Transcript To Obtain a One

Using Statistics To Make
Inferences 4
Summary
Two sample t test
Paired comparisons
Assignment 1
Wednesday, 06 April 2016
3:19 PM
4.11
Summary
To perform and interpret a two
sample t test
Practical
Perform two sample t tests
4.22
Comparison Of Two Sample
Means
The null hypothesis is that the two
populations are identical. In that
case they would have the same
means. That is μ1 = μ2
Notation
n ,n 
1
2
 x , x  s , s 
1
2
1
2
Size of two samples Mean of two samplesStandard deviation
of two samples
Note, assess if s1 and s2 are numerically close.
We will test this later in the course.
4.33
But which is most appropriate, s1 or s2?
Key Equations
Pooled standard deviation
n  1s  n  1s
s
n  1  n  1
x1  x2
t test statistic t 
1 1
s

n1 n2
degrees of freedom
2
1
1
1
2
2
2
2
  n1  1  n2  1
Note the degrees of freedom match the divisor in the
pooled standard deviation. This is consistent with the
equivalent one sample test.
4.44
Confidence Interval
A confidence interval
is constructed
around the estimate
of a population
parameter.
1 1
x1  x2  t ( ) s

n1 n2
with ν = (n1–1)+(n2-1) degrees of freedom
x1  x2 Sample mean of difference
n1 n2 Sample sizes
Degrees of freedom, (n1–1)+(n2-1)
ν
Pooled standard deviation
s
Proportion of occasions that the
α
true mean lies outside the range
Critical value of t from tables
tν
Don’t
forget to
multiply or
divide
before you
add or
subtract
4.5
Confidence Interval

1 1
1 1
 , x1  x2  t ( ) s
 
 x1  x2  t ( ) s
n1 n2
n1 n2 

with ν = (n1–1)+(n2-1) degrees of freedom
x1  x2 Sample mean of difference
n1 n2 Sample sizes
Degrees of freedom, (n1–1)+(n2-1)
ν
Pooled standard deviation
s
Proportion of occasions that the
α
true mean lies outside the range
Critical value of t from tables
tν
4.6
Example 1
In a clinical test the following
scores were obtained for “normal”
and “diseased” patients.
Normal
10.3 11.8 12.6 8.6 9.2 10.1 10.2 7.4
Diseased 10.1 12.7 14.3 13.6 9.8 15.0 11.2 11.4
H0 is that
H1 is that
μ1 = μ2
μ1 ≠ μ2
under a two tail test
4.77
Calculations 1
n1  8 x1  10.025
s1  1.669
n2  8 x2  12.262
s2  1.936
n  1s  n  1s
s
n  1  n  1
2
1
1
1
2
2
2
2
7  1.669  7  1.936

 1.807
77
2
2
4.88
Calculations 1
n1  8
n2  8
x1  10.025
s  1.807
x2  12.262
x1  x2
10.025  12.262
t

 2.476
1 1
1 1
s

1.807 
n1 n2
8 8
  n1  1  n2  1  7  7  14
4.99
Conclusion 1
t  2.476
ν
14
p=0.05
1.761
p=0.025
2.145
  14
p=0.005
2.977
p=0.0025 p=0.0025
3.326
3.787
t14(0.025) = 2.145 and t14(0.005) = 2.977
In an attempt to “estimate” p.
Since 0.01 < p < 0.05 (two tail test) the result is
significant at the 5% level. The null hypothesis is
rejected and the mean levels are apparently
different.
4.10
10
SPSS 1
Analyze > Compare Means > Independent Samples t Test
4.11
11
SPSS 1
Analyze > Compare Means > Independent Samples t Test
Note the need to define the groups (normal/diseased)
4.12
12
SPSS 1
Basic descriptive statistics
Group Statistics
Score
Group
Normal
Diseas ed
N
8
8
Mean
10.025
12.263
Std. Deviation
1.6688
1.9361
Std. Error
Mean
.5900
.6845
4.13
13
SPSS 1
The method described here assumes equal variances.
More of this later in the course.
Note the p value is
less than .05
Independent Samples Test
Levene's Test for
Equality of Variances
F
Score
Equal variances
as sumed
Equal variances
not ass umed
.792
Sig.
.389
t-test for Equality of Means
t
df
Sig. (2-tailed)
Mean
Difference
Std. Error
Difference
95% Confidence
Interval of the
Difference
Lower
Upper
-2.476
14
.027
-2.2375
.9037
-4.1757
-.2993
-2.476
13.702
.027
-2.2375
.9037
-4.1797
-.2953
Note the confidence interval excludes 0
4.14
14
Excel 1
1
2
3
4
5
6
7
8
9
A
Normal
10.3
11.8
12.6
8.6
9.2
10.1
10.2
7.4
B
Diseased
10.1
12.7
14.3
13.6
9.8
15
11.2
11.4
C
D
p value
0.0267
=TTEST(A2:A9,B2:B9,2,2)
foreground
background
Note the final parameters.
2 – two tailed
2 – type – assuming equal variances
The
same p
value as
on the
previous
slide
4.15
15
Boxplot of Normal, Diseased
Boxplot of Normal, Diseased
15
14
13
Data
12
11
10
9
8
7
Normal
Diseased
4.16
16
Example 2
A study of 22 patients suffering from
Parkinsons disease was conducted. An
operation was performed on 8 of them,
while it improved their general condition
it might adversely affect their speech.
In the data a higher value indicates a
greater difficulty in speaking.
Operated 2.6 2.0 1.7 2.7 2.5 2.6 2.5 3.0
Others
1.2 1.8 1.9 2.3 1.3 3.0 2.2 1.3
1.5 1.6 1.3 1.5 2.7 2.0
4.17
17
Calculation 2
n1  8
x1  2.450
s1  0.411
n2  14 x2  1.829
s2  0.557
n  1s  n  1s
s
n  1  n  1
2
1
1
1
2
2
2
2
7  0.411  13  0.557

 0.511
7  13
2
2
4.18
18
Calculation 2
n1  8
x1  2.450
n2  14 x2  1.829
s  0.511
x1  x2
2.450  1.829
t

 2.742
1 1
1 1
s

0.511 
n1 n2
8 14
  n1  1  n2  1  7  13  20
4.19
19
Conclusion 2
t  2.742
ν
20
p=0.05
1.725
p=0.025
2.086
  20
p=0.005
2.845
p=0.0025 p=0.0025
3.153
3.552
t20(0.025) = 2.086 and t20(0.005) = 2.845
In an attempt to “estimate” p.
Since 0.01 < p < 0.05 (two tail test) the
result is significant at the 5% level. The null
hypothesis is rejected, the operation
appears to affect speech.
4.20
20
SPSS 2
The method described here assumes equal variances.
More of this later in the course.
Note the p value is
less than .05
Independent Samples Test
Levene's Test for
Equality of Variances
F
Score
Equal variances
as sumed
Equal variances
not ass umed
1.299
Sig.
.268
t-test for Equality of Means
t
df
Sig. (2-tailed)
Mean
Difference
Std. Error
Difference
95% Confidence
Interval of the
Difference
Lower
Upper
2.748
20
.012
.6214
.2262
.1496
1.0932
2.990
18.461
.008
.6214
.2079
.1855
1.0573
Note the confidence interval excludes 0
4.21
21
Excel 2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
A
Operated
2.6
2
1.7
2.7
2.5
2.6
2.5
3
B
Others
1.2
1.8
1.9
2.3
1.3
3
2.2
1.3
1.5
1.6
1.3
1.5
2.7
2
C
D
p value
0.0124
=TTEST(A2:A9,B2:B15,2,2)
foreground
background
The
same p
value as
on the
previous
slide.
4.22
22
Boxplot of Operated, Others
Boxplot of Operated, Others
3.0
Data
2.5
2.0
1.5
1.0
Operated
Others
4.23
23
Practical
In most cases we identified a difference,
as in the previous two examples. This need
not always be the case.
Note that in the practical we examine two
case studies in one, on waiting times
between eruptions of the Old Faithful
geyser, the evidence (see the next three
slides) suggests the two means are equal.
4.24
24
Practical
Waiting times between eruptions of Old Faithful geyser
for two different periods are given. Is there evidence
that the waiting times tend to be longer for one of the
periods than for the other?
WT1 Minutes between eruptions 1/8 to 5/8/, 1985
WT2 Minutes between eruptions 6/8 to 10/8, 1985
4.25
25
SPSS 3
The method described here assumes equal variances.
More of this later in the course.
Note the p value is
greater than .05
Independent Samples Test
Levene's Test for
Equality of Variances
F
Time
Equal variances
as sumed
Equal variances
not ass umed
2.079
Sig.
.151
t-test for Equality of Means
t
df
Sig. (2-tailed)
Mean
Difference
Std. Error
Difference
95% Confidence
Interval of the
Difference
Lower
Upper
-.590
198
.556
-1.1400
1.9333
-4.9525
2.6725
-.590
196.984
.556
-1.1400
1.9333
-4.9526
2.6726
Note the confidence interval includes 0
4.26
26
Boxplot of WT1, WT2
Boxplot of WT1, WT2
110
100
Data
90
80
70
60
50
40
WT1
WT2
4.27
27
Paired Comparisons
- Example 3
Certain mental tasks are performed
before and after exercise. The scores
for each subject were recorded.
Subject
1
Exercise 46
Relaxed 53
2
38
46
3
62
60
4
54
58
5
42
49
6
37
34
7
55
65
8
52
53
9
41
47
10
39
43
We want to test the difference, so subtract.
Does exercise have an effect?
4.28
28
Paired Comparisons
- Example 3
Certain mental tasks are performed
before and after exercise. The scores
for each subject were recorded.
Subject
Exercise
Relaxed
Difference
1
46
53
7
2
38
46
8
3
62
60
-2
4
54
58
4
5
42
49
7
6
37
34
-3
7
55
65
10
8
52
53
1
9
41
47
6
Perform a one sample t-test on the
difference. No effect implies a zero value.
4.29
29
10
39
43
4
Calculation 3
Differences (d) (recall the sign test
7 8 -2 4 7 -3 10 1 6 4
n = 10
Σd = 7 + 8 + ... + 6 + 4 = 42
Σd2 = 72 + 82 + ... + 62 + 42 = 344
n = 10 Σd = 42 Σd2 = 344
4.30
30
Calculation 3
n = 10 Σd = 42 Σd2 = 344
42
xd 
 4.2
10

1 
d 
d


n
i 1
 i 1 
n
vard 

n

2
n 1
n  10
2
1
344  42 2
10

 18 .6222
10  1
xd  4.20 sd  4.32
4.31
31
Calculation 3
n  10
xd  4.20 sd  4.32
xd  0
4.20
t

 3.074
sd
4.32
10
n
  n 1  9
The
population
value being
tested is
zero.
Exercise
claimed to
have no
effect.
Note the degrees of freedom match the divisor in the
standard deviation (see vard on the previous slide).
4.32
32
Conclusion 3
t  3.074
 9
ν
p=0.05
p=0.025
p=0.005
p=0.0025
p=0.0025
9
1.833
2.262
3.250
3.690
4.297
t9(0.025) = 2.262 and t9(0.005) = 3.250
In an attempt to “estimate” p.
Since 0.01 < p < 0.05 (two tail test) the result is
significant at the 5% level. The null hypothesis is
rejected, the mean performance levels appear to
differ.
4.33
33
SPSS 3
Transform > Compute Variable
4.34
34
SPSS 3
Transform > Compute Variable
4.35
35
SPSS 3
Analyze > Compare Means > One Sample t Test
Note the test value is 0, no difference.
4.36
36
SPSS 3
Note the p value is
less than .05
One-Sample Test
Test Value = 0
difference
t
3.078
df
9
Sig. (2-tailed)
.013
Mean
Difference
4.20000
95% Confidenc e
Int erval of t he
Difference
Lower
Upper
1.1130
7.2870
Note the confidence interval excludes 0
4.37
37
SPSS 3
Or directly as a t test on paired samples
Analyze > Compare Means > Paired Samples t Test
Some additional output are generated
4.38
38
SPSS 3
Paired Samples Statistics
Mean
Pair 1
N
Std. Deviation
Std. Error Mean
Exercise
46.60
10
8.618
2.725
Relaxed
50.80
10
9.016
2.851
Paired Samples Correlations
N
Pair 1
Exercise & Relaxed
Correlation
10
.881
Sig.
.001
4.39
39
SPSS 3
Note the p value is
less than .05
Paired Samples Test
Paired Differences
95% Confidence Interval of the
Difference
Mean
Pair 1
Exercise - Relaxed
-4.200
Std. Deviation
4.315
Std. Error Mean
1.365
Lower
-7.287
Upper
-1.113
t
-3.078
df
Sig. (2-tailed)
9
.013
Note the confidence interval excludes 0
Of course the result is unchanged
4.40
40
Excel 3
1
2
3
4
5
6
7
8
9
10
11
A
Exercise
46
38
62
54
42
37
55
52
41
39
B
Relaxed
53
46
60
58
49
34
65
53
47
43
C
D
p value
0.0132
=TTEST(A2:A11,B2:B11,2,1)
foreground
background
Note the final parameter.
1 – type – paired values
The
same p
value as
on the
previous
slide.
4.41
41
Normal Approximation
For greater than 30 degrees of freedom the
Students t distribution is well approximated by
the standard normal distribution.
You will notice that the final row in most
Students t tables simply give the normal values.
Skip
review
4.42
42
SPSS t Tests
The software offers three options which will
now be reviewed
1. One Sample t Test
2. Independent Samples t Test
3. Paired Samples t Test
4.43
43
One-Sample t Test
The One-Sample t Test procedure tests whether the
mean of a single variable differs from a specified
constant.
4.44
44
One-Sample t Test
Examples
A researcher might want to test whether the
average IQ score for a group of students differs
from 100.
A cereal manufacturer can take a sample of boxes
from the production line and check whether the
mean weight of the samples differs from 1.3 pounds
at the 95% confidence level.
4.45
45
One-Sample t Test
Statistics
For each test variable: mean, standard deviation, and
standard error of the mean.
The average difference between each data value and
the hypothesized test value, a t test that tests that
this difference is 0, and generates a confidence
interval for this difference (you can specify the
confidence level).
4.46
46
One-Sample t Test
Data
To test the values of a quantitative variable against a
hypothesized test value, choose a quantitative
variable and enter a hypothesized test value.
4.47
47
One-Sample t Test
Assumptions
This test assumes that the data are normally
distributed; however, this test is fairly robust to
departures from normality.
4.48
48
One-Sample t Test
To Obtain a One-Sample t Test
From the menus choose:
Analyse
Compare Means
One-Sample t Test
Select one or more variables to be tested against the
same hypothesized value.
Enter a numeric test value against which each sample
mean is compared.
Optionally, click Options to control the treatment of
missing data and the level of the confidence interval.
4.49
49
Independent-Samples t Test
The Independent-Samples t Test procedure
compares means for two groups of cases.
Ideally, for this test, the subjects should be
randomly assigned to two groups, so that any
difference in response is due to the treatment (or
lack of treatment) and not to other factors.
4.50
50
Independent-Samples t Test
This is not the case if you compare average income for
males and females.
A person is not randomly assigned to be a male or
female.
In such situations, you should ensure that differences
in other factors are not masking or enhancing a
significant difference in means.
Differences in average income may be influenced by
factors such as education (and not by sex alone).
4.51
51
Independent-Samples t Test
Example
Patients with high blood pressure are randomly
assigned to a placebo group and a treatment group.
The placebo subjects receive an inactive pill, and the
treatment subjects receive a new drug that is
expected to lower blood pressure. After the
subjects are treated for two months, the twosample t test is used to compare the average blood
pressures for the placebo group and the treatment
group. Each patient is measured once and belongs to
one group.
4.52
52
Independent-Samples t Test
Statistics
For each variable: sample size, mean, standard deviation,
and standard error of the mean.
For the difference in means: mean, standard error, and
confidence interval (you can specify the confidence level).
Tests: Levene's test for equality of variances and both
pooled-variances and separate-variances t tests for
equality of means.
4.53
53
Independent-Samples t Test
Data
The values of the quantitative variable of interest are in a
single column in the data file.
The procedure uses a grouping variable with two values to
separate the cases into two groups. The grouping variable
can be numeric (values such as 1 and 2 or 6.25 and 12.5) or
short string (such as yes and no).
As an alternative, you can use a quantitative variable, such
as age, to split the cases into two groups by specifying a
cut point (cut point 21 splits age into an under-21 group
and a 21-and-over group).
4.54
54
Independent-Samples t Test
Assumptions
For the equal-variance t test, the observations should
be independent, random samples from normal
distributions with the same population variance.
For the unequal-variance t test, the observations
should be independent, random samples from normal
distributions.
The two-sample t test is fairly robust to departures
from normality. When checking distributions
graphically, look to see that they are symmetric and
have no outliers.
4.55
55
Independent-Samples t Test
To Obtain an Independent-Samples t Test
From the menus choose:
Analyse
Compare Means
Independent-Samples t Test.
Select one or more quantitative test variables. A
separate t test is computed for each variable.
Select single groupings variable, and then click Define
Groups to specify two codes for the groups that you
want to compare.
Optionally, click Options to control the treatment of
missing data and the level of the confidence interval.
4.56
56
Paired-Samples t Test
The Paired-Samples t test procedure compares the
means of two variables for a single group.
The procedure computes the differences between
values of the two variables for each case and tests
whether the average differs from 0.
4.57
57
Paired-Samples t Test
Example
In a study on high blood pressure, all patients are
measured at the beginning of the study, given a
treatment, and measured again. Thus, each subject has
two measures, often called before and after measures.
4.58
58
Paired-Samples t Test
Example
An alternative design for which this test is used is a
matched-pairs or case-control study, in which each
record in the data file contains the response for the
patient and also for his or her matched control
subject.
In a blood pressure study, patients and controls
might be matched by age (a 75-year-old patient with
a 75-year-old control group member).
4.59
59
Paired-Samples t Test
Statistics
For each variable: mean, sample size, standard
deviation, and standard error of the mean.
For each pair of variables: correlation, average
difference in means, t tests, and confidence
interval for mean difference (you can specify the
confidence level). Standard deviation and
standard error of the mean difference.
4.60
60
Paired-Samples t Test
Data
For each paired test, specify two quantitative
variables (interval level of measurement or ratio level
of measurement).
For a matched-pairs or case-control study, the
response for each test subject and its matched
control subject must be in the same case in the data
file.
4.61
61
Paired-Samples t Test
Assumptions
Observations for each pair should be made under the
same conditions. The mean differences should be
normally distributed.
Variances of each variable can be equal or unequal.
4.62
62
Paired-Samples t Test
To Obtain a Paired-Samples t Test
From the menus choose:
Analyse
Compare Means
Paired-Samples t Test.
Select one or more pairs of variables
Optionally, click Options to control the treatment of
missing data and the level of the confidence interval.
4.63
63
Read
Read Howitt and Cramer pages 109-133
Read Howitt and Cramer (e-text)
pages 189-201
Read Russo (e-text) pages 151-158
Read Davis and Smith pages 237-264
4.64
64
Practical 4
This material is available from the
module web page.
http://www.staff.ncl.ac.uk/mike.cox
Module Web Page
4.65
65
Practical 4
This material for the practical is
available.
Instructions for the practical
Practical 4
Material for the practical
Practical 4
4.66
66
Assignment
See Stage I Handbook
“Assignments submitted in hard copy only
Some modules may involve assignments that
are completed on paper only and do not
require electronic submission. If the
submission is only in paper format you will
hand it in at the School Office with a cover
sheet attached.”
Collect a cover sheet from reception as you
submit your work.
4.67
67
Whoops!
The Conservatives have been accused of
being “totally out of touch” after claiming
more than half of girls in the most deprived
areas of Britain fell pregnant before their
18th birthday.
However, official figures for those areas
suggested that the number of under-18 girls
who got pregnant was more like 54 per 1,000.
5.4% not 54%!!
Telegraph
15 Feb 2010
4.68
68
Whoops!
Explain yourself!
4.69
69