1. Chi Square

2. What does Chi-square analysis tell us?
If the difference between your observed results and expected results are due to random chance or not.
EX –flip a coin enough and you would expect to get heads 50% of time and tails 50% of time. If this number is way off even after 1,000 tosses then something else besides randomness is occurring

3. How’s it works Calculation to use is X2 = ∑ (observed – expected)2 X2
Table to set up
# of Possible
Outcomes
Observed
Expected X2

4. How it works continued
Determine degrees of freedom – it’s one less than the number of possible outcomes
Ex – expect 4 different phenotypes, degrees of freedom is 3
Ex 2 – expect a cell to be in one of two different phases, degrees of freedom is ___

5. What are you looking for? Support for or against the null hypothesis.
Null Hypothesis= there will be no difference between observed and expected results
p = .05 is the cut off pt for most scienitific data, you’re asking does the data support my NULL hypothesis?
Ex. Is the variable being tested actually making a significant difference? We start by assuming it won’t.

6. How it works
Once you have your degrees of freedom and Chi square answer, go to the distribution table and compare your answer to the critical value in the 0.05 column

7. What?
If X2 answer is lower than critical value – ACCEPT the null hypothesis
If X2 answer is higher than critical value – REJECT the null hypothesis (something is going on other than randomness)
ACCEPT NULL
REJECT NULL

8. Explain please If p = .05 then what does that mean?
the difference between observed and expected results is due to random events 5% of time if your critical factor is equal to or less.
Your NULL hypothesis is supported!
(if higher than critical factor than NULL hypothesis not supported – something going on)

9. Try it P = Purple p = Yellow S = Smooth PpSs x PpSs s = Shrunken
What are the expected results?
Hypothesis? 9:3:3:1 due to independent assort of alleles

10. Results analysis
Observed results: An ear of corn has a total of 381 grains, including 216 Purple & Smooth, 79 Purple & Shrunken, 65 Yellow & Smooth, and 21 Yellow & Shrunken.
Expected? – do the dihybrid p-square if need to

11. Table, Chi Square and Degrees of Freedom
Possible Outcomes
Observed
Expected
Chi Square
X2 = ∑ (observed – expected)2
expected
Purple/
smooth
216
Purple/ shrunken 79
Yellow/ smooth 65
Yellow/ shrunken 21
Df:
SUM of Chi Squares:

12. Hypothesis supported?
2.197 is chi square value – compare to proper critical value

13. Answer 2.197 is less than critical value of 7.82 therefore……
Null hypothesis is ACCEPTED – there is no significant difference between observed data and expected!!!

14. Try this one
Same hypothesis: independent assort therefore 9:3:3:1 outcome is expected
Observed: An ear of corn has a total of 389 grains, including 219 Purple & Smooth, 70 Yellow & Smooth, and 100 white.

15. Possible Outcomes Observed Expected Purple/ smooth Purple/ shrunken
Chi Square
X2 = ∑ (observed – expected)2
expected
Purple/
smooth
Purple/ shrunken
Yellow/ smooth
Yellow/ shrunken

16. Mitosis Lab
State a null hypothesis for the experimental treatment on mitosis rate for onion root tip:
How will we collect data for the Expected results?
How will we collect data for the Observed results?
Hint – keep it simple, the cell is either in_______ or _________.
Helpful video: Bozeman Chi Square

17. Practice makes perfect
Animal Behavior data from the Bozeman Video
Next few slides: Mendelian genetics practice

18. 6. FRUIT FLY GENETICS
Cross two heterozygous wild type fruit flies for the traits below. Predict expected outcomes of offspring using a punnett square. W= normal wings w=vestigial
B = normal body b = black
Go to the next slide for actual results then perform Chi Square to determine if there is a significant statistical difference between the expected and observed results.

19. Actual results of the cross
800 Wild/Wild 55 wild-type wings/ black body 50 vestigial wings/ wild type body 902 vestigial/ black body