Presentation on theme: "Chi Square Analysis The chi square analysis allows you to use statistics to determine if your data “good” or not. In our fruit fly labs we are using laws."— Presentation transcript:
1 Chi Square AnalysisThe chi square analysis allows you to use statistics to determine if your data “good” or not.In our fruit fly labs we are using laws of probability to determine possible outcomes for genetic crosses.How will we know if our fruit fly data is “good”?
2 m & m …Peanut m&m data from Mars™Peanut butter m&m data from Mars™
3 The following formula is used If your hypothesis is supported by datayou are claiming that mating is random and so is segregation and independent assortment.If your hypothesis is not supported by datayou are seeing that the deviation between observed and expected is very far apart…something non-random must be occurring….
4 The test statistic is compared to a theoretical probability distribution In order to use this distribution properly you need to determine the degrees of freedomDegrees of freedom is the number of phenotypic possibilities in your cross minus one.If the level of significance read from the table is greater than .05 or 5% then your hypothesis is accepted and the data is useful
5 The hypothesis is termed the null hypothesis which states that there is no substantial statistical deviation between observed and expected data.
6 Let’s look at a fruit fly cross xBlack body, eyelesswildF1: all wild
8 Analysis of the results Once the numbers are in, you have to determine the cross that you were using.What is the expected outcome of this cross?9/16 wild type: 3/16 normal body eyeless: 3/16 black body wild eyes: 1/16 black body eyeless.
9 Now Conduct the Analysis: To compute the hypothesis value take 10009/16 = 626
10 Now Conduct the Analysis: To compute the hypothesis value take 10009/16 = 626
11 Using the chi square formula compute the chi square total for this cross: ( )2/ 5630 = .07( )2/ 1877 = .01( )2/ 1877 = .20( ) 2/ 626 = .02 2= .30How many degrees of freedom?
12 Using the chi square formula compute the chi square total for this cross: ( )2/ 5630 = .07( )2/ 1877 = .01( )2/ 1877 = .20( ) 2/ 626 = .02 2= .30How many degrees of freedom? 3
13 Accept Hypothesis Reject Hypothesis CHI-SQUARE DISTRIBUTION TABLE Probability (p)Degrees ofFreedom0.950.900.800.700.500.300.200.100.050.010.00110.0040.020.060.150.461.071.642.713.846.6410.8320.210.450.711.392.413.224.605.999.2113.8230.350.581.011.422.373.664.646.257.8211.3416.2741.061.652.203.364.887.789.4913.3818.4751.141.612.343.004.356.067.299.2411.0715.0920.5261.633.073.835.357.238.5610.6412.5916.8122.4672.172.833.824.676.358.389.8012.0214.0718.4824.3282.733.494.595.537.349.5211.0313.3615.5120.0926.1293.324.175.386.398.3410.6612.2414.6816.9221.6727.88103.944.866.187.279.3411.7813.4415.9918.3123.2129.59
14 When reporting chi square data use the following formula sentence…. With degrees of freedom, my chi square value is , which gives me a p value between % and %, I therefore my null hypothesis.This sentence would go in the “reults” section of your formal lab.Your explanation of the significance of this data would go in the “discussion” section of the formal lab.
15 Looking this statistic up on the chi square distribution table tells us the following: the P value read off the table places our chi square number of .30 close to .95 or 95%This means that 95% of the time when our observed data is this close to our expected data, this deviation is due to random chance.We therefore accept our null hypothesis.
16 What is the critical value at which we would reject the null hypothesis? For three degrees of freedom this value for our chi square is > 7.815What if our chi square value was 8.0 with 4 degrees of freedom, do we accept or reject the null hypothesis?Accept, since the critical value is >9.48 with 4 degrees of freedom.
17 How can we use the chi square test in a different sort of lab? If we were testing the effect of light on plant growth, how could the chi square analysis be used?In this case our experimental group is plants grown in the dark and the control plants are grown in the light.
18 What would your expected data be? Control plants grown in the light.What would your observed group be?Plants grown in the dark.Do you expect to accept or reject the null hypothesis?reject
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