Download One and Two tailed tests - Mathematics and Statistics

Comparing Populations
Proportions and means
Most studies will have more than one population.
Example The Salk-vaccine trial 1954
A large study to determine if the Salk vaccine
was effective in reducing the incidence of polio.
Two populations:
1. Individuals vaccinated with the Salk vaccine
2. Individuals vaccinated with a placebo
A double blind study
both individuals vaccinated and MD’s treating
the cases did not know who recieved the vaccine
and who received the placebo
When there are more than one population one
will be interested in making comparisons.
Comparisons are sometimes made through
differences, sometimes through ratios
An important fact:
The sampling distribution of differences of Normal
Random Variables
If X and Y denote two independent normal random
variables, then :
D = X – Y is normal with
mean D   X  Y
standard deviation  D   X2   Y2
This fact allows us to determine the sampling
distribution of differences
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
Consider the statistic:
x1 x2
D  pˆ1  pˆ 2 = n1 n2
This statistic has a normal distribution with
D   pˆ   pˆ  p1  p2
1
2
using the
important fact
 D = pˆ  pˆ   p2ˆ   p2ˆ
1
2


1
p1 1  p1 
n1
2

p2 1  p2 
n2
pˆ1 1  pˆ1  pˆ 2 1  pˆ 2 

n1
n2
Thus
z 
D  D
D

pˆ1  pˆ 2 -  p1  p2 
 pˆ  pˆ
1


2
pˆ1  pˆ 2 -  p1  p2 
p1 1  p1  p2 1  p2 

n1
n1
pˆ1  pˆ 2 -  p1  p2 
pˆ1 1  pˆ1  pˆ 2 1  pˆ 2 

n1
n1
Has a standard normal distribution
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
If p1 = p2 (p say) then the test statistic:
z 
D  D
D

pˆ1  pˆ 2 -  p1  p2 
 pˆ  pˆ
1


2
pˆ1  pˆ 2 -  p1  p2 
p1 1  p1  p2 1  p2 

n1
n2
pˆ1  pˆ 2
1 1
p 1  p    
 n1 n2 

pˆ1  pˆ 2
1 1
pˆ 1  pˆ    
 n1 n2 
has a standard normal distribution.
where
x1  x2
pˆ 
n1  n2
is an estimate of the common
value of p1 and p2.
Thus for comparing two binomial probabilities
p1 and p2
The test statistic
z
pˆ1  pˆ 2
1 1
pˆ 1  pˆ    
 n1 n2 
where
x1
x2
pˆ1 
, pˆ 2 
n1
n2
x1  x2
and pˆ 
n1  n2
The Critical Region
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
(Non smokers)
(Pipe smokers)
x1  x2
117  54
pˆ 

n1  n2 1067  402
171

 0.1164
1469
(Combined)
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 Proportions
Summary
The test for a difference in proportions
z 
pˆ1  pˆ 2
1 1
pˆ 1  pˆ   
 n1 n2 
(The test statistic)
Estimating the difference in proportion by a
confidence interval
pˆ 2  pˆ1  z / 2
pˆ1 1  pˆ1  pˆ 2 1  pˆ 2 

n1
n2
Comparing Means
Comparing Means
Situation
• We have two normal populations (1 and 2)
• Let 1 and 1 denote the mean and standard
deviation of population 1.
• Let 2 and 2 denote the mean and standard
deviation of population 2.
• 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 : 1  2 vs H A : 1  2
or
2. H 0 : 1  2 vs H A : 1  2
or
3. H 0 : 1  2 vs H A : 1  2
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 : 1  2 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
i 1
i
n 1
n
n
y
 x  x 
2
sy 
2


y

y
 i
i 1
m 1
The Alternative
Hypothesis HA
The Critical Region
H A : 1  2
z   z / 2 or z  z / 2
H A : 1  2
z  z
H A : 1  2
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 : 1  2
The exercise group did not have a higher
average reduction in blood pressure
vs
H A : 1  2
The exercise 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 1 denote the mean of population 1.
• Let 2 denote the mean of population 2.
• Objective is to estimate the difference in the
two population proportions d = 1 – 2.
Confidence Interval for d = 1 – 2
ˆ1  ˆ 2  z / 2  ˆ ˆ
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 = 1 – 2
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
The t test
Comparing Means – small samples
Situation
• We have two normal populations (1 and 2)
• Let 1 and 1 denote the mean and standard
deviation of population 1.
• Let 2 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 : 1  2 vs H A : 1  2
or
2. H 0 : 1  2 vs H A : 1  2
or
3. H 0 : 1  2 vs H A : 1  2
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 (equal variances)
Situation
• We have two normal populations (1 and 2)
• Let 1 and  denote the mean and standard
deviation of population 1.
• Let 2 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
i 1
i
n 1
n
n
y
 x  x 
2
sy 
2


y

y
 i
i 1
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 1 = 2 this statistic has a t distribution
with n + m –2 degrees of freedom
The Alternative
Hypothesis HA
The Critical Region
H A : 1  2
t  t / 2 or t  t / 2
H A : 1  2
t  t
H A : 1  2
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 : 1  2
The treated group did not have a lower
average final tumour size.
vs
H A : 1  2
The treated 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
2.32
 x  x 
i 1
2.16
2
i
n 1
 0.3215
n
 1.915
sy 
2


y

y
 i
i 1
m 1
 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,
Confidence intervals for the difference in two
means of normal populations (small sample sizes
equal variances)
(1 – )100% confidence limits for 1 – 2
1 1

n m
 x  y   t / 2 sPooled
where
sPooled 
 n  1 s
and df  n  m  2
2
x
  m  1 s
nm2
2
y
Tests, Confidence intervals
for the difference in
two means of normal populations
(small sample sizes, unequal variances)
t
Consider the statistic
xy
2
s
s
y

n m
2
x
For large sample sizes this statistic has standard
normal distribution.
For small sample sizes this statistic has been shown to
have approximately a t distribution with
s
s 
  
 n m
2
x
df 
2
y
2
1 s 
1 s 

 


n 1 n  m 1 m 
2
x
2
2
y
2
The approximate test for a comparing two means
of Normal Populations (unequal variances)
2 2
2
 sx s y 
Test statistic
  
n m
xy

df 
t
2
2
2 2
2
2
sx s y
1  sx 
1  sy 


 


n m
n 1 n  m 1 m 
Null Hypothesis
H0: 1 = 2
Alt. Hypothesis
H0: 1 ≠ 2
H0: 1 > 2
H0: 1 < 2
Critical Region
t < -t/2 or t > t/2
t > t
t < -t
Confidence intervals for the difference in two
means of normal populations (small samples,
unequal variances)
(1 – )100% confidence limits for 1 – 2
 x  y   t / 2
with
df 
s
s

n m
s
s 
  
 n m
2
x
2
y
2
x
2
y
2
1 s 
1 s 

 


n 1 n  m 1 m 
2
x
2
2
y
2
Testing for the equality of
variances
The F test
Situation:
Let x1, x2, x3, … xn, denote a sample from a
Normal distribution with mean x and standard
deviation x
Let y1, y2, y3, … ym, denote a second independent
sample from a Normal distribution with mean y
and standard deviation y
We want to test for the equality of the two
variances
2
2
 x and  y
i.e.:
Test
H0 :  x2   y2 against H A :  x2   y2
(Two sided alternative)
or
Test
H0 :  x2   y2 against H A :  x2   y2
(one sided alternative)
or
Test
H0 :    against H A :   
2
x
2
y
(one sided alternative)
2
x
2
y
The test statistic (F)
s
F
s
2
x
2
y
or
1 s

F s
2
y
2
x
The sampling distribution of the test statistic
If the Null Hypothesis (H0) is true then the sampling
distribution of F is called the F-distribution with
n1 = n - 1 degrees in the numerator
and
n2 = m - 1 degrees in the denominator
The F distribution
n1 = n - 1 degrees in the numerator
0.7
n2 = m - 1 degrees in the denominator
0.6
0.5
0.4
0.3
0.2

0.1
0
0
1
2
3
F(n1, n2)
4
5
Note: If
s
F
s
2
x
2
y
has F-distribution with
n1 = n - 1 degrees in the numerator
and n2 = m - 1 degrees in the denominator
then
s y2
1
 2
F sx
has F-distribution with
n1 = m - 1 degrees in the numerator
and n2 = n - 1 degrees in the denominator
Critical region for the test:
H0 :  x2   y2 against H A :  x2   y2
(Two sided alternative)
Reject H0 if
or
sx2
F  2  F / 2  n  1, m  1
sy
2
y
2
x
1 s
  F / 2  m  1, n  1
F s
Critical region for the test (one tailed):
H0 :    against H A :   
2
x
2
y
2
x
(one sided alternative)
Reject H0 if
sx2
F  2  F  n  1, m  1
sy
2
y
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
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
2.32
 x  x 
i 1
2.16
2
i
n 1
 0.3215
n
 1.915
sy 
2


y

y
 i
i 1
m 1
 0.3693
We want to test:
H0 :  x2   y2 against H A :  x2   y2
(H0 is assumed for the t-test for comparing
the means )
Using  =0.05 we will reject H0 if
or
sx2
F  2  F0.25  2,5   5.79
sy
2
1 sy
 2  F0.025  5, 2   19.30
F sx
Test statistic:
.3215

F
2
.3693
2
and
0.1033

 0.76
0.1364
1 .3693
0.1364


 1.32
2
F .3215
0.1033
2
Therefore we accept
H0 :   
2
x
2
y
The paired t-test
An example of improved
experimental design
•
Often we are interested in comparing the
effect of two (or more) treatments on some
variable.
Examples:
1. The effect of two diets on weight loss.
2. The effect of two drugs on the drop in
Cholesterol levels.
3. The effects of two method in teaching on
Math Proficiency
•
One possible design is to randomly divide the
available subjects into two groups.
• The first group will receive treatment 1.
• The 2nd group will receive treatment 2.
We then collect data on the two groups
1. Let x1, x2, x3,…, xn denote the data for treatment 1.
2. Let y1, y2, y3,…, ym denote the data for treatment 2.
This design is called the independent sample design.
To test for the equality of treatment means we use the
two sample t test
The test statistic:
xy
t
1 1
sPooled

n m
The Alternative
Hypothesis HA
The Critical Region
H A : 1  2
t  t / 2 or t  t / 2
H A : 1  2
t  t
H A : 1  2
t  t
d.f. = n + m - 2
The matched pair experimental design (The paired
sample experiment)
Prior to assigning the treatments the subjects are grouped
into pairs of similar subjects.
Suppose that there are n such pairs (Total of 2n = n + n
subjects or cases), The two treatments are then randomly
assigned to each pair. One member of a pair will receive
treatment 1, while the other receives treatment 2. The data
collected is as follows:
– (x1, y1), (x2 ,y2), (x3 ,y3),, …, (xn, yn) .
xi = the response for the case in pair i that receives
treatment 1.
yi = the response for the case in pair i that receives
treatment 2.
Let xi = the measurement of the response for the subject
in pair i that received treatment 1.
Let yi = the measurement of the response for the subject
in pair i that received treatment 2.
The data
x1
y1
x2
y2
x3
y3
…
xn
yn
Let di = yi - xi. Then
d1, d2, d3 , … , dn
is a sample from a normal distribution with mean,
d = 2 – 1 , and
standard deviation
 d      2 xy x y
2
x
2
y
Note if the x and y measurements are positively
correlated (this will be true if the cases in the pair are
matched effectively) than d will be small.
To test H0: 1 = 2 is equivalent to testing H0: d = 0.
(we have converted the two sample problem into a single
sample problem).
The test statistic is the single sample t-test on the
differences
d1, d2, d3 , … , dn
namely
d 0
td 
sd n
df = n - 1
d  the mean of the d i' s and
sd  the std. dev. of the d i' s
Example
We are interested in comparing the effectiveness of two
method for reducing high cholesterol
The methods
1. Use of a drug.
2. Control of diet.
The 2n = 8 subjects were paired into 4 match pairs.
In each matched pair one subject was given the
drug treatment, the other subject was given the diet
control treatment. Assignment of treatments was
random.
The data
reduction in cholesterol after 6 month period
Pair
Treatment
Drug treatment
Diet control Treatment
1
30.3
25.7
2
10.2
9.4
3
22.3
24.6
4
15.0
8.9
Differences
Pair
Treatment
Drug treatment
Diet control Treatment
di
d  2.3
1
30.3
25.7
4.6
2
10.2
9.4
0.8
3
22.3
24.6
-2.3
4
15.0
8.9
6.1
sd  3.792
d 0
2.3
td 

 1.213
sd n 3.792 4
t0.025  3.182 for df = n – 1 = 3, Hence we accept H0.
Example 2
In this example the researcher is interested in
the effect of an antidepressant in reducing
depression.
Subjects were given a psychological test
measuring depression (on a scale 0-100) at the
beginning of the study (Pre-score) and after a
period of one month on the anti-depressant
(Post-score).
Did the drug have any effect on reducing
depression?
Table: Prescore (xi), Postscore (yi),
difference (di)
subject
1
Pre
73.7
Post
63.9
d i = diff 9.8
d
d
i
2
3
4
5
6
7
8
9
61.1
60.7
0.4
76.5
72.7
3.8
64.5
50.7
13.8
76.9
67.2
9.7
82.4
66.9
15.5
71.1
62.0
9.1
61.1
44.1
17.0
89.5
90.5
-1.0
i
 d d 
t
i
i
n 1
 8.603
t0.05  1.796 for df  11, thus H 0 is rejected.
d
sd
n
11
12
59.6 58.6 89.3
56.0 69.4 70.8
3.6 -10.8 18.5
 7.450
n
2
sd 
10
 3.00
Comments
• This last example is a matched pair
experiment that occurs frequently.
• You have two observations on the same
subject.
• One observation under 1 condition or treatment
(the Pre score), the other observation under a
second condition (the Post score) (after
treatment)
• The subject is his own matched twin.
• This design is sometimes called a Repeated
Measures design
Example 3
• In this example, one is interested in determining if a
new method of mathematics instruction is an
improvement over the current method.
• To determine this, 20 grade 4 students were selected.
• They were divided into n = 10 matched pairs.
• The students were matched relative to ability.
• One member of each matched pair was instructed
using the new method, the other member using the
current method.
• All students were tested at the end of the instruction
period
The data
Pair
New (x i )
Current (y i )
di = xi - yi
1
2
3
4
5
6
7
8
9
10
90
75
90
88
55
67
94
75
88
87
84
67
90
95
40
68
85
67
86
81
6
8
0
-7
15
-1
9
8
2
6
d 0
d  4.60, sd  6.2218 and t 
 2.338
sd
n
t0.05  1.833, t0.01  2.821 for d . f .  n  1  9
Summary of Tests
One Sample Tests
Situation
Test Statistic
Sample form the Normal
distribution with unknown
mean and known variance
(Testing )
z
Sample form the Normal
distribution with unknown
mean and unknown variance
(Testing )
Testing of a binomial
probability 
Sample form the Normal
distribution with unknown
mean and unknown variance
(Testing )
t
z
n x  0 
H0
  
0
n x   0 
s
pˆ  p0
p0 (1  p0 )
n

n  1s 2
U
 02
  
 
p = p0
 0
HA
  
  
  
  
  
  
p ≠p0 
p >p0
 p0
p <
  0
Critical Region
z < -z/2 or z > z/2
z > z
z <-z
t < -t/2 or t > t/2
t > t
t < -t
z < -z/2 or z > z/2
z > z
z < -z
U   12 / 2 n  1 or
  0
U   2 n  1
  0
U   12 n  1
U   2 / 2 n  1
Two Sample Tests
Situation
Two independent samples
from the Normal distribution
with unknown means and
known variances
(Testing 1 - 2)
Test Statistic
 x1  x2 
z
 12
n1

H0
HA
Critical Region
1   2 1   2 z < -z/2 or z > z/2
 22
1   2 z > z
n2
1   2 z < -z
Two independent samples
from the Normal distribution
with unknown means and
unknown but equal
variances.
(Testing 1 - 2)
t
sp
sp 
Estimation of a the
difference between two
binomial probabilities, p1-p2
x1  x2 
zz 
1   2 1   2 t < -t/2 or t > t/2
df  n  m  2
1 1

n1 n2
1   2 t > t
n  1s12  m  1s22
1   2 t < -t
nm2
pˆˆ11 ˆpˆ2 2
 11 11 
ˆ
pˆˆ (1


)
1  pˆ n1 n2  
 n1 n2 
p11  p22
df  n  m  2
df  n  m  2
p1 
 2 z < -z/2 or z > z/2
1  p2
p11  
p22 z > z
p1  p2
z < -z
Two Sample Tests - continued
Situation
Two independent
Normal samples
with unknown means
and variances
(unequal)
Two independent
Normal samples
with unknown means
and variances
(unequal)
Test statistic
t
x1  x2
H0
HA
1  2 1 ≠ 2
s12 s22

n1 n2
* = df 
t > t df = *
1 < 2
t < - t df = *
F > F(n-1, m -1)
1 < 2
1/F > F(m-1, n -1)
2
2
1 s 
1  s 2y 

 


n2  1  nm2 
n1 1  n 1  m
2
F > F/2(n-1, m -1) or 1/F > F/2(m-1, n -1)
1 > 2
 s1x2 s22y 
  
 n1 nm2 
2
1x
t < - t/2 or t > t/2 df = *
1 > 2
1  2 1 ≠ 2
s12
1 s22
F  2 or

s2
F s12
Critical Region
2
The paired t test
Situation
n matched pair of
subjects are treated
with two treatments.
di = xi – yi has mean
d = 1 – 2
Test statistic
t
H0
HA
Critical Region
1  2 1 ≠ 2
d
sd
n
Independent samples
Treat 1 Treat 2
1 > 2
t > t df = n - 1
1 < 2
t < - t
df = n - 1
Matched Pairs
Treat 1
Pair 1
Pair 2
Pair 3
Possibly equal
numbers
t < - t/2 or t > t/2 df = n - 1
Pair n
Treat 2
Sample size determination
When comparing two or more
populations
Estimating a difference in proportions
using confidence intervals
Confidence Interval for d = p1 – p2 :
pˆ1  pˆ 2  B
where B  z 2
p1 1  p1  p2 1  p2 

n1
n2
Again we want to choose n1 and n2 to set B
at some predetermined level with a fixed
level of confidence 1 – .
There are many solutions for n1 and n2 that will
achieve a specified error bound B with level of
confidence 1 – .
You can make B small by increasing n1 or n2 or a
combination of both.
Some useful practical solutions satisfy
1. Equal sample size: n1 = n2 This would be an
appropriate choice if one researcher was to
collect data from population 1, another was to
collect data from population 2 and you
wanted to equalize the workload.
2. Minimize Total sample size: Choose n1
and n2 so that the required error bound B is
achieved and the total sample size, n1 + n2,
is minimized. This would be an appropriate
choice if a single researcher was to collect
data from both population 1 and population
2 and you wanted to minimize his
workload.
3. Minimize Total Cost of the sample:
Suppose that the study has a fixed cost of
C0$ and that the cost of a single
observation populations 1 and 2 is c1$ and
c2$ repectively,
Then the total cost of the study is:
C0 + n1c1 + n2c2 .
This approach chooses n1 and n2 so that the
required error bound B is achieved and the
total cost, C0 + n1c1 + n2c2, is minimized.
Special solutions - case 1: n1 = n2 = n.
then
n1  n2  n  z / 2
2
p1111 p11  p2 11 p22 
22
BB
Special solutions - case 2: Choose n1 and n2 to
minimize N = n1 + n2 = total sample size
then
z2 / 2
n1  2
B
p 11p  
p11 11  p11  p22 11  p22  

z2 / 2
n2  2
B
p 11p 
p1 11 p11 p2 11 p2  
11
22
11
22


Special solutions - case 3: Choose n1 and n2 to minimize
C = C0 + c1 n1 + c2 n2 = total cost of the study
Note: C0 = fixed (set-up) costs
c1 = cost per unit in population 1
c2 = cost per unit in population 2
then
z2 / 2
n1  2
B


cc2 2
 1p111 1p12p211 2p2
p1111p11
c1c1


z2 / 2
n2  2
B


cc1 1
 1p111 1p1 2p211 2p2 
p221 p22
cc2 2


Determination of sample size (means)
When the objective is to compare the two
means of two Normal populations
Estimating a difference in means using
confidence intervals
Confidence Interval for d = 1 – 2 :
x1  x2  B
where B  z 2

2
1
n1


2
2
n2
Again we want to choose n1 and n2 to set B at
some predetermined level with a fixed level
of confidence 1 – .
The sample sizes required, n1 and n2, to estimate 1 – 2 within
an error bound B with level of confidence 1 –  are:
Equal sample sizes
n  n1  n2  z / 2
2
 1x2   2y2 
2
B
Minimizing the total sample size N = n1 + n2 .
z2 / 2 2
n1  2  1x  1x 2y 
B
z2 / 2 2
n2  2  x2  1x y2
B
Minimizing the total cost C = C0 + c1n1 + c2n2 .
z2 / 2
n1  2
B
 2


z2 / 2  2
c2
c1
1x 2y  n2  2  2y 
1x 2y 
1x 
c1
B 
c2



Some general comments
• If a population is more variable (2 larger)
– more observations should be assigned to the
sample from that population
• If it is less costly to take observations in a
population
– more observations should be assigned to the
sample from that population
Next Topic: Comparing k
populations