Chi Square

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

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

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 ___

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.

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

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

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)

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

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

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:

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

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!!!

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.

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

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

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

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.

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