1 / 30

TESTING HYPOTHESES

TESTING HYPOTHESES. Two ways of arriving at a conclusion. 1. Deductive inference. sample. population. 2. Inductive inference. sample. population. IF YOUR DATA ARE:. 1. Continuous data. 4. Equal variance (F-test). 2. Ratio or interval.

Télécharger la présentation

TESTING HYPOTHESES

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. TESTING HYPOTHESES

  2. Two ways of arriving at a conclusion 1. Deductive inference sample population 2. Inductive inference sample population

  3. IF YOUR DATA ARE: 1. Continuous data 4. Equal variance (F-test) 2. Ratio or interval 5. Conclusions about population based on sample (inductive) 3. Approximately normal distribution sample population 6. Sample size > 10

  4. Imagine the following experiment: 2 groups of crickets Group 1 – fed a diet with extra supplements Group 2 – fed a diet with no supplements Weights Mean = 9.49 Mean = 12.8

  5. What you’re doing here is comparing two samples that, because you’ve not violated any of the assumptions we saw before, should represent populations that look like this: 9.49 12.8 Frequency Weight Are the means of these populations different??

  6. Are the means of these populations different?? To answer this question – use a statistical test A statistical test is just a method of determining mathematically whether you definitively say ‘yes’ or ‘no’ to this question What test should I use??

  7. IF YOU HAVEN’T VIOLATED ANY OF THE ASSUMPTIONS WE MENTIONED BEFORE…… Number of groups compared 2 other than 2 ANOVA T -test Are the means of more than two populations the same? Are the means of two populations the same? Direction of difference specified? Number of factors being tested Yes No 1 2 >2 Does each data point in one data set (population) have a corresponding one in the other data sets? One-tailed Two- tailed Does each data point in one data set (population) have a corresponding one in the other data set? Two way ANOVA Other tests Yes No Yes No Repeated Measures ANOVA One way ANOVA Paired t-test Unpaired t-test

  8. A simple t-test 1. State hypotheses Ho – there is no difference between the means of the two populations of crickets (i.e. the extra nutrients had no effect on weight) H1 – there is a difference between the means of the two populations of crickets (i.e. the extra nutrients had an effect on weight)

  9. A simple t-test 2. Calculate a t-value (any stats program does this for you) (for the truly masochistic) 3. Use a probability table for the test you used to determine the probability that corresponds to the t-value that was calculated.

  10. A simple t-test 2. Calculate a t-value (any stats program does this for you) 3. Use a probability table for the test you used to determine the probability that corresponds to the t-value that was calculated. Data Test statistic Probability

  11. Unpaired t test Do the means of Nutrient fed and No nutrient differ significantly? P value The two-tailed P value is < 0.0001, considered extremely significant. t = 7.941 with 38 degrees of freedom. 95% confidence interval Mean difference = -3.307 (Mean of No nutrient minus mean of Nutrient fed) The 95% confidence interval of the difference: -4.150 to -2.464 Assumption test: Are the standard deviations equal? The t test assumes that the columns come from populations with equal SDs. The following calculations test that assumption. F = 1.192 The P value is 0.7062. This test suggests that the difference between the two SDs is not significant. Assumption test: Are the data sampled from Gaussian distributions? The t test assumes that the data are sampled from populations that follow Gaussian distributions. This assumption is tested using the method Kolmogorov and Smirnov: Group KS P Value Passed normality test? =============== ====== ======== ======================= Nutrient fed 0.1676 >0.10 Yes No nutrient 0.1279 >0.10 Yes

  12. Interpretation of p < .0001? This means that there is less than 1 chance in 10,000 that these two means are from the same population. In the world of statistics, that is too small a chance to have happened randomly and so the Ho is rejected and the H1 accepted

  13. For all statistical tests that you’ll use, it is convention that the minimum probability that two samples can differ and still be from the same population is 5% or p = .05

  14. Nonparametric Statistics (Nominal Data) & Goodness-of-Fit Tests

  15. What happens if you violate any of the assumptions? Step 1 - Panic

  16. What happens if you violate any of the assumptions? Step 1 - Panic Step 2 - It depends on what assumptions have been violated.

  17. Nonparametric Tests These tests are used when the assumptions of t-tests and ANOVA have been violated They are called “nonparametric” because there is no estimation of parameters (means, standard deviations or variances) involved. Several kinds: Goodness-of-Fit tests - when you calculate an expected value Non-parametric equivalents of parametric tests

  18. Goodness-of-Fit Tests Use with nominal scale data e.g. results of genetic crosses Also, you’re using the population to deduce what the sample should look like

  19. Classic example - genetic crosses Do they conform to an “expected’ Mendelian ratio? Back to our little ball creatures - Critterus sphericales Phenotypes: A_B_ A_bb aaB_ aabb Mendelian inheritance -Predict a 9:3:3:1 ratio

  20. -sampled 320 animals

  21. -sampled 320 animals

  22. -sampled 320 animals

  23. -sampled 320 animals

  24. -sampled 320 animals

  25. -sampled 320 animals (o -e)2 e S C2 = = 1.08 + .82 + .82 + 9.8 = 12.52 df = number of classes -1 = 3

  26. X2 = 12.52 Critical value for 3 degrees of freedom at .05 level is 7.82 X2 Table The actual probability of X2 =12.52 and df = 3 is .01 > p > .001 Conclusion: Probability of these data fitting the expected distribution is < .05, therefore they are not from a Mendelian population

  27. A little X2 wrinkle - the Yates correction (o -e)2 e S Formula is C2 = Except of df = 1 (i.e. you’re using two categories of data) Then the formula becomes (|o -e| - 0.5)2 e S C2 =

  28. A second goodness-of-fit test G-test or Log-Likelihood Ratio Use if |o - e | < e e.g. if o is 12 and e is 7 o e o e G = 2 o ln = 4.60517 * o log10 S S

  29. Summary!

  30. All of the parametric tests (remember the big flow chart!) have non-parametric equivalents (or analogues)

More Related