GEOG 442 chi-square lab on lichen field data: Dr. Rodrigue, F/2003

GEOG 442

Biogeography

Lab 5: Chi-Squared Analysis of Your Lichen Field Data

The purpose of this lab is to have you analyze the data you collected on lichens during your field observation in Charmlee Park. Chi-squared analysis will be used to see if there is a significant difference between the two field plots (the vertical rock face and the sunnier horizontal rock face) in terms of the quadrats reporting the three species of lichen (crustose, foliar, and fruticose or "mossy").
It would be helpful if you referred back to Lab 2, in which you first (re?)did a Chi-square dry run. If you are doing this lab in our Geography Lab, please be sure to bring an IBM-formatted 3.5" floppy diskette or a Zip disk -- wouldn't want Woody to purge your half-done lab inadvertantly!

Lab 5a: Getting and Preparing Your Data for Analysis

For your group (vertical, shady rock wall or horizontal, sunny rock surface), please organize your field data into three generously proportioned Chi-square tables:

Crustose and Foliar: Two columns under Crustose, one showing a "+" for "present" and one showing a "-" for absent, and then two rows beside Foliar, again with "+" and "-".
Crustose and Mossy: Two columns under Crustose, one showing a "+" for "present" and one showing a "-" for absent, and then two rows beside Mossy, again with "+" and "-".
Foliar and Mossy: Two columns under Foliar, one showing a "+" for "present" and one showing a "-" for absent, and then two rows beside Mossy, again with "+" and "-".

For each of the four data cells in each of the three tables, please count the number of quadrats that satisfy the "+/-" combinations above (how many quadrats had Crustose present but Foliar absent, for example). Each of the three tables should show total quadrat counts of 100 (10 columns on your PVC pipe sampling frame by 10 rows). If they don't, someone made a mistake and you need to re-examine your classification until you have 100 quadrats accounted for. When you are satisfied that you have all three quadrat counts done properly (they all add up to 100 quadrats), then enter those counts into the upper parts of the appropriate cells as your observed frequencies.
You should have three tables, then, that look kind of like this:
VERTICAL or HORIZONTAL | CRUSTOSE | | | row | present | absent | totals ________________________________________________________________ |(a) |(b) |-e- present | obs = | obs = | | exp = | exp = | FOLIAR ____________________________________________________ |(c) |(d) |-f- absent | obs = | obs = | | exp = | exp = | ________________________________________________________________ |-g- |-h- |-n- 100 column totals | | | | | | VERTICAL or HORIZONTAL | CRUSTOSE | | | row | present | absent | totals ________________________________________________________________ |(a) |(b) |-e- present | obs = | obs = | | exp = | exp = | MOSSY ____________________________________________________ |(c) |(d) |-f- absent | obs = | obs = | | exp = | exp = | ________________________________________________________________ |-g- |-h- |-n- 100 column totals | | | | | | VERTICAL or HORIZONTAL | FOLIAR | | | row | present | absent | totals ________________________________________________________________ |(a) |(b) |-e- present | obs = | obs = | | exp = | exp = | MOSSY ____________________________________________________ |(c) |(d) |-f- absent | obs = | obs = | | exp = | exp = | ________________________________________________________________ |-g- |-h- |-n- 100 column totals | | | | | |

Lab 5b: Your Hypotheses
Examining your three tables, do you think you see a relationship or association between any pairs of species? That is, do you suspect foliar lichens are generally found with crustose lichens? For each of the three, please state, first, your hunch about the association between the species and, then, the null hypothesis (or inverse of your hunch).
Crustose and foliar lichens working hypothesis: _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ Crustose and foliar lichens null hypothesis: _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ Crustose and mossy lichens working hypothesis: _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ Crustose and mossy lichens null hypothesis: _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ Foliar and mossy lichens working hypothesis: _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ Foliar and mossy lichens null hypothesis: _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________
We will reject the null hypothesis if our results are so extreme that there is no more than a five percent chance that we could have gotten them by pure random luck-of-the-draw in placing those quadrat sampling frames. That is (stats refesher), we will use the 0.05 alpha standard. Another way of looking at it is that, by using such an extreme standard, we can have a 95 percent confidence in our conclusion, should we wind up rejecting the null hypotheses and deciding that the associations between each pair of lichen "species" is not random, that there is some sort of relationship between them.

Lab 5c: Doing the Analysis

Compute the marginal totals. That is, sum the observed frequencies in each row and put those sums in the appropriate row total (e or f). Do the same for the frequencies in each column and put those sums in the appropriate column total (g or h). The sum of row totals should equal the sum of column totals. Check that the total number equals 100) in cell n.

Then, create the "expected frequencies" for each data cell (a through d). This is the distribution of cell counts you would expect from your data if there were no association between the two plant species (i.e., random processes were allocating them among the cells). To do this for each data cell, a through d, multiply the row total to its right by the column total below it and then divide the answer by n. Put the answer, rounded to three decimal places of accuracy, in its cell below the actual observed frequency.

Still lost? Okay, okay. In other words, multiply cells e and g and divide the answer by 100. Put the answer, properly rounded, in the lower part of cell a. Similarly, multiply cells e and h and divide by 100, and put that answer in cell b. And so forth.

That done, examine the expected frequencies to make sure you can properly proceed. Chi-square should not be used if any expected frequencies are below 2 (or, irrelevantly in this case, if more than 20 percent of the data cells have fewer than 5 actual cases). You will note that there are no such problems with your contingency table, so you can safely proceed through Chi-square.

Now, move on to the worksheet below for calculating Chi-squared. In the first column, enter the observed frequencies for each data cell (the number in the upper, obs = part of cells a through d).

In the second column, square those OBS frequencies.

In the third column, divide each squared observed frequency by the corresponding expected frequency in the bottom of the appropriate data cell (a through d).

Now, sum the third column and put the answer near the bottom of the spreadsheet (sum(O²/E). Show your work here to three decimal places of accuracy.

Finally, subtract n (or 100) from that sum. This answer is your calculated Chi-squared (X²). Put it at the bottom of the whole spreadsheet, also rounded to three decimal places of accuracy.
________________________________________________________________________ CRUSTOSE VS. FOLIAR DATA CELL | O | O² | O²/E ________________________________________________________________________ (a) | | | ________________________________________________________________________ (b) | | | ________________________________________________________________________ (c) | | | ________________________________________________________________________ (d) | | | ________________________________________________________________________ | sum(O²/E) = ________________________________________________________________________ | sum(O²/E) - n = X² = ________________________________________________________________________ ________________________________________________________________________ CRUSTOSE VS. MOSSY DATA CELL | O | O² | O²/E ________________________________________________________________________ (a) | | | ________________________________________________________________________ (b) | | | ________________________________________________________________________ (c) | | | ________________________________________________________________________ (d) | | | ________________________________________________________________________ | sum(O²/E) = ________________________________________________________________________ | sum(O²/E) - n = X² = ________________________________________________________________________ ________________________________________________________________________ FOLIAR VS. MOSSY DATA CELL | O | O² | O²/E ________________________________________________________________________ (a) | | | ________________________________________________________________________ (b) | | | ________________________________________________________________________ (c) | | | ________________________________________________________________________ (d) | | | ________________________________________________________________________ | sum(O²/E) - n = X² = ________________________________________________________________________

Now, to interpret this X²_calc, you need to compare it with a critical X². To do this, you will need the Chi-squared table in Figure 1. You need your pre-selected alpha level to pick the right column and the degrees of freedom for your 2 x 2 contingency table to choose the right row to enter the table. Degrees of freedom in Chi-squared can be defined as:
DF = (r - 1)(k - 1) where r = number of rows with obs data in them and k = number of columns with obs data in them
So, you will enter the table at the intersection of:
the column headed ________ and the row corresponding to ________ degrees of freedom.
What, then, is your critical Chi-squared value for the association between crustose and foliar lichens?
X²_crit = ________
What is the critical Chi-squared value for the association between crustose and mossy lichens?
X²_crit = ________
What is the critical Chi-squared value for the association between foliar and mossy lichens?
X²_crit = ________

Is your X²_calc ________ greater than or ________ less than the X²_crit for the crustose and foliar lichen association?
Is your X²_calc ________ greater than or ________ less than the X²_crit for the crustose and mossy lichen association?
Is your X²_calc ________ greater than or ________ less than the X²_crit for the foliar and mossy lichen association?

If your actual, calculated Chi-square value is greater than the critical Chi-square, you may safely conclude that your pattern is not just a random one. In other words, there is a statistically significant probability that there is a real association of some sort between your variables (in this case, between species). If the calculated Chi-square value is less than the critical test value, the relationship probably is random.
Can the null hypothesis of random association between crustose and foliar lichens in this study area be rejected in this case?
_____ reject H_o _____ do not reject H_o

Can the null hypothesis of random association between crustose and mossy lichens in this study area be rejected in this case?
_____ reject H_o _____ do not reject H_o

Can the null hypothesis of random association between foliar and mossy lichens in this study area be rejected in this case?
_____ reject H_o _____ do not reject H_o

It's always good etiquette, whenever possible, to calculate the prob-value of a Type I error, to express your faith in the null hypothesis, however, in the off chance that a reader may have compelling reasons to use a different standard of alpha than you chose. I have provided you the needed data in Figure 2 to tell the probability that you could have gotten results as extreme as yours if there is but a random association between each pair of species.
________ prob-value of H_o for crustose and foliar association ________ prob-value of H_o for crustose and mossy association ________ prob-value of H_o for foliar and mossy association

As you may remember from your earlier encounter with Chi-square, the results of the technique are affected by sample size. Chi-square can tell you IF there is a significant association but it's less solid about telling you the STRENGTH of that association. It's a good idea to calculate a measure of strength as well, which, for a 2 x 2 table, could be Yule's Q.
To calculate Yule's Q, multiply data cells a and d and also cells b and c. Then, enter these multiplications into the following formula:
ad - bc Q = _______ ad + bc

So, what is the Q value for the relationship between crustose and foliar lichens? ________
What is the Q value for the relationship between crustose and mossy lichens? ________
What is the Q value for the relationship between foliar and mossy lichens? ________
Remember, Yule's Q can vary from -1 to +1. The closer it is to 0, the weaker the relationship is. The closer it is to -1 or +1, the stronger the relationship is, whether inverse (negative) or direct (positive).

Lab 5d: So, What Does It All Mean?

Please describe the results of your field data collection and lab analysis. Are there any significant associations between any pair of lichen species? If so, what is the direction and strength of that association?
_________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________

Figure 1 Critical Values for Chi-Square (X²_crit)

alpha df 0.100 0.050 0.025 0.010 0.005 1 2.706 3.841 5.024 6.635 7.879 2 4.605 5.991 7.378 9.210 10.597 3 6.251 7.815 9.348 11.345 12.838 4 7.779 9.488 11.143 13.277 14.860 5 9.236 11.070 12.832 15.086 16.750 6 10.645 12.592 14.449 16.812 18.548 7 12.017 14.067 16.013 18.475 20.278 8 13.362 15.507 17.535 20.090 21.955 9 14.684 16.919 19.023 21.666 23.589 10 15.987 18.307 20.483 23.209 25.188 11 17.275 19.675 21.920 24.725 26.757 12 18.549 21.026 23.337 26.217 28.300 13 19.812 22.362 24.736 27.688 29.819 14 21.064 23.685 26.119 29.141 31.319 15 22.307 24.996 27.488 30.578 32.801 16 23.542 26.296 28.845 32.000 34.267 17 24.769 27.587 30.191 33.409 35.718 18 25.989 28.869 31.526 34.805 37.156 19 27.204 30.144 32.852 36.191 38.582 20 28.412 31.410 34.170 37.566 39.997 21 29.615 32.671 35.479 38.932 41.401 22 30.813 33.924 36.781 40.289 42.796 23 32.007 35.172 38.076 41.638 44.181 24 33.196 36.415 39.364 42.980 45.558 25 34.382 37.652 40.646 44.314 46.928 26 35.563 38.885 41.923 45.642 48.290 27 36.741 40.113 43.195 46.963 49.645 28 37.916 41.337 44.461 48.278 50.994 29 39.087 42.557 45.722 49.588 52.335 30 40.256 43.773 46.979 50.892 53.672 40 51.805 55.758 59.342 63.691 66.766 50 63.167 67.505 71.420 76.154 79.490 60 74.397 79.082 83.298 88.379 91.952 70 85.527 90.531 95.023 100.425 104.215 80 96.578 101.879 106.629 112.329 116.321 90 107.565 113.145 118.136 124.116 128.299 100 118.498 124.342 129.561 135.807 140.170

Figure 2: p-Values for X²_calc
X² 1 DF X² 1 DF X² 1 DF X² 1 DF 3.2 .0736 4.4 .0359 5.6 .0180 6.8 .0091 3.3 .0692 4.5 .0339 5.7 .0170 6.9 .0086 3.4 .0652 4.6 .0320 5.8 .0160 7.0 .0082 3.5 .0614 4.7 .0302 5.9 .0151 7.1 .0077 3.6 .0578 4.8 .0285 6.0 .0143 7.2 .0073 3.7 .0544 4.9 .0268 6.1 .0135 7.3 .0669 3.8 .0513 5.0 .0254 6.2 .0128 7.4 .0065 3.9 .0483 5.1 .0239 6.3 .0121 7.5 .0062 4.0 .0455 5.2 .0226 6.4 .0114 7.6 .0058 4.1 .0429 5.3 .0213 6.5 .0108 7.7 .0055 4.2 .0404 5.4 .0201 6.6 .0102 7.8 .0052 4.3 .0381 5.5 .0190 6.7 .0096 >7.8 <.0050

first placed on the web: 11/27/01
last revised: 12/06/03
© Dr. Christine M. Rodrigue