based on the values of
statistics
Estimation
• Definitions
– An estimator of an unknown parameter is a sample
statistic used for this purpose
– An estimate is the value of the estimator after the data
is collected
• The performance of an estimator is assessed by determining
its sampling distribution and measuring its closeness to the
parameter being estimated
Examples of Estimators
The Sample Proportion
Let p = population proportion of interest
or binomial probability of success.
Let
X
no. of succeses
pˆ 

n no. of bimomial trials
= sample proportion or proportion of successes.
Then the sampling distributi on of p̂
is a normal distribution with
mean  pˆ  p
 pˆ 
p(1  p)
n
Sampling distributi on of p̂
30
25
20
15
c
10
5
0
0
0.1
0.2
0.3
 pˆ  p
0.4
0.5
0.6
0.7
0.8
0.9
1
The Sample Mean
Let x1, x2, x3, …, xn denote a sample of size n from a normal
distribution with mean  and standard deviation .
n
Let
x
x
i
i 1
n
 sample mean
Then the sampling distributi on of x
is a normal distribution with
mean  x  
x 

n
0.3
Sampling distributi on of x
0.25
population
0.2
n =5
n = 10
0.15
n = 15
c
n = 20
0.1
0.05
0
80
90
100
x  
110
120
Confidence Intervals
Estimation by Confidence Intervals
• Definition
– An (100) P% confidence interval of an unknown
parameter is a pair of sample statistics (t1 and t2)
having the following properties:
1. P[t1 < t2] = 1. That is t1 is always smaller than t2.
2. P[the unknown parameter lies between t1 and t2] = P.
•
the statistics t1 and t2 are random variables
•
Property 2. states that the probability that the
unknown parameter is bounded by the two
statistics t1 and t2 is P.
Critical values for a distribution
• The a upper critical value for a any distribution
is the point xa underneath the distribution such
that P[X > xa] = a
a
xa
Critical values for the standard Normal
distribution
P[Z > za] = a
a
za
Critical values for the standard Normal
distribution
P[Z > za] = a
Confidence Intervals for a proportion p
Let t1  pˆ  za / 2 pˆ  pˆ  za / 2
 pˆ  za / 2
pˆ 1  pˆ 
n
and t2  pˆ  za / 2 pˆ  pˆ  za / 2
 pˆ  za / 2
pˆ 1  pˆ 
n
p1  p 
n
p1  p 
n
Then t1 to t2 is a (1 – a)100% = P100%
confidence interval for p
z
Logic:
pˆ  p
 pˆ
has a Standard Normal distribution
Then P za  z  za   1  a  P


pˆ  p
and P  za / 2 
 za / 2   1  a
 pˆ


P   za / 2 pˆ  pˆ  p  za / 2 pˆ   1  a
P   za / 2 pˆ  p  pˆ  za / 2 pˆ   1  a
Hence


P pˆ  za / 2 pˆ  p  pˆ  za / 2 pˆ  1  a
Pt1  p  t2   1  a
Thus t1 to t2 is a (1 – a)100% = P100% confidence
interval for p
Example
• Suppose we are interested in determining the success rate
of a new drug for reducing Blood Pressure
• The new drug is given to n = 70 patients with abnormally
high Blood Pressure
• Of these patients to X = 63 were able to reduce the
abnormally high level of Blood Pressure
• The proportion of patients able to reduce the abnormally
high level of Blood Pressure was
X 63
pˆ 

 0.900
n 70
If P = 1 – a = 0.95 then a/2 = .025
Then
and
and za = 1.960
pˆ 1  pˆ 
t1  pˆ  za / 2
n
0.900.10
 (0.90)  (1.960)
70
 (0.90)  .0703  0.8297
pˆ 1  pˆ 
t2  pˆ  za / 2
n
0.900.10
 (0.90)  (1.960)
70
 (0.90)  .0703  0.9703
Thus a 95% confidence interval for p is 0.8297 to
0.9703
Confidence Interval for a Proportion
100P% Confidence Interval for the population proportion:
pˆ  za / 2 pˆ
 pˆ 
p1  p 

n
pˆ 1  pˆ 
n
za / 2  upper a / 2 critical point
of the standard normal distribtio n
Interpretation: For about 100P% of all randomly selected
samples from the population, the confidence interval computed in
this manner captures the population proportion.
Error Bound
For a (1 – a)% confidence level, the
approximate margin of error in a sample
proportion is
Error Bound  za
pˆ 1  pˆ 
n
Factors that Determine the Error Bound
1. The sample size, n.
When sample size increases, margin of error decreases.
2. The sample proportion, p̂ .
If the proportion is close to either 1 or 0 most individuals
have the same trait or opinion, so there
is little natural variability and the margin of error
is smaller than if the proportion is near 0.5.
3. The “multiplier” za/2.
Connected to the “(1 – a)%” level of confindence of the
Error Bound. The value of za/2 for a 95% level of
confidence is 1.96 This value is changed to change the
level of confidence.
Determination of Sample Size
In almost all research situations the
researcher is interested in the question:
How large should the sample be?
Answer:
Depends on:
• How accurate you want the answer.
Accuracy is specified by:
• Specifying the magnitude of
the error bound
• Level of confidence
Error Bound:
B  za / 2
p1  p 
 za / 2
n
pˆ 1  pˆ 
n
• If we have specified the level of confidence then the
value of za/2 will be known.
• If we have specified the magnitude of B, it will also
be known
Solving for n we get:
za2/ 2 p1  p  za2/ 2 p * 1  p *
n

2
2
B
B
Summarizing:
The sample size that will estimate p with an Error
Bound B and level of confidence P = 1 – a is:
za2/ 2 p * 1  p *
n
2
B
where:
• B is the desired Error Bound
• za is the a/2 critical value for the standard normal
distribution
• p* is some preliminary estimate of p.
If you do not have a preliminary estimate of p, use p* = 0.50
Reason
2
za / 2 p * 1  p *
n
For p* = 0.50
B2
n will take on the largest value.
3000
2500
n
2000
1500
1000
500
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
p*
Thus using p* = 0.50, n may be larger than required
if p is not 0.50. but will give the desired accuracy or
better for all values of p.
Example
• Suppose that I want to conduct a survey and want to
estimate p = proportion of voters who favour a downtown
location for a casino:
I know that the approximate value of p is
• p* = 0.50. This is also a good choice for p if one has no
preliminary estimate of its value.
• I want the survey to estimate p with an error bound B = 0.01
(1 percentage point)
• I want the level of confidence to be 95% (i.e. a = 0.05 and
za = z0.05 = 1.960
Then
2

1.960 0.500.50
n
 9604
2
0.01
Confidence Intervals for the mean
of a Normal Population, 
Let
and
t1  x  za / 2 x  x  za / 2

n
t2  x  za / 2 x  x  za / 2

n
Then t1 to t2 is a (1 – a)100% = P100%
confidence interval for 
Logic:
z
x
has a Standard Normal distribution
x
Then P za  z  za   1  a  P


x
and P  za / 2 
 za / 2   1  a
x


Hence
Px  za / 2 x    x  za / 2 x   1  a
Pt1    t2   1  a
Thus t1 to t2 is a (1 – a)100% = P100% confidence
interval for p
Example
• Suppose we are interested average Bone Mass Density
(BMD) for women aged 70-75
• A sample n = 100 women aged 70-75 are selected and
BMD is measured for eahc individual in the sample.
• The average BMD for these individuals is:
x  25.63
• The standard deviation (s) of BMD for these individuals
is:
s  7.82  
If P = 1 – a = 0.95 then a/2 = .025
Then
t1  x  za / 2

 x  za / 2
and za = 1.960
s
n
n
7.82
 25.63  1.960
 25.63  1.53  24.10
100
and
t2  x  za / 2

 x  za / 2
s
n
n
7.82
 25.63  1.960
 25.63  1.53  27.16
100
Thus a 95% confidence interval for  is 24.10 to 27.16
Determination of Sample Size
Again a question to be asked:
How large should the sample be?
Answer:
Depends on:
• How accurate you want the answer.
Accuracy is specified by:
• Specifying the magnitude of
the error bound
• Level of confidence
Error Bound:
B  za / 2

n
• If we have specified the level of confidence then the
value of za/2 will be known.
• If we have specified the magnitude of B, it will also
be known
Solving for n we get:
z 
z s *
n

2
B
B2
2
a/2
2
2
a/2
2
Summarizing:
The sample size that will estimate  with an Error
Bound B and level of confidence P = 1 – a is:
z 
z s *
n

2
2
B
B
2
a/2
2
2
a/2
2
where:
• B is the desired Error Bound
• za is the a/2 critical value for the standard normal
distribution
• s* is some preliminary estimate of s.
Notes:
z 
z s *
n

2
2
B
B
2
a/2
2
2
a/2
2
• n increases as B, the desired Error Bound, decreases
– Larger sample size required for higher level of accuracy
• n increases as the level of confidence, (1 – a), increases
– za increases as a/2 becomes closer to zero.
– Larger sample size required for higher level of
confidence
• n increases as the standard deviation, , of the population
increases.
– If the population is more variable then a larger sample
size required
Summary:
The sample size n depends on:
• Desired level of accuracy
• Desired level of confidence
• Variability of the population
Example
• Suppose that one is interested in estimating the average
number of grams of fat (m) in one kilogram of lean beef
hamburger :
This will be estimated by:
• randomly selecting one kilogram samples, then
• Measuring the fat content for each sample.
• Preliminary estimates of  and  indicate:
– that  and  are approximately 220 and 40 respectively.
• I want the study to estimate  with an error bound 5
and
• a level of confidence to be 95% (i.e. a = 0.05 and za =
z0.05 = 1.960)
Solution

1.960 40
n
2
5
2
2
 245.9  246
Hence n = 246 one kilogram samples are
required to estimate  within B = 5 gms with a
95% level of confidence.
Confidence Intervals
Confidence Interval for a Proportion
pˆ  za / 2 pˆ
 pˆ 
p1  p 

n
pˆ 1  pˆ 
n
za / 2  upper a / 2 critical point
of the standard normal distribtio n
B  za / 2 pˆ  za / 2 pˆ
p 1  p 
n
 za / 2 pˆ
 Error Bound
pˆ 1  pˆ 
n
Determination of Sample Size
The sample size that will estimate p with an Error Bound B
and level of confidence P = 1 – a is:
za2/ 2 p * 1  p *
n
B2
where:
• B is the desired Error Bound
• za is the a/2 critical value for the standard normal
distribution
• p* is some preliminary estimate of p.
Confidence Intervals for the mean
of a Normal Population, 
x  za / 2 x
or x  za / 2
or x  za / 2

n
s
n
x  sample mean
za / 2  upper a / 2 critical point
of the standard normal distribtio n
s  sample standard deviation  
Determination of Sample Size
The sample size that will estimate  with an Error Bound B
and level of confidence P = 1 – a is:
z 
z s *
n

2
2
B
B
2
a/2
2
2
a/2
2
where:
• B is the desired Error Bound
• za is the a/2 critical value for the standard normal
distribution
• s* is some preliminary estimate of s.
Hypothesis Testing
An important area of statistical
inference
Definition
Hypothesis (H)
– Statement about the parameters of the population
• In hypothesis testing there are two hypotheses
of interest.
– The null hypothesis (H0)
– The alternative hypothesis (HA)
Either
– null hypothesis (H0) is true or
– the alternative hypothesis (HA) is true.
But not both
We say that are mutually exclusive and
exhaustive.
One has to make a decision
– to either to accept null hypothesis
(equivalent to rejecting HA)
or
– to reject null hypothesis (equivalent to
accepting HA)
There are two possible errors that can be
made.
1. Rejecting the null hypothesis when it is
true. (type I error)
2. accepting the null hypothesis when it is
false (type II error)
An analogy – a jury trial
The two possible decisions are
– Declare the accused innocent.
– Declare the accused guilty.
The null hypothesis (H0) – the accused is
innocent
The alternative hypothesis (HA) – the accused
is guilty
The two possible errors that can be made:
– Declaring an innocent person guilty.
(type I error)
– Declaring a guilty person innocent.
(type II error)
Note: in this case one type of error may be
considered more serious
Decision Table showing types of Error
H0 is True
H0 is False
Accept H0
Correct
Decision
Type II
Error
Reject H0
Type I
Error
Correct
Decision
To define a statistical Test we
1. Choose a statistic (called the test statistic)
2. Divide the range of possible values for the
test statistic into two parts
• The Acceptance Region
• The Critical Region
To perform a statistical Test we
1. Collect the data.
2. Compute the value of the test statistic.
3. Make the Decision:
• If the value of the test statistic is in
the Acceptance Region we decide to
accept H0 .
• If the value of the test statistic is in
the Critical Region we decide to
reject H0 .
Example
We are interested in determining if a coin is fair.
i.e.
H0 : p = probability of tossing a head = ½.
To test this we will toss the coin n = 10 times.
The test statistic is x = the number of heads.
This statistic will have a binomial distribution with
p = ½ and n = 10 if the null hypothesis is true.
Sampling distribution of x when H0 is true
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
6
7
8
9
10
Note
We would expect the test statistic x to be around 5 if
H0 : p = ½ is true.
Acceptance Region = {3, 4, 5, 6, 7}.
Critical Region = {0, 1, 2, 8, 9, 10}.
The reason for the choice of the Acceptance region:
Contains the values that we would expect for x if the
null hypothesis is true.
Definitions: For any statistical testing
procedure define
1. a = P[Rejecting the null hypothesis when it
is true] = P[ type I error]
. b = P[accepting the null hypothesis when it
is false] = P[ type II error]
In the last example
1. a = P[ type I error] = p(0) + p(1) + p(2) + p(8)
+ p(9) + p(10) = 0.109, where p(x) are
binomial probabilities with p = ½ and n = 10 .
. b = P[ type II error] = p(3) + p(4) + p(5) + p(6)
+ p(7), where p(x) are binomial probabilities
with p (not equal to ½) and n = 10. Note: these
will depend on the value of p.
Table: Probability of a Type II
error, b vs. p
p
0.1
0.2
0.3
0.4
0.6
0.7
0.8
0.9
b
0.070
0.322
0.616
0.820
0.820
0.616
0.322
0.070
Note: the magnitude of b
increases as p gets
closer to ½.
Comments:
1. You can control a = P[ type I error] and b = P[
type II error] by widening or narrowing the
acceptance region. .
2. Widening the acceptance region decreases a
= P[ type I error] but increases b = P[ type II
error].
3. Narrowing the acceptance region increases a
= P[ type I error] but decreases b = P[ type II
error].
Example – Widening the Acceptance
Region
1. Suppose the Acceptance Region includes
in addition to its previous values 2 and 8
then a = P[ type I error] = p(0) + p(1) +
p(9) + p(10) = 0.021, where again p(x)
are binomial probabilities with p = ½ and
n = 10 .
. b = P[ type II error] = p(2) + p(3) + p(4) +
p(5) + p(6) + p(7) + p(8). Tabled values of
are given on the next page.
Table: Probability of a Type II
error, b vs. p
p
0.1
0.2
0.3
0.4
0.6
0.7
0.8
0.9
b
0.264
0.624
0.851
0.952
0.952
0.851
0.624
0.264
Note: Compare these
values with the
previous definition
of the Acceptance
Region. They have
increased,
Example – Narrowing the Acceptance
Region
1. Suppose the original Acceptance Region
excludes the values 3 and 7. That is the
Acceptance Region is {4,5,6}. Then a = P[ type
I error] = p(0) + p(1) + p(2) + p(3) + p(7) +
p(8) +p(9) + p(10) = 0.344.
. b = P[ type II error] = p(4) + p(5) + p(6) .
Tabled values of are given on the next page.
Table: Probability of a Type II
error, b vs. p
p
0.1
0.2
0.3
0.4
0.6
0.7
0.8
0.9
b
0.013
0.120
0.340
0.563
0.563
0.340
0.120
0.013
Note: Compare these
values with the
otiginal definition of
the Acceptance
Region. They have
decreased,
Acceptance
Region
Acceptance
Region
Acceptance
Region
{2,3,4,5,6,7,8}.
{3,4,5,6,7}.
{4,5,6}.
a = 0.021
p
0.1
0.2
0.3
0.4
0.6
0.7
0.8
0.9
b
0.264
0.624
0.851
0.952
0.952
0.851
0.624
0.264
a = 0.109
p
0.1
0.2
0.3
0.4
0.6
0.7
0.8
0.9
b
0.070
0.322
0.616
0.820
0.820
0.616
0.322
0.070
a = 0.344
p
0.1
0.2
0.3
0.4
0.6
0.7
0.8
0.9
b
0.013
0.120
0.340
0.563
0.563
0.340
0.120
0.013
The Approach in Statistical Testing is:
• Set up the Acceptance Region so that a is
close to some predetermine value (the usual
values are 0.05 or 0.01)
• The predetermine value of a (0.05 or 0.01)
is called the significance level of the test.
• The significance level of the test is a =
P[test makes a type I error]
The z-test for Proportions
Testing the probability of success in a
binomial experiment
Situation
• A success-failure experiment has been
repeated n times
• The probability of success p is unknown. We
want to test
– H0: p = p0 (some specified value of p)
Against
– HA:
p  p0
The Data
• The success-failure experiment has been repeated
n times
• The number of successes x is observed.
x
pˆ   the poportion of successes
n
• Obviously if this proportion is close to p0 the Null
Hypothesis should be accepted otherwise the null
Hypothesis should be rejected.
The Test Statistic
• To decide to accept or reject the Null Hypothesis
(H0) we will use the test statistic
z
pˆ  p0
 pˆ

pˆ  p0
p0 1  p0 
n
• If H0 is true we should expect the test statistic z to
be close to zero.
• If H0 is true we should expect the test statistic z to
have a standard normal distribution.
• If HA is true we should expect the test statistic z to
be different from zero.
The sampling distribution of z when H0 is true:
The Standard Normal distribution
0
Reject H0
Accept H0
z
Reject H0
The Acceptance region:
a/2
a/2
Reject H0
 za / 2
0
za / 2
Accept H0
z
Reject H0
PAccept H 0 when true   P za / 2  z  za / 2   1  a
PReject H 0 when true   Pz   za / 2 or z  za / 2   a
• Acceptance Region
 za / 2  z  za / 2
– Accept H0 if:
• Critical Region
– Reject H0 if:
z   za / 2 or z  za / 2
• With this Choice
PType I Error   PReject H 0 when true 
 Pz   za / 2 or z  za / 2   a
Summary
To Test for a binomial probability p
H0: p = p0 (some specified value of p)
Against
HA: p  p0
we
1. Decide on a = P[Type I Error] = the
significance level of the test (usual choices
0.05 or 0.01)
2. Collect the data
3. Compute the test statistic
z
pˆ  p0
 pˆ

pˆ  p0
p0 1  p0 
n
4. Make the Decision
• Accept H0 if:  za / 2  z  za / 2
• Reject H0 if: z   za / 2 or z  za / 2
Example
• In the last election the proportion of the
voters who voted for the Liberal party was
0.08 (8 %)
• The party is interested in determining if that
percentage has changed
• A sample of n = 800 voters are surveyed
We want to test
– H0: p = 0.08 (8%)
Against
– HA: p  0.08 (8%)
Summary
1. Decide on a = P[Type I Error] = the
significance level of the test
Choose (a = 0.05)
2. Collect the data
• The number in the sample that support the
liberal party is x = 92
x 92
pˆ  
 0.115 (11.5%)
n 800
3. Compute the test statistic
pˆ  p0
pˆ  p0
z

 pˆ
p0 1  p0 
n
0.115  0.80

 3.649
0.801  0.80
800
4. Make the Decision za / 2  z0.025  1.960
• Accept H0 if: 1.960  z  1.960
• Reject H0 if: z  1.960 or z  1.960
Since the test statistic is in the Critical
region we decide to Reject H0
Conclude that H0: p = 0.08 (8%) is false
There is a significant difference (a = 5%)
in the proportion of the voters supporting
the liberal party in this election than in the
last election
The one tailed z-test
• A success-failure experiment has been
repeated n times
• The probability of success p is unknown.
We want to test
– H0: p  p0 (some specified value of p)
Against
– HA: p  p0
• The alternative hypothesis is in this case
called a one-sided alternative
The Test Statistic
• To decide to accept or reject the Null Hypothesis
(H0) we will use the test statistic
z
pˆ  p0
 pˆ

pˆ  p0
p0 1  p0 
n
• If H0 is true we should expect the test statistic z to
be close to zero or negative
• If p = p0 we should expect the test statistic z to
have a standard normal distribution.
• If HA is true we should expect the test statistic z to
be a positive number.
The sampling distribution of z when p = p0 :
The Standard Normal distribution
0
Accept H0
z
Reject H0
The Acceptance and Critical region:
a
0
Accept H0
za
z
Reject H0
PAccept H 0 when true   Pz  za   1  a
PReject H 0 when true   P z  za   a
• Acceptance Region
– Accept H0 if:
z  za
• Critical Region
– Reject H0 if:
z  za
• The Critical Region is called one-tailed
• With this Choice
PType I Error   PReject H 0 when true 
 Pz  za   a
Example
• A new surgical procedure is developed for
correcting heart defects infants before the
age of one month.
• Previously the procedure was used on
infants that were older than one month and
the success rate was 91%
• A study is conducted to determine if the
success rate of the new procedure is greater
than 91% (n = 200)
We want to test
– H0: p  0.91 (91%)
Against
– HA: p  0.91 (91%)
p  the success rate of the new procedure
Summary
1. Decide on a = P[Type I Error] = the
significance level of the test
Choose (a = 0.05)
2. Collect the data
• The number of successful operations in the
sample of 200 cases is x = 187
x 187
pˆ  
 0.935 (93.5%)
n 200
3. Compute the test statistic
pˆ  p0
pˆ  p0
z

 pˆ
p0 1  p0 
n
0.935  0.91

 1.235
0.911  0.91
200
4. Make the Decision za  z0.05  1.645
• Accept H0 if: z  1.645
• Reject H0 if:
z  1.645
Since the test statistic is in the Acceptance
region we decide to Accept H0
Conclude that H0: p  0.91 (91%) is true
There is a no significant (a = 5%) increase
in the success rate of the new procedure
over the older procedure
Comments
• When the decision is made to accept H0 is
made it should not be conclude that we have
proven H0.
• This is because when setting up the test we
have not controlled b = P[type II error] =
P[accepting H0 when H0 is FALSE]
• Whenever H0 is accepted there is a
possibility that a type II error has been
made.
In the last example
The conclusion that there is a no significant
(a = 5%) increase in the success rate of the
new procedure over the older procedure
should be interpreted:
We have been unable to proof that the new
procedure is better than the old procedure
An analogy – a jury trial
The two possible decisions are
– Declare the accused innocent.
– Declare the accused guilty.
The null hypothesis (H0) – the accused is
innocent
The alternative hypothesis (HA) – the accused
is guilty
The two possible errors that can be made:
– Declaring an innocent person guilty.
(type I error)
– Declaring a guilty person innocent.
(type II error)
Note: in this case one type of error may be
considered more serious
Requiring all 12 jurors to support a guilty verdict :
– Ensures that the probability of a type I error
(Declaring an innocent person guilty) is
small.
– However the probability of a type II error
(Declaring an guilty person innocent) could
be large.
Hence: When decision of innocence is made:
– It is not concluded that innocence has
been proven
but that
– we have been unable to disprove
innocence
The z-test for the Mean of a
Normal Population
We want to test, , denote the mean
of a normal population
Situation
• A success-failure experiment has been
repeated n times
• The probability of success p is unknown. We
want to test
– H0: p = p0 (some specified value of p)
Against
– HA:
p  p0
The Data
• Let x1, x2, x3 , … , xn denote a sample from a
normal population with mean  and standard
deviation .
• Let
n
x
x
i 1
n
i
 the sample mean
• we want to test if the mean, , is equal to some
given value 0.
• Obviously if the sample mean is close to 0 the
Null Hypothesis should be accepted otherwise the
null Hypothesis should be rejected.
The Test Statistic
• To decide to accept or reject the Null Hypothesis
(H0) we will use the test statistic
z
x  0
x

x  0

 n
x  0

x  0
 n
s
n
• If H0 is true we should expect the test statistic z to
be close to zero.
• If H0 is true we should expect the test statistic z to
have a standard normal distribution.
• If HA is true we should expect the test statistic z to
be different from zero.
The sampling distribution of z when H0 is true:
The Standard Normal distribution
0
Reject H0
Accept H0
z
Reject H0
The Acceptance region:
a/2
a/2
Reject H0
 za / 2
0
za / 2
Accept H0
z
Reject H0
PAccept H 0 when true   P za / 2  z  za / 2   1  a
PReject H 0 when true   Pz   za / 2 or z  za / 2   a
• Acceptance Region
 za / 2  z  za / 2
– Accept H0 if:
• Critical Region
– Reject H0 if:
z   za / 2 or z  za / 2
• With this Choice
PType I Error   PReject H 0 when true 
 Pz   za / 2 or z  za / 2   a
Summary
To Test for a binomial probability p
H0:  = 0 (some specified value of p)
Against
HA :    0
1. Decide on a = P[Type I Error] = the
significance level of the test (usual choices
0.05 or 0.01)
2. Collect the data
3. Compute the test statistic
z n
x  0

x  0
 n
s
4. Make the Decision
• Accept H0 if:  za / 2  z  za / 2
• Reject H0 if: z   za / 2 or z  za / 2
Example
A manufacturer Glucosamine capsules claims
that each capsule contains on the average:
• 500 mg of glucosamine
To test this claim n = 40 capsules were
selected and amount of glucosamine (X)
measured in each capsule.
Summary statistics:
x  496.3 and s  8.5
We want to test:
H 0 :   500
Manufacturers claim is correct
against
H A :   500
Manufacturers claim is not
correct
The Test Statistic
z
x  0
x

x  0

 n
x  0

x  0
 n
s
n
496.3  500
 40
8.5
 2.75
The Critical Region and
Acceptance Region
Using a = 0.05
za/2 = z0.025 = 1.960
We accept H0 if
-1.960 ≤ z ≤ 1.960
reject H0 if
z < -1.960 or z > 1.960
The Decision
Since
z= -2.75 < -1.960
We reject H0
Conclude: the manufacturers’s claim is
incorrect: