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8/6/2019 Hypothesis Testing and Sampling Distributions 2011
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Test StatisticsA Statistic for which the frequency of particular
values is known.
Observed values can be used to test hypotheses.
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Single case: used z score and normal distribution.
But usually have scores from a sample.
Example:
n = 25
M= 106
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In order to determine probability of obtaining a score, ithas to be compared to the appropriate distribution known as the Sampling Distribution.
Distribution of sample means
The distribution of sample means is defined as the set ofmeans from all the possible random samples of a specific size(n) selected from a specific population.
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Characteristics of the Sampling Distribution of the mean
The sampling distribution of means will have the same mean as the population
m =
The sampling distribution of means has a smaller variance. This is because the means ofsamples are less likely to be extreme compared to individual scores.
M= standard deviation of the means = standard error
The shape of the sampling distribution approximates a normal curve if either
the population of individual cases is normally distributed the sample size being considered is 30 or more
Demonstration
Mn
2
2
= Mn
=
http://onlinestatbook.com/stat_sim/sampling_dist/index.htmlhttp://onlinestatbook.com/stat_sim/sampling_dist/index.html8/6/2019 Hypothesis Testing and Sampling Distributions 2011
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IQ Scores - Ranked
71, 76, 76, 77, 79, 80, 81, 82, 83, 83, 84, 84, 84, 85, 85, 86,87, 88, 88, 88, 89, 90, 90, 91, 91, 91, 92, 92, 92, 93, 93, 93,93, 93, 93, 94, 94, 94, 94, 95, 95, 95, 96, 96, 97, 97, 97, 97,97, 97, 97, 98, 99, 99, 99, 99, 100, 100, 100, 100, 101, 101, 101,102, 102, 102, 102, 103, 103, 103, 103, 103, 103, 103, 104, 104, 104,105, 105, 106, 106, 107, 107, 107, 107, 107, 108, 108, 108, 108,
109, 109, 110, 111, 111, 112, 112, 112, 113, 113, 113, 113, 114, 114, 115,115, 115, 117, 118, 120, 121, 121, 121, 123, 123, 125, 125, 126, 131, 136
Sample Size Lowest extreme score Highest extreme score
MM
N = 1 71 71 136 136
N = 2 71+76 73.5 131 + 136 133.5
N = 3 71+76 + 76 74.3 126+ 131 + 136 131.0
N = 4 71+76 =76 + 77 75.0 125 + 126 + 131 + 136 127
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Using a sampling distribution
Sample: n = 25, M= 106
IQ test:
Appropriate Sampling Distribution
M= = 100
IQ: = =
100 15
Mn
= = = =15
25
15
53
zM
M
=
zX
=
zM
M
=
=
= =
106 100
3
6
32 00.
Area = .0228 one tailed.0456 two tailed
Reject Null Hypothesis
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Z Test
zM
M
=
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tDistribution
What happens if we do not have the population standarddeviation?
We can use the sample standard deviation as anestimate.
Problem Cannot use z and the normal distribution to estimate
probability.
Sample variance tends to underestimate the populationvariance
Have to use a slightly different distribution - Students tDistribution
zM
M
=
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tDistribution cont.
Not a single distribution but a family of distributions.
One for each degree of freedom (df).df = n - 1
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One sample ttest
Psychomotor abilities of low-birthweight infants (PDI scores)(Nurcombe et al, 1984)
Sample: n = 56 M= 104.3 s = 12.58
Norms: = 100
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Sample: N = 56 M= 104.3 s = 12.58Norms: = 100
Hypotheses:
H1: 1 2H0: 1 = 2 = 100
tM
sM=
z
M
M
=
s sn
M =
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Sample: N = 56 M= 104.3 s = 12.58Norms: = 100
Hypotheses:
H1: 1 2H0: 1 = 2 = 100
tM
s
M
s
nx
=
=
=
= = 104 3 100
1258
56
4125
16822 45
.
.
.
..
df = 56 - 1 = 55
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Potential Problem
What happens if we do not have a population mean?
Two ways of dealing with it. Use a repeated measures design
Use two independent samples
t MsM
=
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Repeated-Measures ttest
(In SPSS called the Paired-Samples t-test)
Use one sample, but test at two different times.
Occasionally matched samples used.
If treatment has no effect what will the outcome be? There will be little difference between the scores on the first
and second testing.
Allows us to hypothesise what the population mean will be.
Mean of the differences will be 0 (zero). Will not be exactly the same due to sampling error
So have to test whether the mean difference observed issignificantly different from o.
Need a sampling distribution
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MDI 6months (X1)
MDI 24months (X2)
Difference (D)
124 114 -10
94 88 -6
115 102 -13
110 127 17
. . .
. . .
126 114 -12
123 132 9M 111.0 106.71 -4.29
s 13.85 12.95 16.04
n 31 31 31
Example:
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D= =
1 2
0
D = - 4.29 sD= 16.04 n = 31
H0
:
t MsM
=
t D
sD
=
tD
s
D
s
D
s
nD D D
=
=
= 0 0
D
s
n
D
=
=
=
0 4 29 0
1604
31
4 29
2 881 49
.
.
.
..
df = n - 1 (n is number of pairs of observations)
df = 31 - 1 = 30
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Independent samples ttest
When we do not have population parameters, we canuse an independent samples (between subjects) designto get sample data that allows us to evaluate thedifference between two populations using theindependent samples t test.
As with all hypothesis tests, the general purpose of theindependent-measures t test is to determine whether
the sample mean difference obtained indicates a realmean difference between the two populations (ortreatments) or whether the obtained difference issimply the result of sampling error.
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Distribution of Differences between Means
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Variance of the two Distributions of the Means
12
1
2
2
2Nand
N
Standard Error of the Distributions of theMeans
1
1
2
2N
andN
Mean of the Distribution of MeanDifferences
1 2 0 =
Variance of the Distribution of MeanDifference
X XN N1 2
2 1
2
1
2
2
2
=+
Standard Error of the Distribution of MeanDifference
X XN N1 2
1
2
1
2
2
2
= +
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tM
sM
=
tM M
sM M
=
( ) ( )1 2 1 2
1 2
=
+
=
+
( ) ( ) ( )M M
s
n
s
n
M M
s
n
s
n
1 2 1 2
1
2
1
2
2
2
1 2
1
2
1
2
2
2
Degrees of Freedom df = (n1 - 1) + (n2 - 1) = n1 + n2 - 2
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Summary
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Z score z X=
Z test zM
m
=
test - single sample tM
sM
=
t test - repeated measures tD
s
D
sD D
=
= 0
test - two independent samples tM M
sM M
=
( ) ( )1 2 1 2
1 2
tM M
sM M
=
( )1 2
0
1 2
Tests
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Standard Errors
Z test
X
N=
t test - single sample ss
nM =
t test - two matched samples ss
nD
D=
t test - two independentsampless s
n n M M p
1 2
2
1 2
1 1
= +
sn s n s
n np
2 1 1
2
2 2
2
1 2
1 1
2=
+
+
( ) ( )
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Assumptions for the t test
Repeated measures:
Random sampling
The data are measured on at least an interval scale
The data are matched
The distribution of the population scores is normal
Independent samples
Random sampling
The data are measured on at least an interval scale
The participants in the two samples are independent
The distribution of the populations scores are normal
The variances of the two populations are the same (homogeneity of variance)
Robust
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Analysis of Variance
What happens when we have more than two groups?
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s
k
s s s s sj2
1
2
2
2
3
2
4
2
5
2
5=
+ + + +
First Method
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( ) j G
k
2
1
Second Method
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variance between treatments MSbetweenF = =
variance within treatments MSwithin
Systematic variance (treatment effect) + unsystematic variance MSbetweenF = =
unsystematic variance MSwithin
obtained mean differences (including treatment effects) MSbetweenF = =
differences expected by chance (without treatment effects) MSwithin