Names: ___________________________ Research Literacy
___________________________ Rolly-Polly (Pill Bug, Isopod) Chi Square Lab
___________________________
Virtual Lab: http://bioweb.wku.edu/courses/Biol114/Behavior/PIll_bug1.asp
http://www.phschool.com/science/biology_place/labbench/lab11/intro.html
Dr. Gensert
Orientation of Isopods in Response to Moisture
Procedure: Behavior chambers (choice chambers) were set up so that one side was moist and one side was dry (using filter paper cut to the size of a choice chamber). Five isopods were placed on each side of the chamber (total of 10).
1.
Open the virtual lab ( http://bioweb.wku.edu/courses/Biol114/Behavior/PIll_bug1.asp
), read the page and click on the arrow to begin.
2.
Count and record (in a table below) the number of animals on each side of the chamber every 30 seconds for 10 minutes .
3.
Using the table on the back, calculate Chi Square and determine whether or not you can reject the null hypothesis, indicate whether or not your data are “statistically significant”, and the level of significance/certainty.
Null Hypothesis (H o
):
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Hypothesis (H alt
):
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Time (m) # wet # dry Notes/Observations
Table 1.
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0:00 5 5
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1:00
2:00
3:00
4:00
5:00
6:00
7:00
8:00
9:00
10:00 mean
Figure 1. Choice Chamber Set Up. Five pill bugs were placed in the center of each chamber at the start of the experiment.
Figure 2. Sample Choice Chamber Results.
Pill bug positions at the start (A), and possible positions at end of the experiment if pill bugs prefer wet (B) or dry (C) conditions.
When would you use Chi Square?
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_________________________________________________________________________________________________

Names: ___________________________ Research Literacy
___________________________ Rolly-Polly (Pill Bug, Isopod) Chi Square Lab
___________________________
Virtual Lab: http://bioweb.wku.edu/courses/Biol114/Behavior/PIll_bug1.asp
http://www.phschool.com/science/biology_place/labbench/lab11/intro.html
Dr. Gensert
 
Wet
Dry
 
Expected (e) Observed (o)
     o – e (o – e)
2
(o – e)
2
/e
 2
(Chi Square) =
= ________
 2 = _________ df = number of possible outcomes – 1 = ________
Critical Value at p = 0.05 ____________
Conclusion:
(should include the terms: reject/fail to reject the null hypothesis, critical value, p < {probability level}, significant or not significant) _________________________________________________________________________
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