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Chi-Square Tests 3/14/12 Testing the distribution of a single categorical variable : 2 goodness of fit Testing for an association between two categorical variables: 2 test for association Section 7.1, 7.2 Professor Kari Lock Morgan Duke University

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For more information and to sign up, go to www.stat.duke.edu/datafest www.stat.duke.edu/datafest

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NOTE: This week only, my Friday office hours will be from 3:30 – 5:30 pm (NOT 1-3pm as usual) Office Hours This Week

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Homework 6 (due Monday, 3/19) Homework 6 Anonymous Midterm Evaluation (due Monday, 3/19, 5pm) Anonymous Midterm Evaluation Project 1 (due Thursday, 3/22, 5pm) Project 1 To Do

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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

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Rock-Paper-Scissors (Roshambo) Play a game! Can we use statistics to help us win?

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Rock-Paper-Scissors Which did you throw? a)Rock b)Paper c)Scissors

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Rock-Paper-Scissors ROCKPAPERSCISSORS 342440 How would we test whether all of these categories are equally likely?

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Hypothesis Testing 1.State Hypotheses 2.Calculate a statistic, based on your sample data 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 (2) is, as compared to the distribution generated in (3) test statistic

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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

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Test Statistic Why can’t we use the familiar formula to get the test statistic? More than one sample statistic More than one null value We need something a bit more complicated…

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Observed Counts The observed counts are the actual counts observed in the study ROCKPAPERSCISSORS Observed342440

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Expected Counts The expected counts are the expected counts if the null hypothesis were true For each cell, the expected count is the sample size (n) times the null proportion, p i ROCKPAPERSCISSORS Observed342440 Expected33

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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:

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Rock-Paper-Scissors ROCKPAPERSCISSORS Observed342440 Expected33

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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 2)Distributional Theory

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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. 2) What did you get? (a) Rock(b) Paper(c) Scissors 3) Calculate the 2 statistic. 4) Form a distribution. 5) How extreme is the actual test statistic?

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RStudio Chi-Square Distribution

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If each of the expected counts are at least 5, AND if the null hypothesis is true, then the 2 statistic follows a 2 –distribution, with degrees of freedom equal to df = number of categories – 1 Rock-Paper-Scissors: df = 3 – 1 = 2

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Chi-Square Distribution

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Chi-Square p-values 1.Applet: http://surfstat.anu.edu.au/surfstat-home/tables/chi.php 2.RStudio: pchisq(ts,df,lower.tail=FALSE) 3.TI-83: 2 nd DISTR 7: 2 cdf( lower bound, upper bound, df Because the 2 is always positive, the p-value (the area as extreme or more extreme) is always the upper tail

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t-distribution p-values and t * 1.Applet: http://surfstat.anu.edu.au/surfstat-home/tables/t.php 2.RStudio: p-value: pt(ts,df) (use lower.tail=TRUE or FALSE, and double if two-tailed) t* for 95% CI: qt(0.975,df) 3.TI-83: p-value: 2 nd DISTR 5: tcdf( lower bound, upper bound, df t*: Approximate with a normal (invnorm)

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Goodness of Fit A 2 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

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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. 2.Calculate the expected counts for each cell as np i. Make sure they are all greater than 5 to proceed. 3.Calculate the 2 statistic: 4.Compute the p-value as the area in the tail above the 2 statistic, for a 2 distribution with df = (# of categories – 1) 5.Interpret the p-value in context.

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Mendel’s Pea Experiment 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. 26: 1–32) 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)

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Mendel’s Pea Experiment Mate SSYY with ssyy: 1 st Generation: all Ss Yy Mate 1 st Generation: 2 nd Generation Second Generation S, Y: Dominant s, y: Recessive PhenotypeTheoretical Proportion Round, Yellow9/16 Round, Green3/16 Wrinkled, Yellow3/16 Wrinkled, Green1/16

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Mendel’s Pea Experiment PhenotypeTheoretical Proportion Observed Counts Round, Yellow9/16315 Round, Green3/16101 Wrinkled, Yellow3/16108 Wrinkled, Green1/1632 Let’s test this data against the null hypothesis of each p i equal to the theoretical value, based on genetics

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Mendel’s Pea Experiment PhenotypeTheoretical Proportion Observed Counts Round, Yellow9/16315 Round, Green3/16101 Wrinkled, Yellow3/16108 Wrinkled, Green1/1632 The 2 statistic is made up of 4 components (because there are 4 categories). What is the contribution of the first cell (315)? (a) 0.016(b) 1.05(c) 5.21(d) 107.2

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Mendel’s Pea Experiment Phenotypepipi Observed Counts Expected Counts Round, Yellow 9/16315 556 9/16 = 312.75 Round, Green 3/16101 556 3/16 = 104.25 Wrinkled, Yellow 3/16108 556 3/16 = 104.25 Wrinkled, Green 1/1632 556 1/16 = 34.75

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Mendel’s Pea Experiment Compare to a 2 distribution with 4 – 1 = 3 degrees of freedom http://surfstat.anu.edu.au/surfstat-home/tables/chi.php p-value = 0.9254

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Mendel’s Pea Experiment p-value = 0.9254 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

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Two Categorical Variables The statistics behind a 2 test easily extends to two categorical variables A 2 test for association (often called a 2 test for independence) tests for an association between two categorical variables

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Rock-Paper-Scissors Did reading the example on Rock- Paper-Scissors in the textbook influence the way you play the game? Did you read the Rock-Paper-Scissors example in Section 6.3 of the textbook? a) Yes b) No

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Rock-Paper-Scissors RockPaperScissorsTotal Read Example210517 Did Not Read Example32123480 Total34243997 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

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Expected Counts Maintains row and column totals, but redistributes the counts as if there were no association between the variables RockPaperScissorsTotal Read Example210517 Did Not Read Example32123480 Total34243997

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Chi-Square

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Chi-Square Statistic OBSERVEDRockPaperScissorsTotal Read Example210517 Did Not Read Example32123480 Total34243997 EXPECTEDRockPaperScissorsTotal Read Example4.216.8417 Did Not Read Example28.0419.7932.1680 Total34243997

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Rock-Paper-Scissors The smallest expected count is 5.96, which is greater than 5, so it is okay to use the chi-square distribution to find the p-value. p-value = 0.0005 We have strong evidence that there is an association between reading the RPS example in the textbook and the first RPS throw. (This means you are paying attention and using what you learn in the textbook! )

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Chi-Square Test for Association 1.H 0 : The two variables are not associated H a : The two variables are associated 2.Calculate the expected counts for each cell: Make sure they are all greater than 5 to proceed. 3.Calculate the 2 statistic: 4.Compute the p-value as the area in the tail above the 2 statistic, for a 2 distribution with df = (r – 1) (c – 1) 5.Interpret the p-value in context.

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Metal Tags and Penguins Let’s return to the question of whether metal tags are detrimental to penguins. Source: Saraux, et. al. (2011). “Reliability of flipper-banded penguins as indicators of climate change,” Nature, 469, 203-206. SurvivedDiedTotal Metal Tag33134167 Electronic Tag68121189 Total101255356 Is there an association between type of tag and survival? (a) Yes(b) No

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Metal Tags and Penguins Calculate expected counts (shown in parentheses). First cell: Calculate the chi-square statistic: p-value: SurvivedDiedTotal Metal Tag33134167 Electronic Tag68121189 Total101255356 SurvivedDiedTotal Metal Tag33 (47.38)134 (119.62)167 Electronic Tag68 (53.62)121 (135.38)189 Total101255356 df = (2 – 1) × (2 – 1) = 1 p-value = 0.0007 There is strong evidence that metal tags are detrimental to penguins.

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Metal Tags and Penguins Difference in proportions test: z = -3.387 (we got something slightly different due to rounding) p-value = 0.0007 (two-sided) Chi-Square test: 2 = 11.47 p-value = 0.0007 Note: (-3.387) 2 = 11.47

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Two Categorical Variables with Two Categories Each When testing two categorical variables with two categories each, the 2 statistic is exactly the square of the z-statistic The 2 distribution with 1 df is exactly the square of the normal distribution Doing a two-sided test for difference in proportions or doing a 2 test will produce the exact same p-value.

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Summary The 2 test for goodness of fit tests whether one categorical variable differs from a hypothesized distribution The 2 test for association tests whether two categorical variables are associated For both, you calculate expected counts for each cell, compute the 2 statistic as and find the p-value as the area above this statistic on a 2 distribution

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Lecture 11. The chi-square test for goodness of fit.

Lecture 11. The chi-square test for goodness of fit.

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