Transcript One and Two tailed tests

Hypothesis Testing
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 .
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 Test Statistic
z
pˆ  p0
 pˆ

pˆ  p0
p0 1  p0 
n
The Acceptance and Critical Region
• Accept H0 if:  z / 2  z  z / 2
• Reject H0 if: z   z / 2 or z  z / 2
Two-tailed critical region
One-tailed critical regions
These are used when the alternative hypothesis (HA) is
one-sided
i.e.
H 0 : p  p0 and H A : p  p0
The Acceptance and Critical Region
• Accept H0 if: z  z
• Reject H0 if: z  z
or if
H 0 : p  p0 and H A : p  p0
• Accept H0 if:  z  z
• Reject H0 if: z   z
One-tailed critical regions
H 0 : p  p0 and H A : p  p0
The Acceptance and Critical Region
Accept H0 if: z  z , Reject H0 if: z  z
One-tailed critical regions
H 0 : p  p0 and H A : p  p0
The Acceptance and Critical Region
Accept H0 if: z   z, Reject H0 if: z   z
Comments
• Whether you use a one-tailed or a two-tailed
tests is determined by the choice of the
alternative hypothesis HA
• The alternative hypothesis, HA, is usually the
research hypothesis. The hypothesis that the
researcher is trying to “prove”.
Examples
1. A person wants to determine if a coin should
be accepted as being fair. Let p be the
probability that a head is tossed.
H 0 : p  12 vs H A : p 
1
2
One is trying to determine if there is a difference
(positive or negative) with the fair value of p.
2. A researcher is interested in determining if a
new procedure is an improvement over the
old procedure. The probability of success for
the old procedure is p0 (known). The
probability of success for the new procedure
is p (unknown) .
H 0 : p  p0 vs H A : p  p0
One is trying to determine if the new procedure
is better (i.e. p > p0) .
2. A researcher is interested in determining if a
new procedure is no longer worth
considering. The probability of success for the
old procedure is p0 (known). The probability
of success for the new procedure is p
(unknown) .
H 0 : p  p0 vs H A : p  p0
One is trying to determine if the new procedure
is definitely worse than the one presently being
used (i.e. p < p0) .
The z-test for the Mean of a
Normal Population
We want to test, m, denote the mean
of a normal population
The Situation
• Let x1, x2, x3 , … , xn denote a sample from a
normal population with mean m and standard
deviation .
• Let
n
x
x
i 1
n
i
 the sample mean
• we want to test if the mean, m, is equal to some
given value m0.
• Obviously if the sample mean is close to m0 the
Null Hypothesis should be accepted otherwise the
null Hypothesis should be rejected.
The Test Statistic
z
x  m0
x

x  m0

n
x  m0

n

x  m0
n
s

The Acceptance and Critical Region
This depends on H0 and HA
Two-tailed critical region
H 0 : m  m0 and H A : m  m0
• Accept H0 if:  z / 2  z  z / 2
• Reject H0 if: z   z / 2 or z  z / 2
One-tailed critical regions
H 0 : m  m0 and H A : m  m0
H 0 : m  m0 and H A : m  m0
• Accept H0 if: z  z
• Accept H0 if: z   z
• Reject H0 if: z  z
• Reject H0 if: z   z
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 : m  
Manufacturers claim is correct
against
H A : m  
Manufacturers claim is not
correct
The Test Statistic
z
x  m0
x

x  m0

 n
x  m0

x  m0
 n
s
n
496.3  500
 40
8.5
 2.75
The Critical Region and
Acceptance Region
Using  = 0.05
z/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:
“Students” t-test
Recall: The z-test for means
The Test Statistic
z
x  m0
x
x  m0
x  m0



s
n
n
Comments
• The sampling distribution of this statistic is the
standard Normal distribution
• The replacement of  by s leaves this
distribution unchanged only the sample size n
is large.
For small sample sizes:
The sampling distribution of
x  m0
t
s
n
Is called “students” t distribution with
n –1 degrees of freedom
Properties of Student’s t
distribution
• Similar to Standard normal distribution
– Symmetric
– unimodal
– Centred at zero
• Larger spread about zero.
– The reason for this is the increased variability introduced
by replacing  by s.
• As the sample size increases (degrees of freedom
increases) the t distribution approaches the standard
normal distribution
0.4
0.3
0.2
0.1
-4
-2
2
4
t distribution
standard normal distribution
The Situation
• Let x1, x2, x3 , … , xn denote a sample from a
normal population with mean m and standard
deviation . Both m and  are unknown.
• Let
n
x
x
i 1
n
n
s
i
 the sample mean
 x  x 
i 1
2
i
n 1
 the sample standard deviation
• we want to test if the mean, m, is equal to some
given value m0.
The Test Statistic
x  m0
t
s
n
The sampling distribution of the test
statistic is the t distribution with n-1
degrees of freedom
The Alternative
Hypothesis HA
The Critical Region
H A : m  m0
t  t / 2 or t  t / 2
H A : m  m0
t  t
H A : m  m0
t  t
t and t/2 are critical values under the t
distribution with n – 1 degrees of
freedom
Critical values for the t-distribution
 or /2
0
t
t / 2 or t
Critical values for the t-distribution are
provided in tables. A link to these tables are
given with today’s lecture
Look up 
Look up df
Note: the values
tabled for df = ∞
are the same
values for the
standard normal
distribution
Example
• Let x1, x2, x3 , x4, x5, x6 denote weight loss
from a new diet for n = 6 cases.
• Assume that x1, x2, x3 , x4, x5, x6 is a sample
from a normal population with mean m and
standard deviation . Both m and  are
unknown.
• we want to test:
H 0 : m  0 New diet is not effective
versus
HA : m  0
New diet is effective
The Test Statistic
x  m0
t
s
n
The Critical region:
Reject if
t  t
The Data
1
2.0
2
1.0
3
1.4
4
-1.8
5
0.9
6
2.3
The summary statistics:
x  0.96667 and s  1.462418
The Test Statistic
x  m0
0.96667  0
t

 1.619
1.462418
s
n
6
The Critical Region (using  = 0.05)
Reject if
t  t0.05  2.015 for 5 d.f.
Conclusion: Accept H0:
Confidence Intervals
Confidence Intervals for the mean of a Normal
Population, m, using the Standard Normal
distribution
x  z / 2

n
Confidence Intervals for the mean of a Normal
Population, m, using the t distribution
x  t / 2
s
n
The Data
1
2.0
2
1.0
3
1.4
4
-1.8
5
0.9
6
2.3
The summary statistics:
x  0.96667 and s  1.462418
Example
• Let x1, x2, x3 , x4, x5, x6 denote weight loss
from a new diet for n = 6 cases.
The Data:
1
2.0
2
1.0
3
1.4
4
-1.8
5
0.9
6
2.3
The summary statistics:
x  0.96667 and s  1.462418
Confidence Intervals (use  = 0.05)
x  t0.025
s
n
1.462418
0.96667  2.571
6
0.96667 1.535
 0.57 to 2.50
Comparing Populations
Proportions and means
Sums, Differences, Combinations of R.V.’s
A linear combination of random variables, X, Y, . . .
is a combination of the form:
L = aX + bY + …
where a, b, etc. are numbers – positive or negative.
Most common:
Sum = X + Y
Difference = X – Y
Simple Linear combination of X, bX + a
Means of Linear Combinations
If
L = aX + bY + …
The mean of L is:
Mean(L) = a Mean(X) + b Mean(Y) + …
Most common:
Mean( X + Y) = Mean(X) + Mean(Y)
Mean(X – Y) = Mean(X) – Mean(Y)
Mean(bX + a) = bMean(X) + a
Variances of Linear Combinations
If X, Y, . . . are independent random variables and
L = aX + bY + … then
Variance(L) = a2 Variance(X) + b2 Variance(Y) + …
Most common:
Variance( X + Y) = Variance(X) + Variance(Y)
Variance(X – Y) = Variance(X) + Variance(Y)
Variance(bX + a) = b2Variance(X)
Combining Independent Normal Random Variables
If X, Y, . . . are independent normal random variables,
then L = aX + bY + … is normally distributed.
In particular:
X + Y is normal with mean m X  mY
standard deviation
 X2   Y2
X – Y is normal with mean m X  mY
standard deviation
 X2   Y2
Comparing proportions
Situation
• We have two populations (1 and 2)
• Let p1 denote the probability (proportion) of
“success” in population 1.
• Let p2 denote the probability (proportion) of
“success” in population 2.
• Objective is to compare the two population
proportions
We want to test either:
1. H 0 : p1  p2 vs H A : p1  p2
or
2. H 0 : p1  p2 vs H A : p1  p2
or
3. H 0 : p1  p2 vs H A : p1  p2
The test statistic:
z
pˆ1  pˆ 2
 pˆ  pˆ
1
2

pˆ1  pˆ 2
pˆ1 1  pˆ1  pˆ1 1  pˆ1 

n1
n1
Where:
A sample of n1 is selected from population 1
resulting in x1 successes
A sample of n2 is selected from population 2
resulting in x2 successes
x1
pˆ1 
n1
and
x2
pˆ 2 
n2
 pˆ 
Logic:
1
p1 1  p1 
n1
 pˆ  pˆ    
1
2
pˆ1
2

 pˆ 
2
p2 1  p2 
n1
2
pˆ 2
p1 1  p1  p2 1  p2 

n1
n2
1 1
 p1  p    if p1  p2  p
 n1 n2 
1 1
 pˆ 1  pˆ   
 n1 n2 
The Alternative
Hypothesis HA
The Critical Region
H A : p1  p2
z   z / 2 or z  z / 2
H A : p1  p2
z  z
H A : p1  p2
z   z
Example
• In a national study to determine if there was an
increase in mortality due to pipe smoking, a
random sample of n1 = 1067 male nonsmoking
pensioners were observed for a five-year period.
• In addition a sample of n2 = 402 male pensioners
who had smoked a pipe for more than six years
were observed for the same five-year period.
• At the end of the five-year period, x1 = 117 of the
nonsmoking pensioners had died while x2 = 54 of
the pipe-smoking pensioners had died.
• Is there a the mortality rate for pipe smokers
higher than that for non-smokers
We want to test:
H 0 : p1  p2 vs H A : p1  p2
The test statistic:
z
pˆ1  pˆ 2
 pˆ  pˆ
1
2

pˆ1  pˆ 2
1 1
pˆ 1  pˆ   
 n1 n2 
Note:
x1
117
pˆ1 

 0.1097
n1 1067
x2
54
pˆ 2 

 0.1343
n2 402
x1  x2
117  54
pˆ 

n1  n2 1067  402
171

 0.1164
1469
The test statistic:
z 

pˆ1  pˆ 2
1 1
pˆ 1  pˆ   
 n1 n2 
0.1097  .1343
1 
 1
0.11641  0.1164 


 1067 402 
 1.315
We reject H0 if:
z  z  z0.05  1.645
Not true hence we accept H0.
Conclusion: There is not a significant ( =
0.05) increase in the mortality rate due to
pipe-smoking
Estimating a difference proportions using
confidence intervals
Situation
• We have two populations (1 and 2)
• Let p1 denote the probability (proportion) of
“success” in population 1.
• Let p2 denote the probability (proportion) of
“success” in population 2.
• Objective is to estimate the difference in the
two population proportions d = p1 – p2.
Confidence Interval for d
100P% = 100(1 – ) % :
= p1 – p2
pˆ1  pˆ 2  z / 2  pˆ1  pˆ 2
pˆ1  pˆ 2  z / 2
pˆ1 1  pˆ1  pˆ 2 1  pˆ 2 

n1
n2
Example
• Estimating the increase in the mortality rate
for pipe smokers higher over that for nonsmokers d = p2 – p1
pˆ1 1  pˆ1  pˆ 2 1  pˆ 2 
pˆ 2  pˆ1  z / 2

n1
n2
0.10971  0.1097 0.13431  0.1343
0.1343  0.1097  1.960

1067
0.0247  0.0382
 0.0136 to 0.0629
 1.36% to 6.29%
402
Comparing Means
Situation
• We have two normal populations (1 and 2)
• Let m1 and 1 denote the mean and standard
deviation of population 1.
• Let m2 and 2 denote the mean and standard
deviation of population 1.
• Let x1, x2, x3 , … , xn denote a sample from a
normal population 1.
• Let y1, y2, y3 , … , ym denote a sample from a
normal population 2.
• Objective is to compare the two population means
We want to test either:
1. H 0 : m1  m2 vs H A : m1  m2
or
2. H 0 : m1  m2 vs H A : m1  m2
or
3. H 0 : m1  m2 vs H A : m1  m2
Consider the test statistic:
z

xy
 xy
xy

 
2
x
xy

2
1
n


2
2
m

2
y
xy
2
x
2
y
s
s

n m
H 0 : m1  m2 is true
If:
z
xy

2
1
n


2
2
m

xy
2
x
2
y
s
s

n m
• will have a standard Normal distribution
• This will also be true for the approximation
(obtained by replacing 1 by sx and 2 by sy) if
the sample sizes n and m are large (greater than
30)
Note:
n
n
x
x
i 1
i
n
sx 
y
i 1
m
i 1
i
n 1
n
n
 yi
 x  x 
2
sy 
y
i 1
 y
2
i
m 1
The Alternative
Hypothesis HA
The Critical Region
H A : m1  m2
z   z / 2 or z  z / 2
H A : m1  m2
z  z
H A : m1  m2
z   z
Example
• A study was interested in determining if an
exercise program had some effect on reduction of
Blood Pressure in subjects with abnormally high
blood pressure.
• For this purpose a sample of n = 500 patients with
abnormally high blood pressure were required to
adhere to the exercise regime.
• A second sample m = 400 of patients with
abnormally high blood pressure were not required
to adhere to the exercise regime.
• After a period of one year the reduction in blood
pressure was measured for each patient in the
study.
We want to test:
H 0 : m1  m2
The exercize group did not have a higher
average reduction in blood pressure
vs
H A : m1  m2
The exercize group did have a higher
average reduction in blood pressure
The test statistic:
z
xy

 xy
xy

 
2
x
xy

2
1
n


2
2
m

2
y
xy
2
x
2
y
s
s

n m
Suppose the data has been collected and:
n
n
x
x
i 1
i
n
 10.67
sx 
 x  x 
y
i 1
m
i
i 1
n 1
n
n
 yi
2
 7.83
sy 
y
i 1
i
 3.895
 y
m 1
2
 4.224
The test statistic:
z
xy
2
x
2
y
s
s

n m

10.67  7.83
3.895
2
500

4.224 

2.84

 10.4
0.273765
2
400
We reject H0 if:
z  z  z0.05  1.645
True hence we reject H0.
Conclusion: There is a significant ( = 0.05)
effect due to the exercise regime on the
reduction in Blood pressure
Estimating a difference means using
confidence intervals
Situation
• We have two populations (1 and 2)
• Let m1 denote the mean of population 1.
• Let m2 denote the mean of population 2.
• Objective is to estimate the difference in the
two population proportions d = m1 – m2.
Confidence Interval for
d = m1 – m2
mˆ1  mˆ 2  z / 2  mˆ  mˆ
1
x  y  z / 2
2
x
2
2
y
s
s

n m
Example
• Estimating the increase in the average
reduction in Blood pressure due to the
excercize regime d = m1 – m2
x  y  z / 2
2
x
2
y
s
s

n m

3.895
10.67  7.83  1.960 
2
500
2.84  1.96(.273765)
2.84  0.537
2.303 to 3.337

4.224 

2
400
Comparing Means – small samples
Situation
• We have two normal populations (1 and 2)
• Let m1 and 1 denote the mean and standard
deviation of population 1.
• Let m2 and 2 denote the mean and standard
deviation of population 1.
• Let x1, x2, x3 , … , xn denote a sample from a
normal population 1.
• Let y1, y2, y3 , … , ym denote a sample from a
normal population 2.
• Objective is to compare the two population means
We want to test either:
1. H 0 : m1  m2 vs H A : m1  m2
or
2. H 0 : m1  m2 vs H A : m1  m2
or
3. H 0 : m1  m2 vs H A : m1  m2
Consider the test statistic:
z

xy
 xy
xy

 
2
x
xy

2
1
n


2
2
m

2
y
xy
2
x
2
y
s
s

n m
If the sample sizes (m and n) are large the
statistic
t
xy
2
x
2
y
s
s

n m
will have approximately a standard
normal distribution
This will not be the case if sample
sizes (m and n) are small
The t test – for comparing means –
small samples
Situation
• We have two normal populations (1 and 2)
• Let m1 and  denote the mean and standard
deviation of population 1.
• Let m2 and  denote the mean and standard
deviation of population 1.
• Note: we assume that the standard deviation
for each population is the same.
1 = 2 = 
Let
n
n
x
x
i 1
i
n
sx 
y
i 1
m
i 1
i
n 1
n
n
 yi
 x  x 
2
sy 
y
i 1
 y
2
i
m 1
The pooled estimate of .
Note: both sx and sy are estimators of .
These can be combined to form a single
estimator of , sPooled.
sPooled 
n  1sx2  m  1s 2y
nm2
The test statistic:
xy
t
s
2
Pooled
n

s
2
Pooled
m
xy

1 1
sPooled

n m
If m1 = m2 this statistic has a t distribution
with n + m –2 degrees of freedom
The Alternative
Hypothesis HA
The Critical Region
H A : m1  m2
t  t / 2 or t  t / 2
H A : m1  m2
t  t
H A : m1  m2
t  t
t / 2 and t
are critical points under the t distribution with
degrees of freedom n + m –2.
Example
• A study was interested in determining if
administration of a drug reduces cancerous
tumor size.
• For this purpose n +m = 9 test animals are
implanted with a cancerous tumor.
• n = 3 are selected at random and
administered the drug.
• The remaining m = 6 are left untreated.
• Final tumour sizes are measured at the end
of the test period
We want to test:
H 0 : m1  m2
The treated group did not have a lower
average final tumour size.
vs
H A : m1  m2
The exercize group did have a lower
average final tumour size.
The test statistic:
xy
t
1 1
sPooled

n m
Suppose the data has been collected and:
drug treated
untreated
1.89
2.08
1.79
1.28
1.29
1.75
n
x
 xi
n
 1.657
i 1
n
sx 
n
y
y
i 1
m
1.90
i
 x  x 
sy 
i
i 1
n 1
y
i 1
2.16
2
n
 1.915
2.32
i
 0.3215
 y
m 1
2
 0.3693
The test statistic:
sPooled 
n  1sx2  m  1s 2y
nm2
20.3215  50.3693
 0.3563
7
2

2
1.657  1.915
 .258
t

 1.025
.252
1 1
0.3563 
3 6
We reject H0 if:
t  t   t0.05  1.895
with d.f. = n + m – 2 = 7
Hence we accept H0.
Conclusion: The drug treatment does not
result in a significant ( = 0.05) smaller final
tumour size,