Need a clear paper including this requirement
You will need entries for:
- Descriptive statistics: Mean, Median, Mode, Standard deviation, variance, range (100 pts)
- z test (100)
- T-Tests (multiple types) (100)
- ANOVA (One and Two)
- Notes on creating graphs
- Discussion of errors and probability distributions.
- Suggestion: a glossary for all the terms. (Bonus 100)
Outline for each entry:
- The written description of the test.
- Why you would use this test.
- with clear variable definitions
- can reference formulas
- Clear handwritten example calculations
- Code for R with an example
- Can come from homework or a project.
- Explain what the results mean. (accept or reject the hypothesis. p-value meaning, etc)
· Chi-Square Test (ꭓ2)
· Definition – a test to determine the various deviations expected by chance, if the hypothesis is true.
· Say you have a known distribution that you sample from. You expect to get the same distribution within your sample. This test determines the deviation from the known distribution in your sample. Genetics is a common field this is used in. There is a known probability for genotypes within a population. Chi-square can be used to determine if your sample deviates from this known genotypes distribution.
· Where O = the observed frequency
· E = expected frequency
· You have plants with red, yellow, and orange petals with the following genotypes and probabilities:
· RR = Red – 25%
· Rr = Orange – 50%
· rr = Yellow – 25%
· In a population of 320 you expect an observation of
· RR = 80
· Rr = 160
· rr = 80
· IF there is incomplete dominance.
· You grow these plants in your garden, this is what you see:
· RR = 65
· Rr = 189
· rr = 66
· ꭓ2 = 10.51
· Using the Critical Values of Chi Square distribution. A df of 2, p would be between 0.01 or 0.005.
· My null hypothesis is that the Observed and Expected outcomes would be equal if there is incomplete dominance.
· If my null hypothesis is true, deviations this large should only be expected 0.5-1% of the time.
· In this case, the hypothesis will be rejected because this is too unlikely that you would get these results with incomplete dominance.
· R code and example
· chisq.test(x, p = rep(1/length(x), length(x))
· x = data set
· p = probabilities for each data point in the data set
> genetest <- c(65, 189, 66)
> chisq.test(genetest, p = c(1/4, 1/2, 1/4))
Chi-squared test for given probabilities
X-squared = 10.519, df = 2, p-value = 0.005199
· Conclusion is that the probability is so small of this happening, the expected outcome is not correct. You reject the null hypothesis.