# Chi-Square Tests. Comments on Projects. Comments on Projects. The Big Picture. Comments on Projects. Population 12/23/2012. Sample and Population

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1 STAT 101 Dr. Kari Lock Morgan 10/30/1 Chi-Square Tests SECTIONS 7.1, 7. Testing the distribution of a single categorical variable : goodness of fit (7.1) Testing for an association between two categorical variables: test for association (7.) Comments on Projects Random sampling is different than convenience sampling Even if your initial sample selected is representative, if people choose not to respond this introduces bias Generally a good idea to present the response rate Comments on Projects Sample is whatever your variable is measured on (rows of your dataset) The type of inference we have learned can only generalize from sample to population (and only if a random sample) We have not learned how to account for uncertainty in the data values themselves Sample and Population Was your sample the same as your population? (Did you have data on all the cases in your population?) a) Yes b) No Comments on Projects The Big Picture If you have data on the whole population, inference is not necessary! Population Sampling Statistical Inference Sample 1

2 Multiple Categories So far, we ve learned how to do inference for categorical variables with only two categories Today, we ll learn how to do hypothesis tests for categorical variables with multiple categories Play a game! (Roshambo) Can we use statistics to help us win? Which did you throw first? a) Rock b) Paper c) Scissors ROCK PAPER SCISSORS How would we test whether all of these categories are equally likely? Hypothesis Testing 1. State Hypotheses. Calculate a statistic, based on your sample data test statistic 3. Create a distribution of this statistic, as it would be observed if the null hypothesis were true 4. Measure how extreme your test statistic from () is, as compared to the distribution generated in (3) Hypotheses Let p i denote the proportion in the i th category. H 0 : All p i s are the same H a : At least one p i differs from the others OR H 0 : Every p i = 1/3 H a : At least one p i 1/3

3 Test Statistic Why can t we use the familiar formula sample statistic null value SE to get the test statistic? More than one sample statistic More than one null value We need something a bit more complicated Observed Counts The observed counts are the actual counts observed in the study ROCK PAPER SCISSORS Observed Expected Counts The counts are the counts if the null hypothesis were true ROCK PAPER SCISSORS Observed Expected For each cell, the count is the sample size (n) times the null proportion, p i = np i Chi-Square Statistic A test statistic is one number, computed from the data, which we can use to assess the null hypothesis The chi-square statistic is a test statistic for categorical variables: observed - ROCK PAPER SCISSORS Observed Expected (368.3)(18.3)(378.3) What Next? We have a test statistic. What else do we need to perform the hypothesis test? A distribution of the test statistic assuming H 0 is true How do we get this? Two options: 1) Simulation ) Distributional Theory 3

4 Simulation Let s see what kind of statistics we would get if rock, paper, and scissors are all equally likely. 1) Take 3 scraps of paper and label them Rock, Paper, Scissors. Fold or crumple them so they are indistinguishable. Choose one at random. ) What did you get? (a) Rock (b) Paper (c) Scissors 3) Calculate the statistic. 4) Form a distribution. 5) How extreme is the actual test statistic? Simulation Chi-Square ( ) Distribution If each of the counts are at least 5, AND if the null hypothesis is true, then the statistic follows a distribution, with degrees of freedom equal to Chi-Square Distribution df = number of categories 1 : df = 3 1 = p-value using distribution Chi-Square p-values 1. StatKey simulation or theoretical. RStudio: tail.p( chisquare,stat,df,tail= upper ) 3. TI-83: nd DISTR 7: cdf( lower bound, upper bound, df Because the is always positive, the p-value (the area as extreme or more extreme) is always the upper tail 4

5 Goodness of Fit A test for goodness of fit tests whether the distribution of a categorical variable is the same as some null hypothesized distribution The null hypothesized proportions for each category do not have to be the same Chi-Square Test for Goodness of Fit 1. State null hypothesized proportions for each category, p i. Alternative is that at least one of the proportions is different than specified in the null.. Calculate the counts for each cell as np i. Make sure they are all greater than 5 to proceed. 3. Calculate the statistic: observed - 4. Compute the p-value as the area in the tail above the statistic, for a distribution with df = (# of categories 1) 5. Interpret the p-value in context. In 1866, Gregor Mendel, the father of genetics published the results of his experiments on peas He found that his experimental distribution of peas closely matched the theoretical distribution predicted by his theory of genetics (involving alleles, and dominant and recessive genes) Source: Mendel, Gregor. (1866). Versuche über Pflanzen-Hybriden. Verh. Naturforsch. Ver. Brünn 4: 3 47 (in English in 1901, Experiments in Plant Hybridization, J. R. Hortic. Soc. 6: 1 3) S, Y: Dominant s, y: Recessive Mate SSYY with ssyy: 1 st Generation: all Ss Yy Mate 1 st Generation: nd Generation Second Generation Phenotype Round, Yellow 9/16 Round, Green 3/16 Wrinkled, Yellow 3/16 Wrinkled, Green 1/16 Theoretical Proportion Phenotype Theoretical Proportion Observed Counts Round, Yellow 9/ Round, Green 3/ Wrinkled, Yellow 3/ Wrinkled, Green 1/16 3 Let s test this data against the null hypothesis of each p i equal to the theoretical value, based on genetics H0 : p1 9 /16, p 3/16, p3 3/16, p4 1/16 H : At least one p is not as specified in H a i 0 Phenotype Theoretical Proportion Observed Counts Round, Yellow 9/ Round, Green 3/ Wrinkled, Yellow 3/ Wrinkled, Green 1/16 3 The statistic is made up of 4 components (because there are 4 categories). What is the contribution of the first cell (315)? (a) (b) 1.05 (c) 5.1 (d)

6 Phenotype p i Observed Counts Expected Counts Round, Yellow 9/ /16 = Round, Green 3/ /16 = Wrinkled, Yellow 3/ /16 = Wrinkled, Green 1/ /16 = n 556 = 556 observed = 0.47 Two options: o Simulate a randomization distribution o Compare to a distribution with 4 1 = 3 df p-value = 0.95 Does this prove Mendel s theory of genetics? Or at least prove that his theoretical proportions for pea phenotypes were correct? a) Yes b) No You can never accept the null hypothesis!!! The null hypothesis can only be rejected with statistically significant results. A high p-value doesn t let you conclude anything. Two Categorical Variables The statistics behind a test easily extends to two categorical variables A test for association (often called a test for independence) tests for an association between two categorical variables Did reading the example on Rock- Paper-Scissors in the textbook influence the way you play the game? Did you read the example in Section 6.3 of the textbook? a) Yes b) No Rock Paper Scissors Total Read Example Did Not Read Example Total H 0 : Whether or not a person read the example in the textbook is not associated with how a person plays rock-paper-scissors H a : Whether or not a person read the example in the textbook is associated with how a person plays rock-paper-scissors 6

7 = Expected Counts row total column total sample size Chi-Square Statistic observed - Calculate the count for your cell. Note: Maintains row and column totals, but redistributes the counts as if there were no association = Randomization Test Chi-Square Statistic observed - If the counts are all at least 5, this can be compared to a distribution with (r 1)(c 1) degrees of freedom, where r = number of rows, c = number of columns Chi-Square Distribution df = (r 1)(c 1) = ( 1)(3 1) = Chi-Square Test for Association 1. H 0 : The two variables are not associated H a : The two variables are associated. Calculate the counts for each cell: row total column total = sample size Make sure they are all greater than 5 to proceed. 3. Calculate the statistic: observed - 4. Compute the p-value as the area in the tail above the statistic, for a distribution with df = (r 1) (c 1) 5. Interpret the p-value in context. 7

8 Metal Tags and Penguins Let s return to the question of whether metal tags are detrimental to penguins. Survived Died Total Metal Tag Electronic Tag Total Is there an association between type of tag and survival? (a) Yes (b) No Source: Saraux, et. al. (011). Reliability of flipper-banded penguins as indicators of climate change, Nature, 469, Metal Tags and Penguins Survived Died Total Metal Tag 33 (47.38) 134 (119.6) 167 Electronic Tag 68 (53.6) 11 (135.38) 189 Total Calculate counts (shown in parentheses). First cell: row total column total sample size 356 Calculate the chi-square statistic: p-value: df = ( 1) ( 1) = 1 p-value = Statistics: There strong Unlocking evidence the Power that of metal Data tags are detrimental to penguins. Lock 5 Metal Tags and Penguins Difference in proportions test: z = (we got something slightly different due to rounding) p-value = (two-sided) Chi-Square test: = p-value = Note: (-3.387) = Two Categorical Variables with Two Categories Each When testing two categorical variables with two categories each, the statistic is exactly the square of the z-statistic The distribution with 1 df is exactly the square of the normal distribution Doing a two-sided test for difference in proportions or doing a test will produce the exact same p-value. Summary The test for goodness of fit tests whether one categorical variable differs from a hypothesized distribution The test for association tests whether two categorical variables are associated For both, you calculate counts for each cell, compute the statistic as observed - and find the p-value as the area above this statistic on a distribution To Do Read Chapter 7 Do Homework 6 (due Tuesday, 11/6) 8

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