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

expected

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)

REJECT NULLACCEPT 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

SUM of Chi Squares: Df:

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 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