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WINKS
Online Manual
Chapter
4 Part 6
Frequency
and Crosstabulation Procedures
Frequencies
Analysis
In
the Frequencies analysis option, WINKS "counts" the occurrence of each
data value for a single variable or field and displays that information in a
table.
For
example, in the EXAMPLE database file, one of the fields (variables) in this
database is STATUS -- referring to socioeconomic status. Suppose you want to
know how the total data set is divided up into the five levels of STATUS. To
perform this analysis follow these steps:
Step
1: Open the
database named EXAMPLE.
Step
2 From the
Analyze menu, “Crosstabulations, Frequencies, Chi Square,” then
“Frequencies.”
Step
3: Select
STATUS from the field list and click “Add”. Optionally, you can select a
Group field (GROUP) -- then multiple frequency tables will be created, one for
each group. Click Ok.
Step
4: WINKS
will count the data in each of the five categories of STATUS and display the
results in the viewer.
---------------------------------------------------------------------------
Frequency Tables C:\WINKS46P\EXAMPLE.DBF
---------------------------------------------------------------------------
Number of records in database = 50
------ GROUP = A
Frequency Table for STATUS
Cumulative Cumulative
STATUS Frequency Percent Frequency Percent
-----------------------------------------------------------
2 1 9.09 1 9.09
3 3 27.27 4 36.36
4 1 9.09 5 45.45
5 6 54.55 11 100.0
------ GROUP = B
Frequency Table for STATUS
Cumulative Cumulative
STATUS Frequency Percent Frequency Percent
-----------------------------------------------------------
1 2 6.9 2 6.9
2 6 20.69 8 27.59
3 2 6.9 10 34.48
4 4 13.79 14 48.28
5 15 51.72 29 100.0
------ GROUP = C
Etc...
Performing
a Goodness of Fit Analysis
A
goodness-of-fit test of a single population is a test to determine if the
distribution of observed frequencies in the sample data closely matches the
expected number of occurrences under a hypothetical distribution of the
population. The data observations must be independent and each data value can be
counted in one and only one category. It is also assumed that the number of
observations is fixed. The hypotheses being tested are
Ho:
The population distribution follows the hypothesized distribution.
Ha: The population does not follow the hypothesized distribution.
A
Chi-Square statistic is calculation and a decision can be made based on the
p-value associated with that statistic. A low p-value indicates rejection of the
null hypothesis. That is, a low p-value indicates that the data do not follow
the hypothesized, or theoretical, distribution.
For
example, data for this test comes from the Zar (1974), page 46. According to a genetic theory, crossbred
pea plants show a 9:3:3:1 ratio of yellow smooth, yellow wrinkled, green smooth,
green wrinkled offspring. Out of 250
plants, under the theoretical ratio (distribution) of 9:3:3:1, you would expect
about
(9/16)x250=140.625
yellow smooth peas,
(3/16)x250=46.875 yellow
wrinkled peas
(3/16)x250=46.875 green
smooth peas
(1/16)x250=15.625 green
wrinkled peas
After
growing 250 of these pea plants, you observe that
152
have yellow smooth peas
39 have yellow wrinkled peas
53 have green smooth peas
6 have green
wrinkled peas
To
perform this analysis, use the following steps:
Step
1: From the
Analyze menu, select “Crosstabulations, Frequencies, Chi Square” then choose
the "Goodness-of-Fit" option.
Step
2: You will
be prompted to enter the number of categories. In this case, enter 4 for the
four categories of peas (yellow smooth, yellow wrinkled, green smooth, green
wrinkled).
Step
3: A dialog
box will appear allowing you to enter the observed data and either the expected
values or ratios. In this example, click (select) on the check box labeled
“Check to enter ratios rather than expected values.” Enter the observed
values from the table above:
152,
39, 53, 6
and
the ratios
9,3,3,1
Press
Tab to move from cell to cell.
Click Calculate and the calculated chi-square statistic in this case is 8.97 and
the p-value is 0.031 is displayed at the bottom of the dialog box. Click Exit
and the output will be displayed on the viewer.
---------------------------------------------------------------------------
Goodness of Fit
---------------------------------------------------------------------------
Obs. Exp.
|-------|----------|
1 | 152| 140.625|
|-------|----------|
2 | 39| 46.875|
|-------|----------|
3 | 53| 46.875|
|-------|----------|
4 | 6| 15.625|
|-------|----------|
250 250.
Calculated CHI SQUARE = 8.97
Degrees of freedom = 3 Appx p = 0.031
At
a 0.05 level of significance, this p-value indicates that there is enough
evidence to reject the null hypothesis that the observed values follow the
theoretical distribution. That is, the test (at the 0.05 significance level)
suggests that a 9:3:3:1 ratio of yellow smooth:yellow wrinkled:green
smooth:green wrinkled peas is not an appropriate distribution for the population
from which these data are taken.
Note:
you can perform several analyses in the Goodness of Fit dialog box, and each
will be displayed in the viewer when you select End.
Performing a Crosstabulation Analysis (Chi-square)
Crosstabulations
can be used to perform a chi-square test for independence or a chi-square test
for homogeneity. A two-way table is constructed that displays the number of
counts for each category. It must be possible to assume that the data
observations are independent and that each data value can be counted in one and
only one category. It is also assumed that the number of observations is fixed.
WINKS allows you to enter data for a two-way table from the keyboard or from a
database.
Entering
Data from the Keyboard
When
you choose to enter the two-way table from the keyboard (“Chi-Square table for
Keyboard” option), WINKS will ask you the size of the table (number of rows
and columns). A blank table will be presented on the screen, and you will then
be prompted to enter a number in each cell of the table.
Entering
Data from a Database
If
you choose to enter the information from a database, you will be prompted to
indicate what tables are to be calculated. Select one or more fields for the
“Data field” (top right hand list box) and select one or more fields for the
“By Var” field (bottom right hand side list box). For example if you select
fields A and B in the first box, and C in the second box, the tables A x C and B
x C will be calculated.
For
all tables, you will be prompted to specify what output options you want
included in the output tables:
-
Frequencies
only
-
Include
Expected Values
-
Include
Expected Values and Percents
-
Include
Expected Values, Chi-Contribution and Percents
-
Include
Percents
-
Include
Expected Values and Chi-Contribution
Percents
display row, column and total percent of the number in each cell of the table.
The contribution to Chi-Square shows how much a particular cell contributed to
the size of the Chi-Square statistic. This often comes in handy when you are
trying to discover what may have caused a table to result in a low p-value (a
high Chi-Square statistic).
For
a test for independence, a contingency table looks at two categorical variables
from a single sample of one population and tests whether the two variables are
related in some way, (e.g., are sex and hair color related?) The hypotheses
being tested are:
Ho:
The variables are independent of each other. (There is no association between
them).
Ha: The variables are not independent of each other.
A
Chi-Square statistic is calculated, with (r-1)(c-1) degrees of freedom where r
is the number of rows and c the number of columns. A low p-value indicates
rejection of the null hypothesis.
WINKS
reports both the chi-square statistic and the p-value. If the expected value in
one or more cells is less than 5, the chi-square test may not be valid. A
warning to this effect appears on the screen if appropriate. In the case of a 2 by 2 table, Fisher's Exact
Test and the chi-square with Yates' correction are also performed and results
displayed. Note: Tables as large as 15 columns by 100 rows may be created by
reading data from a database. If there are more categories than this, WINKS
combines remaining categories in a group called REST. To prevent his, you might
combine some groups.
Data
for this example are observations of the number of beetles and bugs on the
upper and lower sides of leaves (Zar,1974, page 292).
2
by 2 Contingency Table Data
Beetles
Bugs
Upper
Leaf 12
7
Lower
Leaf 2
8
To
perform this analysis, follow these steps:
Step
1: From the
Analyze menu, select “Crosstabulations, Frequencies, Chi Square” and choose
"Crosstabulations, Chi-Square - From Keyboard" option.
Step
2: You will
be prompted to give the size of the table. When asked for the number of rows and
columns, type 2, 2 and press Enter. You will then be prompted to select output
options. For this example, just select Frequencies. An empty table will appear .
Enter the counts for each category into the appropriate cell, and choose
Calculate. Preliminary results will appear on the status bar a the bottom of the
screen. You can perform calculations on several tables, and all results will
appear in the viewer when you select Exit.
---------------------------------------------------------------------------
Crosstabulations and Chi Square
---------------------------------------------------------------------------
2-Way Contingency Table
FREQUENCY| | | TOTAL
------------------------
| 12| 7| 19
------------------------
| 2| 8| 10
------------------------
TOTAL 14 15 29
48.3 51.7 100.0
WARNING - Some Expected values less than 5. Chi-Square may not be valid.
Statistic DF Value p-value
-----------------------------------------------------------------
Chi-Square 1 4.887 0.028
Yates' Chi-Square 1 3.312 0.069
Fisher's Exact Test (one-tail) 0.033
(two-tail) 0.111
Phi Coefficient .411
Cramer's V .411
Contingency Coefficient .380
Relative Risk 3.158
Odds Ratio 6.857
Sensitivity .857
Specificity .533
Sensitivity and Specficity calculations are based on a
table where the cells are in the following pattern:
TP FP
FN TN
T=True, F=False, P=Positive, N=Negative
Step
3: In the
viewer, the calculated chi-square statistic in reported as 4.89 with a p-value
of 0.028. The chi-square with Yates correction is 3.31 with a p-value of 0.069
and the Fisher Exact Test (two tail) has a p-value of 0.050. Because one of the
cells produces an expected value less than 5, WINKS gives a warning that the
chi-square analysis for this data may not be valid. Given this warning, it is
best to rely on the Fisher's Exact Test for making a decision.
A
low p-value indicates rejection of the null hypothesis. At a 0.05 significance
level, the Fisher's Exact Test p-value of 0.050 indicates (on the borderline)
that there is enough evidence to reject the null hypothesis of independence of
the two variables and to conclude that leaf side and type of insect are not
independent. In this case it appears that beetles prefer the upper sides of
leaves and bugs are about split in their preference. In the case of the Yates
results, this decision is marginal.
Notes on 2x2 Table
Statistics
Four new statistics have been recently added to the 2 x 2 crosstabulation
option in the Crosstabulation, Frequencies and Chi Square analysis. These are
Relative Risk, Odds Ratio, Specificity and Sensitivity.
RELATIVE RISK is given by the formula
a/(a +
b)
RR = ----------
c/(c + d)
where the two by two table is
Factor 1
+ -
---------
+|a |
b|
Factor 2 ---------
-|c |
d|
---------
ODDS RATIO is calculated by
a / b
OR = -----
c / d
SENSITIVITY is calculated by
a
SEN = -----
a + c
Sensitivity is a measure of the ability to call positive those
patients/subjects that have the disease/condition.
SPECIFICITY is calculated by
d
SP = -----
b + d
Note: Sensitivity and Specificity calculations are only available on the
crosstabulations with data entered from the keyboard. Results are based on a
table where the cells are in the following pattern:
TP | FP
————————
FN | TN
Where T=True, F=False, P=Positive, and N=Negative.
Specificity is a measure of the ability to call negative patients/subjects
that do not have the disease or condition. For more information on these four
statistics, reference any biostatistics or epidemiology text. One example is Basic
Biostatistics in Medicine and Epidemiology, A. A. Rimm,
Appleton-Century-Crofts, 1980.
Entering data from a Database
When
entering crosstabulation data from a database follow these steps:
Step
1: Open a
database. Data to be used may be numeric or character, but they should be
categorical such as hair color, sex, etc.
Step
2: From the Analyze menu, select the “Crosstabulations,
Frequencies and Chi-Square” then select “Crosstabulations - Chi Square”
Step
3: Select
which tables to calculate, and what options to display in the output.
Step
4: Results
will be displayed in the viewer.
Example: Crosstabulation as a test for homogeneity
A
crosstabulations analysis may also be used to perform a chi-square test for
homogeneity. In this case the two-way table looks at variables in samples from
two populations and tests whether
the variables follow the same distribution for both populations, that is,
whether the populations are homogeneous.
The
test statistic and procedure are the same as for the chi-square test for
independence. The data are organized so that the populations being tested for
homogeneity are represented as values (groups) of a grouping variable. The
hypotheses being tested in this case are:
Ho:
The populations are homogeneous.
Ha: The populations are not homogeneous.
To
use the crosstabulations procedure to test for homogeneity, let the rows of the
table represent the categories of the variable and the columns the different
populations. (Of course, for test purposes it doesn't matter whether the rows or
columns represent the populations.)
If
entering the data from the keyboard, simply enter the totals for each category.
If creating a database, you need a field for the categorical variable you wish
to use in the test for homogeneity, and a grouping variable which identifies the
population from which each observation comes. Each record represents one
observation. There may be other fields in the database for other variables in
these same populations and you can do separate crosstabulation analyses on them.
Suppose
you want to check whether the ratio of men and women is the same in three
different departments of a company. You obtain the following data:
Dept 1 Dept 2
Dept 3
Men
10 45
15
Women
8
22
4
The
calculated chi-square statistic in this case is 2.30 with a p-value of 0.317.
A decision can be made using this p-value of the test. A low p-value
(less than the chosen significance level) is usually taken to indicate rejection
of the null hypothesis. At a 0.05 significance level, the p-value of 0.317
indicates that there is not enough evidence to reject the null hypothesis of
homogeneity of the three departments. That is, you cannot conclude that the
departments are significantly different with respect to sex of employees based
on this test.
McNemar's
Test
McNemar's
test is appropriate for use with paired, dichotomous data. This test is
sometimes called a test for related samples or a test for the significance of
changes. It is useful for comparing paired or related observations in which the
response is dichotomous, that is, the response is one of only two possible
outcomes. McNemar's test is the 2 by 2 version of Cochran's Q test described in
the section on non-parametric tests. The test assumes that any pair of
observations is independent of any other pair of observations, although clearly
the observations within a pair are not independent of each other.
For
example, you may wish to know if a certain advertisement has an effect on the
impression consumers have of a product. You could select a group of people and
check their impressions of the product before viewing the advertisement, and
again after viewing the ad. You record the reactions of each person as
"favorable" or "unfavorable" both before and after seeing
the ad. Thus, there are four categories of before-after responses:
“favorable-favorable",
"favorable-unfavorable",
"unfavorable-favorable",
"unfavorable-unfavorable"
You
want to know if there is a difference in the reactions before and after viewing
the advertisement. It is possible to use McNemar's test if you have only the
totals for each of the four categories, or if you have the record of each
individual response. The hypotheses being tested are:
Ho:
The proportions (of "favorable" and "unfavorable"
reactions) in the two groups (before and after) are the same.
Ha: The proportions in the two groups are not the same.
In
other words, you are testing whether there is a change in the number of people
who react favorably to the product after seeing the advertisement. The test
statistic used is:
Q
= (B - C)2/(B + C)
where
B and C are the number of "favorable-unfavorable" and
"unfavorable-favorable" reactions. This test statistic approaches a
chi-square distribution with one degree of freedom. The chi-square statistic is
useful if the number of pairs of observations is at least 10.
WINKS
displays both the chi-square statistic and the p-value of the test. The p-value
associated with this test can be used to make a decision. A small p-value (less
than the chosen significance level, e.g., 0.05) is usually taken to indicate
rejection of the null hypothesis.
For
example, in the test of the effect of an advertisement, suppose 20 people
participated with the following results (listed on the next page), where 1 is
the code for "favorable" and 0 for "unfavorable". For
McNemar's test, the data must be coded as 0 or 1, representing the two possible
responses. To perform this analysis, follow these steps:
Step
1: Open the
database named MCNEMAR.DBF. If you choose to create the database yourself, you
can choose the New Database option from the File menu, then use the pre-defined
database structure option named “For paired t-tests or McNemar’s Test.”
Step
2: From the
Analyze menu, select “Crosstabulations, Frequencies and Chi-Square” then
choose "McNemar's test."
You
will be prompted to choose the two fields (groups) you wish to compare. In this
case, there are only two fields. Select
BEFORE and AFTER. Then select which options you want to appear on the output
table. WINKS will perform the calculations and display the results in the
viewer.
The
chi-square statistic in this case is 0.57, and the p-value is 0.450. The p-value
is large, so the null hypothesis of equal proportions is not rejected. That is,
there is not enough evidence to say that there is a difference in the reactions
before and after viewing the advertisement.
Comparison of Proportions
To compare proportions select Proportion Comparison in the Crosstabs menu.
You will be prompted to enter two proportions (or two observed counts) and the
sample size associated with each proportion. For example, for the first
proportion enter 10 observations out of 20 and for the second enter 15
observations out of 75. When you select Calculate then Exit, the following
results are displayed:
Proportion(1) = 0.5 Z = 2.707
Proportion(2) = 0.2 P = 0.007 (two tail)
This tells you that the z-statistic for the comparison of these proportions
is 2.707, and the p-value associated with the test is 0.007. Thus, there is a
significant difference in these proportions.
Continue to Chapter
4. Part 7. (Life Table and Survival Analysis.)
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