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Chi-Square (χ²)

Chi-Square (χ²) . By: Anastasiya Biloblotska & Cheryl Li Period 1+2 - AP Biology. What is χ²?. - A type of statistical test called “goodness of fit” test

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Chi-Square (χ²)

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  1. Chi-Square (χ²) By: Anastasiya Biloblotska & Cheryl Li Period 1+2 - AP Biology

  2. What is χ²? - A type of statistical test called “goodness of fit” test - Purpose: To determine if any differences between our observed measurements and expected are due to chance (sample error) or some other reason(s) - Generally used when we deal with discrete data (count data, non-continuous data)

  3. Formula for χ² *found on Appendix A!!

  4. So what do we do with χ²? - Now, we determine the probability of getting this χ² value by using the Chi-Square table. - The probability value (p-value) will tell us what the chances are that the differences in our data are due simply to chance alone.

  5. Chi-Square Distribution Table

  6. What we need: - Observed data (o) - Expected data (e) - Degree(s) of freedom (# of categories - 1) - χ² - Null hypothesis (there should be no change between observed and expected)

  7. Let’s Do Some Breeding Null Hypothesis: The turtles were produced by a monohybrid cross (Aa x aa) resulting in an expected ratio of 1:1 Aa aa - Green shell - Circular shell - Eats shrimp - Long appendages - Red shell - Oval shell - Eats shellfish - Short appendages F1 aa Aa - Half of the turtles are green shelled, eat shrimp, and have long appendages - Half of the turtles are red shelled, eat shellfish, and have short appendages

  8. F1 Generation . - Normal green shell without variation Eggs: 17 Aa - Normal red shell without variation Eggs: 15 aa

  9. Analysis

  10. Calculations • Using the Chi-Square formula we calculated the chi-square total: • (15-16)2/16 = .0625 • (17-16)2/16 = .0625 • X2 = .125 • How many degrees of freedom? 1 • The Null Hypothesis was supported with X2= .125 and a df of 1. The differences between the observed and expected results could be due to chance alone and are nonsignificant. We can say that the turtles were produced by a monohybrid cross (Aa x aa).

  11. Squeak! 2 black-eyed mice were crossed and the result was 64 black-eyed mice and 36 red-eyed mice. [Black (B), Red (b)] Null hypothesis: The black-eyed parents were Bb x Bb. Does the X² analysis support your hypothesis?

  12. Show your work! Hint: Make a Punnett Square first to determine the expected values.What are the chances of getting black-eyed mice? Red-eyed mice? B b B b BB Bb Bb bb

  13. Results - Black-eyed mice (BB, Bb): ¾ - Red-eyed mice (bb): ¼ X² = 1.61 + 4.84 X² = 6.45df = 2 - 1 df = 1

  14. Conclusion Since X² = 6.45 with df = 1, we reject the null hypothesis. There’s a 1% chance that the difference between the observed and expected data is due to random chance. The differences between the observed and expected results could not be due to random chance alone and therefore are significant. It could be due to mutations, environmental conditions, etc.

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