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Transcript
```Research Methods I
T-tests
Different Kinds of T-tests

One sample t-test
 When

population variance is not known
Two sample t-test for independent
samples
 Compare
two sample means
 Samples are independent of each other

Two sample t-test for dependent samples
 Compare
two sample means
 Samples are paired / dependent
One-Sample T-test


One-sample t-test is being used when µ is
known but σ is not known.
For the one-sample t-test σ is being substituted
with σ(hat) otherwise it is exactly the same as
the z-test
 Same
formula as for standard deviation but we divide
by n-1
 This is especially important when the sample size is
small; with large samples it makes little difference
whether s or σ is being used.

In order to calculate t: t = xbar - µ / σhat / √n
T-distribution



The sampling distributions of z and t are different.
For z-tests the sampling distribution of sample
means is a normal curve
For t-tests the sampling distribution changes
according to sample size (above n=120 almost like
z distribution)
 The
smaller n the flatter the curve, the larger the tails
and the higher the critical t-value in order to reject the
null hypothesis.


For n > 120 tcritical is 1.658 while zcritical is 1.645
For n = 6 tcritical is 2.015 while zcritical is still 1.645
 Table
on page 251 gives critical t-values for each
sample size (better degrees of freedom: n-1)
Degrees of Freedom





The critical value for t depends on the degrees of
freedom (df)
Degrees of freedom depend on the sample size
df = n-1
All tables of critical values of significance tests
Degrees of freedom tell us how many numbers are
free to vary.
 If
we have x1 + x2 + x3 = 10 only two of the three numbers
are free to vary the last one has to be a certain way to
get the fixed result: here there are two degrees of
freedom
Two-sample t-tests

In two-sample t-tests two sample means are
being compared.
 What
is the probability that the difference in the means
that is observed is due to chance?



Population parameters are no longer known
Normal distribution of population must be
assumed unless sample size is very large
There are basically two kinds of t-tests
 Dependent
 Independent
Dependent vs Independent Samples
Independent samples:
 Assumes
that composition of one sample is independent
from composition of other sample.
 Both samples reflect different populations
 H0: µ1= µ2
H1: µ1≠ µ2 or µ1< µ2 or µ1> µ2
 Sample size (n), sample mean, and sample variance
have to be known for each sample
 If two random samples are collected and one has had a
treatment exposure only the treatment exposure should
make the difference between the two groups.
Dependent vs Independent Samples
Dependent samples:
 Members
of one sample are matched or
paired in some way with members of the other
sample
 Examples are pairs of twins, husband and
wife, repeated measures experiment (get a
person’s measure before and after a
treatment)
T-test for Independent Samples

T-test for independent samples is based on
distribution of differences between means.
 It
is like the t-distribution except using the differences
between means

In order to calculate t we get the difference
between the two sample means, and divide it by
the standard deviation of the sampling
distribution (the standard error)
 However,
standard error for this distribution is based
on differences of means and is therefore different.
T-test for Independent Samples
1. Write out null and alternative hypothesis
for the original problem.
2. For each sample determine its sample
size, its mean, and its variance
3. To determine which t formula to use fothe
F test for homogeneity of variance
4. Perform the appropriate F test
T-tests for Independent Samples

The standard error for our distribution is
based on variances of two samples.
 If
both variances are close in magnitude we
can calculate the pooled estimate of common
variance. This is based on a weighted
average of our two sample variances.
 If both variances are not close in magnitude
we have to use a different formula for the ttest
F-test for homogeneity of variance

The F-test for homogeneity of variance
tells us whether the variance between
both samples is similar or different
1.
2.
3.
4.
5.
Write out the null and alternative hypothesis
for the F test
Calculate F and its two degrees of freedom
Compare the F value with Fcritical
If your F value is smaller than Fcritical then
assume equal population variances.
If your F value is larger than Fcritical then
assume unequal population variances.
F-test for homogeneity of variance


H0 : variance of sample 1 is equal to variance of
sample 2
H1 : variance of sample 1 is not equal to
variance of sample 2
F



tests are always non-directional
F = larger of the two sample variances divided
by the smaller of the two sample variances
Degrees of freedom for both numerator and
denominator: df = n – 1
F-table gives critical values
 n1
= numerator degrees of freedom
SAS for Paired T-tests










data new;
set work.julia20;
diff = final_exam - mid_term_exam;
run;
proc print;
var diff;
run;
proc means n mean stderr t prt;
var diff;
run;
N
12
The MEANS Procedure
Analysis Variable : diff
Mean
Std Error t Value Pr > |t|
-4.0000000
1.1742180
-3.41
0.0059
For dependent sample t-test we run a proc means with the
options n, mean, stderr, t, and prt
These statistics will be computed for the difference variable
T will give the t-value and its probability, testing the null
hypothesis that the variable DISS comes from a
population whose mean is zero.
The mean gives the average difference score. If p<.05 we can
say that the two groups are significantly different from one
another.
ANOVA:
Analysis of Variation
The basic ANOVA situation
Two variables: 1 Categorical, 1 Quantitative
Main Question: Do the (means of) the quantitative variables depend on which
group (given by categorical variable) the individual is in?
If categorical variable has only 2 values:
• 2-sample t-test
ANOVA allows for 3 or more groups
An example ANOVA situation
Subjects: 25 patients with blisters
Treatments: Treatment A, Treatment B, Placebo
Measurement: # of days until blisters heal
Data [and means]:
• A: 5,6,6,7,7,8,9,10
[7.25]
• B: 7,7,8,9,9,10,10,11
[8.875]
• P: 7,9,9,10,10,10,11,12,13
[10.11]
Are these differences significant?
Informal Investigation
Graphical investigation:
• side-by-side box plots
• multiple histograms
Whether the differences between the groups are significant depends on
• the difference in the means
• the standard deviations of each group
• the sample sizes
ANOVA determines P-value from the F statistic
Side by Side Boxplots
13
12
11
days
10
9
8
7
6
5
A
B
treatment
P
What does ANOVA do?
At its simplest (there are extensions) ANOVA
tests the following hypotheses:
H0: The means of all the groups are equal.
Ha: Not all the means are equal
doesn’t say how or which ones differ.
 Can follow up with “multiple comparisons”

Note: we usually refer to the sub-populations as
“groups” when doing ANOVA.
Assumptions of ANOVA

each group is approximately normal
 check
this by looking at histograms and/or
normal quantile plots, or use assumptions
 can handle some nonnormality, but not
severe outliers

standard deviations of each group are
approximately equal

rule of thumb: ratio of largest to smallest
sample st. dev. must be less than 2:1
Normality Check
We should check for normality using:
• histograms for each group
• normal quantile plot for each group
With such small data sets, there really isn’t a really good way to check
normality from data, but we make the common assumption that physical
measurements of people tend to be normally distributed.
Standard Deviation Check
Variable
days
treatment
A
B
P
N
8
8
9
Mean
7.250
8.875
10.111
Compare largest and smallest standard deviations:
• largest: 1.764
• smallest: 1.458
• 1.458 x 2 = 2.916 > 1.764
Note: variance ratio of 4:1 is equivalent.
Median
7.000
9.000
10.000
StDev
1.669
1.458
1.764
Notation for ANOVA
• n = number of individuals all together
• I = number of groups
• = mean for entire data set is
x
Group i has
• ni = # of individuals in group i
• xij = value for individual j in group i
• = mean for group i
• si = standard deviation for group i
xi
How ANOVA works (outline)
ANOVA measures two sources of variation in the data and
compares their relative sizes
• variation BETWEEN groups
• for each data value look at the difference between
its group mean and the overall mean
x i  x 
2
• variation WITHIN groups
• for each data value we look at the difference
between that value and the mean of its group
x
 xi 
2
ij
The ANOVA F-statistic is a ratio of the Between Group Variaton
divided by the Within Group Variation:
Between MSG
F

Within
MSE
A large F is evidence against H0, since it indicates that there is more
difference between groups than within groups.
Minitab ANOVA Output
Analysis of Variance for days
Source
DF
SS
MS
treatment
2
34.74
17.37
Error
22
59.26
2.69
Total
24
94.00
F
6.45
P
0.006
R ANOVA Output
treatment
Residuals
Df Sum Sq Mean Sq F value Pr(>F)
2
34.7
17.4
6.45 0.0063 **
22
59.3
2.7
How are these computations
We want to measure the amount of variation due to BETWEEN group
variation and WITHIN group variation
For each data value, we calculate its contribution to:
• BETWEEN group variation:
• WITHIN group variation:
x  x 
2
i
( xij  xi )
2
An even smaller example
Suppose we have three groups
• Group 1: 5.3, 6.0, 6.7
• Group 2: 5.5, 6.2, 6.4, 5.7
• Group 3: 7.5, 7.2, 7.9
We get the following statistics:
SUMMARY
Groups
Column 1
Column 2
Column 3
Count
Sum Average Variance
3
18
6 0.49
4 23.8 5.95 0.176667
3 22.6 7.533333 0.123333
Excel ANOVA Output
ANOVA
Source of Variation SS
Between Groups 5.127333
Within Groups
1.756667
Total
6.884
df
MS
F
P-value F crit
2 2.563667 10.21575 0.008394 4.737416
7 0.250952
9
1 less than number
of groups
1 less than number of individuals
(just like other situations)
number of data values number of groups
(equals df for each
Computing ANOVA F statistic
data
group
5.3
1
6.0
1
6.7
1
5.5
2
6.2
2
6.4
2
5.7
2
7.5
3
7.2
3
7.9
3
TOTAL
TOTAL/df
group
mean
6.00
6.00
6.00
5.95
5.95
5.95
5.95
7.53
7.53
7.53
WITHIN
difference:
data - group mean
plain
squared
-0.70
0.490
0.00
0.000
0.70
0.490
-0.45
0.203
0.25
0.063
0.45
0.203
-0.25
0.063
-0.03
0.001
-0.33
0.109
0.37
0.137
1.757
0.25095714
overall mean: 6.44
BETWEEN
difference
group mean - overall mean
plain
squared
-0.4
0.194
-0.4
0.194
-0.4
0.194
-0.5
0.240
-0.5
0.240
-0.5
0.240
-0.5
0.240
1.1
1.188
1.1
1.188
1.1
1.188
5.106
2.55275
F = 2.5528/0.25025 = 10.21575
Minitab ANOVA Output
Analysis of Variance for days
Source
DF
SS
MS
treatment
2
34.74
17.37
Error
22
59.26
2.69
Total
24
94.00
F
6.45
P
0.006
# of data values - # of groups
1 less than # of
groups
(equals df for each group
1 less than # of individuals
(just like other situations)
Minitab ANOVA Output
Analysis of Variance for days
Source
DF
SS
MS
treatment
2
34.74
17.37
Error
22
59.26
2.69
Total
24
94.00
 (x
ij
 xi )
obs
2
(x
ij
 x)
(x
2
obs
obs
SS stands for sum of squares
• ANOVA splits this into 3 parts

F
6.45

P
0.006
i
 x)
2
Minitab ANOVA Output
Analysis of Variance for days
Source
DF
SS
MS
treatment
2
34.74
17.37
Error
22
59.26
2.69
Total
24
94.00
MSG = SSG / DFG
MSE = SSE / DFE
F = MSG / MSE
(P-values for the F statistic are in Table E)
F
6.45
P
0.006
P-value
comes from
F(DFG,DFE)
So How big is F?
Since F is
Mean Square Between / Mean Square Within
= MSG / MSE
A large value of F indicates relatively more
difference between groups than within groups
(evidence against H0)
To get the P-value, we compare to F(I-1,n-I)-distribution
• I-1 degrees of freedom in numerator (# groups -1)
• n - I degrees of freedom in denominator (rest of df)
Connections between SST, MST,
and standard deviation
If ignore the groups for a moment and just compute the standard deviation
of the entire data set, we see
s
2
x



ij
x
n 1
  SST
2
DFT
 MST
So SST = (n -1) s2, and MST = s2. That is, SST and MST measure the
TOTAL variation in the data set.

Connections between SSE, MSE,
and standard deviation
Remember:
si
2

x


 xi 
2
ij
ni  1
SS[ Within Group i ]

dfi
So SS[Within Group i] = (si2) (dfi )
This means that we can compute SSE from the standard deviations and sizes
(df) of each group:
SSE  SS[Within]   SS[Within Group i ]
  s (ni  1)   s (dfi )
2
i
2
i
Pooled estimate for st. dev
One of the ANOVA assumptions is that all groups have the same
standard deviation. We can estimate this with a weighted average:
2
2
2
(n
1)s

(n
1)s

...
(n
1)s
1
2
2
I
I
s2p  1
nI
(df1)s  (df 2 )s  ... (df I )s
s 
df1  df 2  ... df I
2
p
SSE
s 
 MSE
DFE
2
p

2
1
2
2
2
I
so MSE is the pooled
estimate of variance
In Summary
SST   (x ij  x )  s (DFT)
2
2
obs
SSE   (x ij  x i )   si (df i )
2
2
obs
groups
SSG   (x i  x) 
2
obs
 n (x
i
i
 x)
2
groups
SS
MSG
SSE SSG  SST; MS 
; F
DF
MSE
2
R
Statistic
R2 gives the percent of variance due to between
group variation
SS[Between ] SSG
R 

SS[Total ]
SST
2
We will see R2 again when we study regression.
Where’s the Difference?
Once ANOVA indicates that the groups do not all appear to have the same
means, what do we do?
Analysis of Variance for days
Source
DF
SS
MS
treatmen
2
34.74
17.37
Error
22
59.26
2.69
Total
24
94.00
Level
A
B
P
N
8
8
9
Pooled StDev =
Mean
7.250
8.875
10.111
1.641
StDev
1.669
1.458
1.764
F
6.45
P
0.006
Individual 95% CIs For Mean
Based on Pooled StDev
----------+---------+---------+-----(-------*-------)
(-------*-------)
(------*-------)
----------+---------+---------+-----7.5
9.0
10.5
Clearest difference: P is worse than A (CI’s don’t overlap)
Multiple Comparisons
Once ANOVA indicates that the groups do not all
have the same means, we can compare them two
by two using the 2-sample t test
• We
need to adjust our p-value threshold because we are
doing multiple tests with the same data.
•There are several methods for doing this.
• If we really just want to test the difference between one
pair of treatments, we should set the study up that way.
Tuckey’s Pairwise Comparisons
Tukey's pairwise comparisons
Family error rate = 0.0500
Individual error rate = 0.0199
95% confidence
Use alpha = 0.0199 for
each test.
Critical value = 3.55
Intervals for (column level mean) - (row level mean)
A
B
-3.685
0.435
P
-4.863
-0.859
B
These give 98.01%
CI’s for each pairwise
difference.
-3.238
0.766
98% CI for A-P is (-0.86,-4.86)
Only P vs A is significant
(both values have same sign)
Tukey’s Method in R
Tukey multiple comparisons of means
95% family-wise confidence level
diff
lwr
upr
B-A 1.6250 -0.43650 3.6865
P-A 2.8611 0.85769 4.8645
P-B 1.2361 -0.76731 3.2395
```
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