Please bring this handout with you every time you do data analysis in the computer lab.
This handout is also available on my website.
JMP Instructions
A worked example
Your biology instructor has encouraged you to investigate some aspects of a flowering plant wingstem. You have decided to investigate whether wingstem grows better in forested areas
than in grasslands. You randomly choose wingstem from each of these two habitats for
laboratory analysis, ending up with 22 plants - 12 from forest and 10 from grassland. In an
attempt to impress your instructor you also note whether there is a trillium plant within 1 meter
of the wingstem plant - there seemed to be quite a few trillium plants near wingstem plants in
the forested habitat.
Entering data in the table
Each column must be one variable. For the purposes of this example, you have decided to
weigh each plant (“wet weight” means it hasn’t had a chance to dry out), and to measure the
length of its stem. If you are comparing forests and grasslands, then one column will be for
“habitat” (forest or grassland), the 2nd column will be “wet weight”, the 3rd column will be “stem
length”, and your final column will be “trillium nearby?”. You can enter these in any order you’d
like.
Make sure each column is assigned an appropriate “modeling type” (continuous, nominal, or
ordinal data) as follows:
In the Columns box, you should see each of your heading names with a symbol (button). Click
on the button to get a pop-up menu and change the variable type. Most of the variables you
use for analysis should be classified as either continuous (data that you measure in some
manner) or nominal (data that you apply a name or category-type to). We may discuss
ordinal data at a later time. Thus make sure that the “habitat” and “trillium nearby” columns
have an “N” button (for nominal), and the “wet weight” and “stem length” columns have a “C”
button (for continuous). The most recent version of jmp replaces the “N” with a bar graph, and
the “C” with a line graph.
Your data file will look something like this:
1
Which test to use?
The answer to this question depends on (1) which questions you ask and on (2) the types of
variables you are considering for asking your questions. Let’s consider a few examples.
Example I. Do wingstem plants from grasslands tend to weigh more than wingstem plants
from forested regions?
To answer this question you would ask whether the mean “wet weight” of wingstem plants from
grasslands is significantly greater than the mean “wet weight” of wingstem from forested
regions.
2
There are two steps to this process:
a. Identify your dependent and independent variables
dependent variable - the variable you believe might be influenced or modified by some
treatment or exposure
independent variable - the variable you believe may be influencing or affecting the dependent
variable
In this case we believe that habitat (the independent variable) may be influencing “wet weight”
(the dependent variable). Note that it doesn’t make sense to look at it the other way around!
b. Decide whether each of your variables is nominal or continuous. If the dependent variable
is continuous, and the independent variable is nominal, you will use a t-test for cases when
your independent variable has two categories (such as this case where we have either forest
or grassland for our “habitat” variable), or an analysis of variance (ANOVA) in cases where
your independent variable has three or more categories.
The table below summarizes all of the possible combinations. At this point we are confining
the discussion to the first row of the table.
Dependent
variable
Continuous
Independent
variable
Nominal
Type of comparison
Test statistics
Means
Continuous
Continuous
Correlation or Regression
Nominal
Nominal
Nominal
Continuous
Frequencies or Proportions or
Percentages
Discriminant Analysis or
Logistic Regression
t (for t-test)
F (for ANOVA (analysis
of variance))
r (correlation coefficient)
R2
Contingency or X2
(chi-square)
Will not be covered
Table 1. Types of variables, types of comparisons and test statistics commonly used for data
analysis
You already knew that you were comparing means - now you know you will do a t-test.
t-tests
Please do the following:
1. Open or create your data file
2. .Analyze menu  Fit Y by X
3. You should see a window as follows:
3
Remember that in this case, our dependent variable is “wet weight”, and we are hypothesizing
that it is being influenced by the independent variable - “habitat”. Also make sure you
understand that “wet weight” is continuous and “habitat” is nominal (you can confirm by looking
at the little buttons in the select columns window). If you check out table 1, you will confirm
that you will be comparing the “wet weight” mean for the forested habitat with the “wet weight”
mean for the “grassland” habitat, and using a t-test to test for whether the means are
statistically significantly different from each other.
4. Highlight “wet weight” and click on “y,response” in the “fit y by x” window. (In JMP language,
“y, response” is the same as dependent variable)
5. Highlight “habitat” and click on “x-factor” in the “fit y by x” window. (In JMP language, xfactor is the same as independent variable).
6. Click OK. The following will come up on your screen (hopefully)
4
Oneway Analysis of “wet weight” in grams By habitat
8
wet weight in grams
7
6
5
4
3
2
1
0
forest
grassland
habitat
7. Click on the red triangle in the title bar to get a pop-up menu. Select “Means and Std Dev”.
Below the graph you will see calculations of the means, Standard deviations, Standard Error of
the Mean and confidence intervals for each of your independent variables. In the graph some
blue lines have been added among your black data points. The center blue line is the mean.
It is connected by a vertical blue line to a pair of short blue horizontal lines that represent the
mean-plus-one-standard-error and the mean-minus-one-standard-error. Further away from
the mean are two longer horizontal blue lines that represent the mean-plus-one-standarddeviation and the mean-minus-one-standard-deviation.
5
8. Click the red triangle again and select “ t test”. You will see the addition of another graph
and other calculations.
Don’t worry about the graph for now. There are two important numbers in this output.
T Ratio = t = -1.81115. The negative sign doesn’t matter, and you should usually report t-vales
down to two decimal points, so in your report you would state that t = 1.81
Prob > |t| = p value = 0.0866. Please refer to the end of this handout for information on
interpreting p-values.
Example II - Linear Correlation
Do taller plants tend to weigh more than shorter plants. A more formal way of stating this in
hypothesis form is as follows:
Hypothesis: There is a positive correlation between “stem length” and “wet weight”.
In this case we believe that “stem length” (the independent variable) may be influencing “wet
weight” (the dependent variable). In this case you need to do some serious thinking to
convince yourself that “stem length” is the factor that would influence “wet weight”, and not the
other way around. Because we hypothesize that “stem length” is influencing “wet weight”,
“stem length” is the independent variable and “wet weight” is the dependent variable.
Both of the variables are continuous. If both variables are continuous, refer to Table 1 and
note that the correct analysis is correlation or linear regression (these two analyses are
functionally equivalent).
JMP Steps
1. Analyze menu  Fit Y by X
2. Click on your dependent variable - ““wet weight” in grams” (which should have a C beside
it). then on “Y, Response”. Click on the independent variable - “stem length” (cm) - (which
should also have a C beside it) and then on “X, Factor.” Click OK.
You will see a scatter plot of the data.
6
3. Click on the red triangle in the title bar and select “ Fit Line.” You will see a red line appear
through your data points. That is the best straight line that fits your data (it minimizes the sum
of the distances from each data point to the line).
A bunch of output will also appear below the graph.
7
For now I am concerned with two parts of this output.
A. Under “summary of fit” you will note the RSquare = 0.880328. If you were reporting your
results you would report that R2 = 0.88. In general, the closer R2 is to 1, the stronger is your
correlation. The graph looks impressively linear, so it is no surprise to have a high R 2 value.
B. Under “Parameter Estimates” you will note that if you go across from “stem length” and
down from Prob > |t| you will find a value of <.0001. That tells us that the correlation between
“stem length” and “wet weight” is very strong.
Finally to determine if the correlation is positive or negative, you will need to look at the graph
and see if the slope of the line is positive or negative.
Example III - Contingency or Chi-square (X2) analysis - for this problem, use the made-up data
table from the previous lab (the wingstem data table with 22 cases)
Table 1. Wingstem data table from previous lab.
Is trillium more likely to be associated with wingstem in forested regions than in grasslands?
Again you could state this more formally:
Hypothesis: There is a higher frequency of trillium near wingstem in forested regions than in
grassland regions.
8
You are asking whether habitat type (forest or grassland) influences whether trillium is likely to
be nearby (yes or no). Thus both variables are nominal. The dependent variable is “trillium
nearby?” and the independent variable is “habitat type”.
Refer to Table 2 and note that in cases in which both variables are nominal, you will compare
frequencies or percentages using a contingency or chi-square test.
Dependent
variable
Continuous
Independent
variable
Nominal
Type of comparison
Test statistics
Means
Continuous
Continuous
Correlation or Regression
Nominal
Nominal
Nominal
Continuous
Frequencies or Proportions or
Percentages
Discriminant Analysis or
Logistic Regression
t (for t-test)
F (for ANOVA (analysis
of variance))
r (correlation coefficient)
R2
Contingency or X2
(chi-square)
Will not be covered
Table 2. Types of variables, types of comparisons and test statistics commonly used for data
analysis
JMP steps
1. Analyze menu  Fit Y by X
2. Click on your dependent variable - “trillium nearby?’ (which should have an N beside it) then
on “Y, Response.” Click on the independent variable (which should have an N beside it), and
then on “X, Factor”. Click OK. This is what you will see if you are a lucky person.
9
3. You will see a “Mosaic Plot,” which shows, for each independent variable, the percentage of
total observations in each category of the Y axis.
4. Below the mosaic plot you will see a “Contingency table.” The upper left square in the table
explains to you what each % of the table refers to (check with me to get a more complete
explanation).
5. Below the contingency table are the test statistics. For our purposes there are three
numbers which are relevant.
10
A. DF stands for degrees of freedom: In the “source” table you should note that the model DF
= 1. Your instructor may elect to explain degrees of freedom to you.
B. In the “Test” table, note that the Likelihood Ratio ChiSquare value =- 3.005, and the
Prob>ChiSq = 0.083.
OK, how might you go about reporting this? I would recommend something like the following:
Eight out of 12 (66.67%) of the wingstem from the forested region had trillium nearby, while 3
out of 10 (30%) of the wingstem from the grassland had trillium nearby. These differences in
frequency are not statistically significantly different (X2 = 3.005, df = 1, p = 0.083).
CHI-Square (contingency table) analysis - the easy way
The data table you made allowed you to test the hypothesis that there is a higher frequency of
trillium near wingstem in forested regions than in grassland regions. The data lean in that
direction, as follows:
66.67% of the wingstem had trillium nearby in the forested region
30% of the wingstem had trillium nearby in the grassland
Please check the contingency table output to make sure you see where these percentages
came from.
Based on these data, your conclusion should be that the p value of 0.083 is not low enough to
support your research hypothesis. I am also hoping that your conclusion will also be that you
need more data to adequately test the hypothesis. Unfortunately for you, your instructor felt
the same way, and sent you back out to collect more data.
The new data look like this:
Forested habitat
Grassland habitat
Trillium nearby?
Yes
no
21
41
37
18
One problem with your previous data table was that it was very cumbersome. You had to type
in yes or no for each case, which was OK when it was 22 times, but would be a real drag when
you had to do it for these data (58 yeses and 59 nos)
Rejoice - there is an easy way. Just enter the table that looks like this:
11
Then do the following steps:
1. Analyze menu  Fit Y by X
2. Click on your dependent variable - “trillium nearby?’ (which should have an N beside it) then
on “Y, Response.” Click on the independent variable - habitat (which should have an N beside
it), and then on “X, Factor”. Click on “how often”, which is your weighting variable, and click on
“weight”. Click OK.
The output will be exactly the same format as before, but the actual numbers will change as a
result of your valiant data collection. You will get a graph that looks like this:
and some data that look like this:
12
How would you report these findings.
Do the results support or fail to support your hypothesis?
Example IV. Analysis of Variance (ANOVA)
Let’s go back to the case in which the dependent variable is continuous and the independent
variable is nominal. Just to remind you, you could organize your understanding of the possible
cases with the following, now famous, table 1.
Dependent
variable
Continuous
Independent
variable
Nominal
Type of comparison
Test statistics
Means
Continuous
Continuous
Correlation or Regression
Nominal
Nominal
Nominal
Continuous
Frequencies or Proportions or
Percentages
Discriminant Analysis or
Logistic Regression
13
t (for t-test)
F (for ANOVA (analysis
of variance))
r (correlation coefficient)
R2
Contingency or X2
(chi-square)
Will not be covered
Table 1. Types of variables, types of comparisons and test statistics commonly used for data
analysis
You will notice that there are two possibilities: t-test and ANOVA. Which do you use?
The answer is very simple. If you’re comparing the means for two categories, like when you
compared the “wet weight” of wingstem in forested vs. the “wet weight” of wingstem in
grassland habitats, then you use a t-test. When you are comparing the means for three or
more categories, then you use an ANOVA.
From what we know about mushrooms, we could propose the following:
Hypothesis: Wetter and shadier habitats will tend to have a higher species richness of
mushrooms than dryer sunnier habitats.
Prediction: If the hypothesis is true, experimental quadrats in forests along the river will have
the greatest mean number of species, while experimental quadrats in dry grasslands will have
lowest mean species richness. Forest away from the river (deep forest) and grasslands along
the river would have intermediate values. (The hypothesis does not make distinguishing
predictions between the latter two categories.)
Notice that we have four categories: Deep forest, river forest, dry grassland and river
grassland.
Here are the made-up data in a JMP table:
14
Table 2. Made-up data of number of different species of mushrooms collected from five 10 X
10 M quadrats in each of four habitats.
Notice that I have a column called “replicate”. Each replicate (or replication) is a sample of the
category. Because we set up five quadrats in each habitat, we have five replicates for each of
the four habitats. I put the replicate numbers in the data table because it facilitates discussion.
For example I could say “replicate 3 of river forest”, and you would know I’m referring to the
quadrat with 12 different species of mushrooms. Numbering your replicates is very useful for
complex data tables. But for simple data tables you probably don’t need to worry about it.
OK. So we want to test our Prediction that experimental quadrats in river forest will have the
greatest mean number of species, while experimental quadrats in dry grasslands will have
lowest mean species richness. Deep forest and river grasslands will have intermediate values.
Please do the following:
1. Open or create your data file
2. .Analyze menu  Fit Y by X
3. Select “species richness” as your Y,response, and “habitat” as your X, factor. Click OK.
4. You will get a graph that shows the data points for each category.
15
Figure 1. Data points for each of four habitat categories.
5. Then click on the red triangle to the left of “oneway”, drag down to means/anova, and you
should see a new window with a great deal of data on it.
You will also that the graph has been redrawn, with the gray horizontal bar representing the
mean number of species overall, the longer green horizontal bar representing the mean for
each category, and the top and bottom of the diamond shape representing the 95% confidence
intervals. Based on these samples we are 95% confident that the true mean for that category
falls between the top tip and bottom tip of the diamond.
We won’t worry about the meaning of the two shorter green horizontal bars for now.
16
Here are the essential data from this table:
A. In the “Summary of Fit” table, the overall mean of response = 3.8. You won’t report that, but
it’s nice to know.
B. In the Analysis of Variance (ANOVA) table, you will note the DF (degrees of freedom) for
habitat = 3, and DF for error = 16. The F-ratio is 11.90, and P = 0.0002. You would report
that as follows: “F3,16 = 11.90, P = 0.0002.”
17
So what does this mean? Because the P-value is so low (much less than 0.05), we can
support our hypothesis that species richness is significantly different for different habitats.
C. In the “Means for Oneway Anova” table, you will see the means and standard errors
reported for each category. Beware! Those standard errors are for the entire sample, not
for the individual categories. To get the standard error for each category, click on the red
triangle and drag to “Means and Std Dev.” You’ll get the following:
These are the correct values for means, standard deviations, and standard errors. Your report
should include (either in the text of the paper, in a figure, or in a table) the means for each
category, and either the standard error of the mean, or the standard deviation.
The frustrating thing is that while we now know that habitat influences mushroom species
richness, we still don’t know which category is significantly different from which. In other words
we still haven’t fully tested the predictions of our hypothesis.
There’s only one more step.
6. Go back to that little red triangle, drag to “compare means”, then drag to all “pairs Tukey
HSD”. You will get the box below:
18
The bottom portion of this output is of relevance. You will notice that “Levels not connected by
same letter are significantly different.” (for our purpose levels = categories). For example the
mean (7.6) for River forest has an A next to it. Thus its mean is significantly greater than the 2
categories that don’t have an A next to them (both grassland categories), but it isn’t
significantly greater than Deep forest (which also has an A next to it). Similarly Deep forest’s
mean is significantly greater than Dry grassland’s mean, but not River grassland (Because
Deep forest and River grassland both share the letter B).
So that’s how we will be comparing means in this class when there are more than two
categories for a nominal independent variable.
Notice that I used the term significantly greater rather than significantly different - if the means
are different I want to know which is greater than which.
Appendix - HYPOTHESIS TESTING
We used a t-test to address the question of whether the mean “wet weight” of wingstem plants
from grasslands is significantly greater than the mean “wet weight” of wingstem from forested
regions.
The t-test, and all of the statistics we have been discussing, allow you to ascribe a confidence
value to the conclusions you will draw based on the analysis. In this case we are formally
testing the hypothesis that wingstem from grasslands on average weigh more than wingstem
from forested regions. We can (and should!) compare the means and say that the answer is
“yes”, based on the sample we collected. But we’re still not sure whether the same conclusion
can be drawn for the entire population of wingstem plants, in part because we haven’t weighed
very many plants. Are we justified in concluding that there is truly a significant difference in
‘“wet weight”” based on our sample? If so, how comfortable do we feel about this conclusion?
The answer to this question is that we will never be able to say for sure wingstem in
grasslands weigh more on average than wingstem in forested regions unless we test the entire
population. Fortunately for us, statisticians have developed techniques that allow us to make
inferences from relatively small samples to much larger populations.
The hypothesis we wish to test is called the research hypothesis (HA). In this case HA is that
there truly is a difference in mean wingstem “wet weight” between the two habitats . To
establish some credibility for the research hypothesis, we first define a null hypothesis (Ho)
that directly opposes HA. In this case Ho is that in reality, there is no difference in mean
wingstem “wet weight” weight between the two habitats.
To support HA, we attempt to show that the sample provides evidence that Ho is unlikely.
How unlikely? It is up to the researcher to decide how unlikely Ho must be before it can be
rejected. This is known as the critical value (). The most common  is .05. This means that
if Ho is true, we would expect to see such a difference between the mean wingstem ““wet
weight”s” in the two habitats in only one of every 20 samples.
Do the data allow us to reject Ho? The t- test allows us to compute the probability of how
often we would expect to have a difference as great as we observe for two habitats by chance
19
alone (i.e. if Ho were true). The calculation of the frequency with which we would expect to
see such a difference if Ho is true is the p-value. If p =.01, we will expect to see the observed
difference between the two groups (in other words between the ““wet weight”s” of wingstems in
forest vs. grassland) due to chance alone, only one out of 100 times. That’s very unlikely, so
as scientists, we can feel pretty good claiming that the Ho is probably wrong, and therefore
that the HA is supported. In a formal sense, we can then compare the p-value to ; if p <  as
it is when p = .01), then we can reject the Ho. This provides support for HA, but does not
prove it (recall we can prove our research hypothesis only by measuring the entire population).
Also beware! If we cannot reject Ho, that does not mean that Ho is correct. It is possible that
we would be able to reject Ho if we had a larger sample size.
The key realization for you to make is that the higher the value of t, the lower your p-value.
The lower your p-value, the more confident you are about rejecting Ho and supporting HA.
In the case presented at the beginning of this handout, the p value = 0.0866. This is greater
than our critical value of 0.05. Thus we are unable to reject H0, and fail to support the research
hypothesis. However, it is still possible that in the natural world, wingstem on average have
higher mean “wet weight” in grasslands than in forested regions. Our data are not sufficient to
support this hypothesis. A big however is that we really were much too lazy and should have
collected many more plants (in other words, our sample size was pretty pitiful).
One wonderful aspect of this approach to hypothesis testing is that all of the tests you use will
generate a p-value. The p-value will be interpreted in the same manner regardless of whether
you are using a t-test, ANOVA, linear regression, or chi-square test.
If p << 0.05, you have supported your research hypothesis
If P >> 0.05, you have failed to support your research hypothesis.
But if P is in the neighborhood of 0.05, you are in the uncomfortable zone, in that it is
not clear whether you have support for your research hypothesis. The reality is that there
really is nothing magical about a critical value of 0.05, so maybe we should say that between
0.04 and 0.08 we have some weak support for our research hypothesis. We’ll talk about this
later in the year.
20