1. Chi-Square Test

2. Chi-Square (χ 2 ) Test Used to determine if there is a significant difference between the expected and observed data Null hypothesis: There is NO statistically significant difference between expected & observed data Any differences are due to CHANCE alone

3. Chi-Square (χ 2 ) Formula

4. How to use the Chi-Square Test 1.Determine null hypothesis All frequencies are equal –OR– Specific frequencies given already 2.Use formula to calculate χ 2 value: n = # of categories, e = expected, o = observed 3.Find critical value using table (Use p=0.05). degrees of freedom (df) = n – 1 4.If χ 2 < Critical Value, then ACCEPT null hypothesis. Differences in data are due to chance alone. If χ 2 > Critical Value, REJECT the null hypothesis: Differences in data are NOT due to chance alone!

6. Sample Problem yellow You buy a package of M&Ms from the factory store and find the following: 20 brown, 20 blue, 20 orange, 20 green, and 20 yellow M&Ms. yellow According to the M&M website, each package of candy should have 13% brown, 24% blue, 20% orange, 16% green, 13% red, and 14% yellow M&Ms. You realize you are missing Red M&M’s in your package! Is this acceptable, or did something happen in the factory during the packaging process? Use the Chi-Square Test to answer this question.

7. Consider this story.... Two tigers at a zoo are bred together and they have four cubs.

8. Two of the four cubs are albino tigers. Based on that, Kristin hypothesizes that both of the parents must be carrying a recessive gene for albinism. The cross would look like: A a x A a Who fell into the bleach? At least they have a future in the circus..... Don't hate me because I'm beautiful

9. If Kristin's hypothesis is accurate the punnett square would look like..

10. This AP Bio student is unconvinced. If your hypothesis is correct, then only ONE of the four kittens should be an albino. You are so dumb...you are really really dumb....

11. But isn't 1/4 pretty close to 2/4...maybe the difference is just due to chance.... Once I flipped a coin four times I got heads 3 times. Sometimes it just happens that way. Maybe you just got lucky and got an extra white kitten..

12. The only way to solve this problem and the argument is to do a statistical analysis. We call this type of analysis a CHI SQUARE The purpose is to determine whether the results are statistically significant. What are the odds that your tigers are Aa x Aa? Or could other factors be at work here? I am so going to win this argument!

13. Here's how to do a chi square. Summed for all classes means that you are looking at all the traits you observed - in this case, orange and white.

14. To apply the formula, plug in your "observed" and "expected" numbers....this will give you I do not like math!

15. 1.33? Is that good or bad? Who is right? Who is wrong? What time is it? = 1.33 To determine if this number is good or not, you must look at a chi square chart. "Degrees of freedom" is one less than the original number of classes you looked at, which was 2 (orange & white) So we will look at the first row (DoF = 1)

16. 1.33 is between the 20% and 30% columns Basically this means that the difference you observed between orange and white cubs can be expected to occur more than 20% of the time, just due to chance. Scientists use 5% as the cut-off percent to reject a hypothesis. Results are always better with a large sample size.

17. If you find that you have a "poor fit", that means that you probably need to reject the hypothesis. In the tiger cub case, we did not have a poor fit. Well obviously, I was right. You can run and tell that..

18. Poor fit. Emily thinks she gets it now. So she looks at another case. She breeds two black mice together and finds that over the course of 3 years, the parents produce 330 brown mice, and 810 black mice. She hypothesizes that the parents are Bb (heterozygous). How can she prove this with a chi square?

19. Online Chi Square Calculator at http://www.graphpad.com/quickcalcs/chisquared1.cfmhttp://www.graphpad.com/quickcalcs/chisquared1.cfm -- just plug in the observed and expected values