The paired sample experiment

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The paired sample experiment

• The paired t test

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• Frequently one is interested in comparing the
  effects of two treatments (drugs, etc) on a
  response variable.
• The two treatments determine two different
  populations
• Popn 1 cases treated with treatment 1.
• Popn 2 cases treated with treatment 2
• The response variable is assumed to have a normal
  distribution within each population differing
  possibly in the mean (and also possibly in the
  variance)

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• Two independent sample design
• A sample of size n cases are selected from
  population 1 (cases receiving treatment 1) and a
  second sample of size m cases are selected from
  population 2 (cases receiving treatment 2).
• The data
• x1, x2, x3, , xn from population 1.
• y1, y2, y3, , ym from population 2.
• The test that is used is the t-test for two
  independent samples

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The test statistic (if equal variances are
assumed)
where
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• 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.
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• Let di yi - xi. Then
• d1, d2, d3 , , dn
• Is a sample from a normal distribution with mean,
• md m2 m1 , and
• variance

standard deviation
Note if the x and y measurements are positively
correlated (this will be true if the cases in the
pair are matched effectively) than sd will be
small.
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• To test H0 m1 m2 is equivalent to testing H0
  md 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
df n - 1
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Example

• We are interested in comparing the effectiveness
  of two method for reducing high cholesterol

• The methods
• Use of a drug.
• 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.
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The datareduction in cholesterol after 6 month
period
Pair Pair Pair Pair
Treatment 1 2 3 4
Drug treatment 30.3 10.2 22.3 15.0
Diet control Treatment 25.7 9.4 24.6 8.9
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Differences
Pair Pair Pair Pair
Treatment 1 2 3 4
Drug treatment 30.3 10.2 22.3 15.0
Diet control Treatment 25.7 9.4 24.6 8.9
di 4.6 0.8 -2.3 6.1
for df n 1 3, Hence we accept H0.
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Nonparametric Statistical Methods
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• Many statistical procedures make assumptions
• The t test, z test make the assumption that the
  populations being sampled are normally
  distributed. (True for both the one sample and
  the two sample test).

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• This assumption for large sample sizes is not
  critical.
• (Reason The Central Limit Theorem)
• The sample mean, the statistic z will have
  approximately a normal distribution for large
  sample sizes even if the population is not normal.

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• For small sample sizes the departure from the
  assumption of normality could affect the
  performance of a statistical procedure that
  assumes normality.
• For testing, the probability of a type I error
  may not be the desired value of a 0.05 or 0.01
• For confidence intervals the probability of
  capturing the parameter may be the desired value
  (95 or 99) but a value considerably smaller

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• Example Consider the z-test
• For a 0.05 we reject the hypothesized value of
  the mean if z lt -1.96 or z gt 1.96

Suppose the population is an exponential
population with parameter l. (m 1/l and s 1/l)
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Actual population
Assumed population
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Suppose the population is an exponential
population with parameter l. (m 1/l and s
1/l) It can be shown that the sampling
distribution of
is the Gamma distribution with
Use mgfs
The distribution of is not the normal
distribution with
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Sampling distribution of
Actual distribution
n 2
Distribution assuming normality
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Sampling distribution of
Actual distribution
n 5
Distribution assuming normality
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Sampling distribution of
Actual distribution
n 20
Distribution assuming normality
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Definition

• When the data is generated from process (model)
  that is known except for finite number of unknown
  parameters the model is called a parametric
  model.
• Otherwise, the model is called a non-parametric
  model

Statistical techniques that assume a
non-parametric model are called non-parametric.
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The sign test

• A nonparametric test for the central location of
  a distribution

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We want to test
H0 median m0
against
HA median ? m0
(or against a one-sided alternative)
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• The assumption will be only that the distribution
  of the observations is continuous.
• Note for symmetric distributions the mean and
  median are equal if the mean exists.
• For non-symmetric distribution, the median is
  probably a more appropriate measure of central
  location.

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The Sign test

1. The test statistic

S the number of observations that exceed m0
Comment If H0 median m0 is true we would
expect 50 of the observations to be above m0,
and 50 of the observations to be below m0,
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If H 0 is true then S will have a binomial
distribution with p 0.50, n sample size.
50
50
median m0
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If H 0 is not true then S will still have a
binomial distribution. However p will not be
equal to 0.50.
m0 gt median
p lt 0.50
p
median
m0
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m0 lt median
p gt 0.50
p
median
m0
p the probability that an observation is
greater than m0.
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Summarizing If H0 is true then S will have a
binomial distribution with p 0.50, n sample
size.
n 10
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The critical and acceptance region
n 10
Choose the critical region so that a is close to
0.05 or 0.01. e. g. If critical region is
0,1,9,10 then a .0010 .0098 .0098 .0010
.0216
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e. g. If critical region is 0,1,2,8,9,10 then a
.0010 .0098 .0439.0439 .0098 .0010 .1094
n 10
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• If one cant determine a fixed confidence region
  to achieve a fixed significance level a , one
  then randomizes the choice of the critical region
• In the example with n 10, if the critical
  region is 0,1,9,10 then a .0010 .0098
  .0098 .0010 .0216
• If the values 2 and 8 are added to the critical
  region the value of increases to 0.216 2(.0439)
  0.0216 0.0878 0.1094
• Note 0.05 0.0216 0.3235(.0878)
• Consider the following critical region
• Reject H0 if the test statistic is 0,1,9,10
• If the test statistic is 2,8 perform a
  success-failure experiment with p Psuccess
  0.3235, If the experiment is a success Reject Ho.
• Otherwise we accept H0.

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Example

• Suppose that we are interested in determining if
  a new drug is effective in reducing cholesterol.
• Hence we administer the drug to n 10 patients
  with high cholesterol and measure the reduction.

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The data
Let S the number of negative reductions 2
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If H0 is true then S will have a binomial
distribution with p 0.50, n 10.
We would expect S to be small if H0 is false.
n 10
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• Choosing the critical region to be 0, 1, 2 the
  probability of a type I error would be
• a 0.0010 0.0098 0.0439 0.0547
• Since S 2 lies in this region, the Null
  hypothesis should be rejected.
• Conclusion There is a significant positive
  reduction (a 0.0547) in cholesterol.

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If n is large we can use the Normal approximation
to the Binomial. Namely S has a Binomial
distribution with p ½ and n sample
size. Hence for large n, S has approximately a
Normal distribution with mean and standard
deviation
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Hence for large n,use as the test statistic (in
place of S)
Choose the critical region for z from the
Standard Normal distribution.
i.e. Reject H0 if z lt -za/2 or z gt za/2
two tailed ( a one tailed test can also be set up.