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WINKS
Online Manual
Chapter
4 Part 3
t-Tests
and ANOVA
Independent
Group t-test Example
Independent
group analysis is appropriate when observations are taken from groups in which
subjects in one group do not appear in another group. That is, the observations
within as well as between groups are independent of one another. A t-test is
performed when there are two groups, and an
ANOVA is performed when there are three or more
groups being compared. When performing a t-test or ANOVA on two or more
independent groups, you are testing the hypotheses:
Ho:
The difference in the means of the groups is zero.
Ha: The difference in the means
of the groups is not zero.
For
a two-sample t-test, two t-statistics are calculated, one for the case in which
the variances of the two samples are equal and the other for use in the case of
unequal variances. WINKS performs a test of the hypothesis that the variances
are equal, that is, a test to determine if the variances are equal, and reports
a p-value. If this p-value is small (e.g., less than 0.05), the hypothesis of
equal variances is rejected and you use the t-statistic for unequal variances.
If the p-value is large, use the t-statistic for equal variances.
Since
the observations are all independent of one another, each observation is entered
as an individual record in the database. The number of data records must be the
same as the total number of observations. Each record includes the response
value of one observation and a number or character to indicate to which
treatment group it belongs. That is, there will be two fields (variables), one
in which to record the response and one in which to indicate the group.
Independent
Group t-test Example
The data used here are heights of 13 plants grown using two different
fertilizers. Suppose you want to
know if there is a difference in the average heights of plants in the two
treatment groups.
Data
for independent group t-test (fertilizer study)
Present
Fertilizer Newer
Fertilizer
46.2 cm
51.3 cm
55.6
52.4
53.3
54.6
44.8
52.2
55.4
64.3
56.0
55.0
48.9
In
order to enter this data into a database, you must assign group numbers (or
letters) such as Present = 1 and Newer = 2, or you could use P and N (if the
variable is of the character type). Follow these steps to perform this analysis:
Step
1: Since the
observations are independent, the database will include thirteen records (one
for each plant) and two fields (one for the response and one for the group
indicator. Create a database (File/New Database) using the pre-defined structure
named "For Independent Group t-test or ANOVA." This will create a
database with the fields GROUP and OBS. Or, use the database on disk named
FERTILIZ.DBF and disk to step 3.
Step
2: Enter the
data will be in two columns, like this:
Rec.
No. GROUP
OBS (HEIGHT)
1
1
46.2
2
1
55.6
3
1
53.3
4
1
44.8
5
1
55.4
6
1
56.0
7
1
48.9
8
2
51.3
9
2
52.4
10
2
54.6
11
2
52.2
12
2
64.3
13
2
55.0
Step
3: Choose the t-tests and ANOVA option from the Analyze menu,
then choose the "Independent
group (t-test/ ANOVA)" option.
Step
4: Select
GROUP as the Group variable and OBS (Height)
as the data (response) variable and click Ok. The results appear in the viewer.
The means for each group (1=Present, 2=Newer) are displayed. A test for equality
of variance (F=1.02) is also
performed to see if the variances of the two
groups can be considered equal. This is necessary for deciding which
t-statistic and p-value to use for the test on means. A p-value for the equal
variances test is displayed. A large p-value (e.g., greater than 0.05)
indicates that you can consider the variances to be equal. In this case,
p=0.4807, large enough to consider the variances to be equal. If the variances
are equal, according to this test, you use the "Equal variances"
t-statistic. Otherwise, use the "Unequal variances" result. In this
case, the two t-statistics are identical at -1.32.
The t-test is performed with 11 degrees of freedom, and the p-value
associated with the test is 0.213. A large p-value (greater than the
significance level, e.g., 0.05) is usually interpreted to mean that there is no
significant difference in the means -- the
null hypothesis of equal means is not rejected.
That is, there is not enough evidence to conclude that the average height
of plants grown with the newer fertilizer is significantly different from the
average height of plants grown with the present fertilizer.
Step
5: Click the
Graph button to display a graphical comparison. This plot shows a box plot
comparison. This is essentially the same as a by-group plot.
One-Way
ANOVA Example
When
more than two independent groups are compared with respect to one variable,
one-way or single factor analysis of variance techniques are appropriate. When
performing an ANOVA on two or more independent groups, you are testing the
hypotheses:
Ho:
The difference in the means of the groups is zero.
Ha: The difference in the means
of the groups is not zero.
This
example uses data for hogs which have been randomly assigned to four groups,
with each group being given a different feed. The response is weight gain.
Data
for Independent Group ANOVA
Gp
1
Gp 2
Gp 3
Gp 4
60.8
78.7
92.6
86.9
67.0
77.7
84.1
82.2
54.6
76.3
90.5
83.7
61.7
79.8
90.3
The
database to analyze this data is similar to the one used for the t-test example,
differing only with respect to the number of groups. In fact, this one-way ANOVA
is an extension of the t-test when there are three or more groups. To perform
this analysis, use these steps:
Step
1: Create a
database (File/New Database) using the pre-defined structure named "For
Independent Group t-test or ANOVA." This will create a database with the
fields GROUP and OBS. Or, use the database on disk named HOGDATA.DBF and disk to
step 3.
The
groups will be numbered 1,2,3,4 according to the type of feed used. The response
field will be OBS.
Step
2: Enter the
data. The data, as entered into the database will look like this:
NO
GROUP OBS
(WEIGHT)
1
1
60.8
2
1
67.0
3
1
54.6
4
1
61.7
5
2
78.7
6
2
77.7
etc...
14
4
83.7
15
4
90.3
Step
3: Choose the t-tests and ANOVA option from the Analyze menu,
then choose the "Independent
group (t-test/ ANOVA)" option.
Step
4: Select
GROUP as the Group field and OBS (Weight) as the data field and click Ok. The
results appear in the viewer:
---------------------------------------------------------------------------
Independent Group Analysis C:\WINKS\HOGDATA.DBF
---------------------------------------------------------------------------
Grouping variable is GROUP
Analysis variable is OBS
Group Means and Standard Deviations
1: mean = 61.025 s.d. = 5.0822 n = 4
2: mean = 78.125 s.d. = 1.4886 n = 4
3: mean = 89.0667 s.d. = 4.4276 n = 3
4: mean = 85.775 s.d. = 3.5976 n = 4
Analysis of Variance Table
Source S.S. DF MS F Appx P
---------------------------------------------------------------------------
Total 1923.41 14
Treatment 1761.24 3 587.08 39.82 <.001
Error 162.17 11 14.74
Error term used for comparisons = 14.74 with 11 d.f.
Critical q
Newman-Keuls Multiple Comp. Difference P Q (.05)
------------------------------------------------------------------------
Mean(3)-Mean(1) = 28.0417 4 13.523 4.256 *
Mean(3)-Mean(2) = 10.9417 3 5.277 3.82 *
Mean(3)-Mean(4) = 3.2917 2 1.587 3.113
Mean(4)-Mean(1) = 24.75 3 12.892 3.82 *
Mean(4)-Mean(2) = 7.65 2 3.985 3.113 *
Mean(2)-Mean(1) = 17.1 2 8.907 3.113 *
Homogeneous Populations, groups ranked
Gp Gp Gp Gp
1 2 4 3
------
---
---
This is a graphical representation of the Newman-Keuls multiple comparisons
test. At the 0.05 significance level, the means of any two groups
underscored by the same line are not significantly different.
Step
5: The
results of this test are summarized in the p-value. In this case, the small
p-value (p<0.001) means that there is a significant difference between
groups. This is taken as evidence of a difference between feeds, a difference
not due to chance. The ANOVA tells you only that there is a difference among the
feeds. In order to find out which groups are significantly different from which
others, examine the multiple comparison results. The Newman-Keuls multiple
comparison test (or whichever you specified at setup) will describe which of the
means are significantly different from which others (at the 0.05 significance
level).
In
the multiple comparison table, comparisons marked with an asterisk “*” are
significantly different at the 0.05 significance level (alpha-level). The group
numbers are given in increasing order of the value of their group means. That
is, Group 1 has the smallest mean, Group 3 the largest. At the 0.05 significance
level, the means of any two groups underscored by the same line are not
significantly different.
1)
The mean for group 1 (feed 1) is statistically significantly less than the means
for all other groups.
2)
The mean for group 2 (feed 2) is significantly greater than the mean for group
1, and significantly less than the means of groups 4 and 3.
3)
The means for groups 4 and 3 are not significantly different from each
other, but they are both significantly greater than the means of groups 1
and 2.
You
can conclude that feeds 3 and 4 are better than feeds 1 and 2, but there is not
enough evidence to say that either feed 3 or 4 is the best overall.
Step
6 Click on
the Graph button to view the graphical comparison: This is the same graph as
described in the section "By Group Plots."
Using the Option button, you can
use this graph to show comparison in terms of error bars, box plots, etc.
Independent
group test from summary data
Consider
the fertilizer comparison problem described in the previous independent group
t-test example. Suppose you don't have all the data. That is,
you aren't given the response of each plant, but you are given the means,
standard deviations and sizes of the two groups. You can still perform an
analysis.
Summary
data for fertilizer study
Group
Mean
St. Dev. Sample Size
1
51.457 4.748
7
2
54.967 4.794
6
Since
you are not given individual data values, you don't have to create a database.
Follow these steps to perform an analysis.
Step
1 From the
Analyze menu select t-tests and ANOVA. Then select "Ind. Group from
summary data."
Step
2 You will
be asked for the number of groups. Enter 2. Then you will be asked to enter the
mean, standard deviation and sample size for each of the groups. In this case,
for group 1 enter 51.457,4.748,7 and press Enter. For group 2 enter
54.967,4.794,6.
Note:
Press Tab to move from field to field. Press Enter when you have entered all the data.
Step
3 The
interpretation of the results is the same as in the independent group t-test
example using this same data.
Note:
You can also perform an Independent group one-way ANOVA using summary data. For
example, perform an ANOVA using the summary data from the
"HOGDATA" example.
Paired t-test (Repeated Measures)
Repeated
measures are observations taken on the same or related subjects over time or in
differing circumstances. Examples would be weight loss, or reaction to a drug
across time. Repeated measures may also be matched subjects. As in the
independent groups module, a t-test is performed when there are two groups (two
repeated measures), and an analysis of variance is performed by WINKS if there
are more than two groups. The ANOVA determines if there is a difference in the
means across groups or repeated measures. A multiple comparison procedure
further identifies where the differences lie.
In
a database for paired or repeated measures data, each record represents one
subject (e.g., person, animal). There must be one field for each repeated
measure (each treatment group). For paired data, there are two groups, hence two
fields. Thus, in each record, there is a field in which to enter data from each
observation (treatment) on that subject. This repeated measures (paired)
analysis requires that all values be available for each subject and any subject
with missing values is eliminated from the analysis. That is, a data record must
have a value for each field, or it will be eliminated. The hypotheses being
tested with a paired t-test or a repeated measures ANOVA is:
Ho:
There is no difference among means of the groups (repeated measures).
Ha: There is a difference among means of the groups.
For
comparing matched or paired data (not independent) from two groups, a paired
t-test is used. For this test both groups must have the same number of data
values, i.e., the two samples must be the same size. A paired t-test is actually
a single sample t-test on the mean of the differences between the two data
values in each pair.
For
example, in a before/after study, the difference "after minus before"
(or "before minus after") is taken for each pair and the average of
these differences calculated. If this average of differences is found by a
single sample t-test to be
significantly different from zero, the conclusion is that there is a change.
The
data in this example are before and after weights for eight persons on a diet.
Notice that in this case, both data values are taken from the SAME entity
(person). Follow these steps to perform this analysis:
Step
1: Create a
database with two fields (BEFORE and AFTER) and eight records, one for each
person. Since the observations are paired, not independent, the database
reflects this by having each record contain a pair of observations.
Use the pre-defined database structure named "Paired t-test or
McNemar’s Test." This will create a database with the fields REP1 and
REP2. The REP1 will be used for Before and REP2 will be used for After. Of
course, you can choose to create a custom database and enter a structure
containing the fields named BEFORE and AFTER. (Or, open the database named DIET,
and skip to step 3.) The data for the paired t-test is:
Person
Before
After
1
162
168
2
170
136
3
184
147
4
164
159
5
172
143
6
176
161
7
159
143
8
170
145
Step
2: Enter the
data for the eight records. The database should look similar to the listing of
the data above.
Step
3: Choose
the t-tests and ANOVA option from the Analyze menu. Then choose the
"Paired/Rep. Measure (t-test/ANOVA) option.
Step
4: Select
REP1 (BEFORE) as the first field and REP2 as the second field and click Ok.
Step
5: The
results are shown in the viewer. A portion of the output is shown below:
Means and standard deviations for 2 repeated measures:
1)REP1: mean = 169.625 s.d. = 8.07001
2)REP2: mean = 150.25 s.d. = 11.04213
Mean Difference = 19.375 s.d.(difference) = 14.78356
95% C.I. about Mean Difference is (7.01367, 31.73633)
Paired t-test
--------------
Hypotheses:
Ho: The mean difference between pairs is 0.
Ha: The mean difference between pairs is not 0.
Calculated t = 3.70687 with 7 D.F. p = 0.0076 (two-sided)
Since p <= 0.05, at the 0.05 significance level you have evidence
to reject the null hypothesis and conclude that the mean difference
between pairs is not 0.
For a one-sided test, you must adjust the p-value according to
the direction of your alternative hypothesis.
The
means and standard deviations for each group are reported, but more importantly,
the mean difference between BEFORE and AFTER measurements is given. The
statistical procedures are performed on this average difference. A 95%
confidence interval for the mean difference is given, as well as a
calculated t-statistic and a p-value. These results are interpreted like
those of a single sample t-test with null hypothesis: mean=0, and alternative
hypothesis: mean <> 0. The calculated t-statistic is 3.70687. The test is
performed with 7 degrees of freedom, and the p-value associated with the test is
0.0076. A small p-value such as this is usually interpreted to indicate
rejection of the null hypothesis and leads to the conclusion that the average
difference in BEFORE and AFTER weights is not zero, i.e., there is evidence of a
significant (at the 0.05 level) change of weight in these eight subjects on
average.
Step
6: Click
Graph to display a graphical comparison of the results.
One-way
repeated measures ANOVA
For
more than a pair of repeated measures on the same subject, a one-way repeated
measures analysis of variance is appropriate. The data in this example are
repeated measures of reaction times of five persons after being treated with
four drugs in randomized order.
One-way
repeated measures ANOVA data
Drug1
Drug2
Drug 3
Drug 4
31
29
17
35
15
17
11
23
25
21
19
31
35
35
21
45
27
27
15
31
To
perform this analysis, follow these steps:
Step
1: Create a
database with four fields (DRUG1, DRUG2, DRUG3 and DRUG4) Use the pre-defined
database structure named "Paired t-test or McNemar’s Test." Or, open
the database named DRUG.DBF and disk to step 3.
Step
2: For the
first record, enter the data for the first person 31,29,17,35. The second record
will contain 15,17,11,23 and so forth. After you finish entering your data, the
database should look similar to the data listing above.
Step
3: Choose
the t-tests and ANOVA option from the Analyze menu. Then choose the
"Paired/Rep. Measure (t-test/ANOVA) option.
Step
4 Select
DRUG1, DRUG2, DRUG3 and DRUG4 as the fields to use and click Ok.
Step
5 The
results are displayed in the viewer: Partial output is shown below:
---------------------------------------------------------------------------
Repeated Measures Analysis Summary C:\WINKS\DRUG.DBF
---------------------------------------------------------------------------
Number of repeated measures is 4
Number of subjects read in 5
Means and standard deviations for 4 repeated measures:
1)DRUG1: mean = 26.6 s.d. = 7.53658
2)DRUG2: mean = 25.8 s.d. = 7.01427
3)DRUG3: mean = 16.6 s.d. = 3.84708
4)DRUG4: mean = 33.0 s.d. = 8.0
Repeated Measures Analysis of Variance
Source -----S.S.----- --DF-- MS F Appx p
---------------------------------------------------------------------------
Between Subject 648.00 4
Within Subject 775.00 15
Rep. Factor 683.80 3 227.93 29.99 <.001
Error 91.2 12 7.60
---------------------------------------------------------------------------
Total 1423.00 19
Error term used for comparisons = 7.6 with 12 d.f.
Critical q
Newman-Keuls Multiple Comp. Difference P Q (.05)
------------------------------------------------------------------------
Mean( 4)-Mean( 3) = 16.4 4 13.302 4.199 *
Mean( 4)-Mean( 2) = 7.2 3 5.84 3.773 *
Mean( 4)-Mean( 1) = 6.4 2 5.191 3.082 *
Mean( 1)-Mean( 3) = 10.0 3 8.111 3.773 *
Mean( 1)-Mean( 2) = 0.8 2 .649 3.082
Mean( 2)-Mean( 3) = 9.2 2 7.462 3.082 *
Homogeneous Populations, repeated measures ranked
Gp 1 refers to DRUG1
Gp 2 refers to DRUG2
Gp 3 refers to DRUG3
Gp 4 refers to DRUG4
The
results of this ANOVA are summarized in the p-value. In this case, the small
p-value (The p<0.001 on the “Repeated Factor" line in the ANOVA
table.) means that there is a statistically significant difference in the mean
response times for the four drugs.
In
the multiple comparison table, comparisons marked with an asterisk “*” are
significantly different at the 0.05 significance level (alpha-level). A
graphical description of the multiple comparisons is given by:
Gp Gp Gp Gp
3 2 1 4
----------
-----
-----
This
tells you that groups 2 and 1 (DRUGS 2 and 1) are not significantly different at the 0.05 significance level. The
Group 3 (DRUG 3) mean is significantly smaller than all other groups' and the
Group 4 (DRUG 4) mean is significantly larger than all others.
Step
6 : Click on
the Graph button to view a comparison graph.
Single Sample t-test Analysis
The
single sample analysis allows you to choose a single variable, and test a
hypothesis that the mean differs from an hypothesized mean. You must enter the
hypothesized population mean. The hypotheses you are testing in this case are:
Ho:
The mean equals the hypothesized value.
Ha: The mean does not equal the hypothesized value.
Follow
these steps to perform the analysis:
Step
1: Choose
the Open Database option from the FILE pull-down menu, then choose the EXAMPLE
database from the displayed list. Or, create a database with at least one
numeric field.
Step
2 From
Analyze, choose the t-test and ANOVA option. Then, choose the “Single sample
t-test” option.
Step
3: Choose
the field you wish to use. In this case, choose the field TIME1.
Step
4: WINKS
will display the mean, standard deviation and sample size of the field TIME1,
and will ask you for the hypothesized mean. In this case, you want to know if
the mean is 20. Enter 20 as the hypothesized mean. Choose Calculate and End.
Step
5:
The results of the analysis will be displayed.
---------------------------------------------------------------------------
Single Sample t-test C:\WINKS\EXAMPLE.DBF
---------------------------------------------------------------------------
Variable Name is TIME1
N = 50 Missing or Deleted = 0
Mean = 21.268 St. Dev (n-1) = 1.71695
Null Hypothesis: mean(POPULATION) = 20
Calculated t = 5.22 with 49 D.F. p = < 0.001 (2 - sided test)
95% C.I. about Mean is (20.77995, 21.75605)
A
small p-value, such as the one in this case p<0.001, supports rejection of
the null hypothesis. That is, you can conclude that the mean of TIME1 in the
population is statistically significantly different from 20 based on this test
procedure.
Note:
Even if you don't have all the data values, you can test whether the mean of a
population is equal to some hypothesized value if you know the mean, standard
deviation and sample size of the data values in a sample. In this case, choose
the Single sample t-test - from summary data option.
Dunnett's Test
Dunnett's
test is a multiple comparison procedure following a one-way ANOVA either from
database data or from summary data that compares a control mean with the other
means in the analysis. Data entry is the same as for a One-Way Independent Group
ANOVA. You will be asked to specify which field in represents the control group,
and multiple comparisons will be performed accordingly.
Continue to Chapter
4. Part 4. (Non-Parametric Procedures.)
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