Lecture 9. Selected material from: Ch. 12 The analysis of categorical data and goodness of fit tests

Transcription

1 Lecture 9 Selected material from: Ch. 12 The analysis of categorical data and goodness of fit tests

2 Univariate categorical data Univariate categorical data are best summarized in a one way frequency table. Example: Births and the Lunar cycle A common urban legend is that more babies than expected are born during certain phases of the lunar cycle, especially near the full moon. The paper The Effect of the Lunar Cycle on Frequency of Births and Birth Complications (American Journal of Obstetrics and Gynecology, 2005) reported data from a random sample of births that occurred during 24 lunar cycles. Perform a test of whether the lunar cycle has an effect on births at the.05 statistical level. 2

3 Example: Births and the Lunar cycle Data from 24 lunar cycles: Lunar phase Number of days Number of births New moon Waxing crescent First quarter Waxing gibbous Full moon Waning gibbous Last quarter Waning crescent

4 Example: Births and the Lunar cycle Notation: Lunar phase Number of days Proportion of births New moon 24 1 Waxing crescent First quarter 24 3 If there is no relationship between births and the lunar phase, the proportion of births in each lunar phase should just be: Waxing gibbous Full moon 24 5 Waning gibbous Last quarter 24 7 Waning crescent Total = 699 days Number of days in phase, Total number of days i. e., 24 / , / , /

5 Example: Births and the Lunar cycle Lunar phase Number of days Under H 0 ( i = #days/total days) New moon 24 1 =.0343 Waxing crescent =.2175 First quarter 24 3 =.0343 Waxing gibbous =.2132 Full moon 24 5 =.0343 Waning gibbous =.2146 Last quarter 24 7 =.0343 Waning crescent =.2175 Total = 699 days Total = 1.0 Null hypothesis H 0 Alternative hypothesis H a : H 0 not true, i.e., at least one of the proportions does not equal its null hypothesized value. 5

6 Example: Births and the Lunar cycle Data from 24 lunar cycles: Lunar phase Number of days Number of births New moon Waxing crescent First quarter Under H 0 : 1 =.0343 Expected number Waxing gibbous of births during the Full moon new moon = Waning gibbous , = Last quarter Waning crescent Total = 222,784 births 6

7 General formulation H 0 : 1 = hypothesized proportion for category 1 2 = hypothesized proportion for category 2 k = hypothesized proportion for category k H a : H 0 is not true, so at least one of the true category proportions differs from the corresponding hypothesized value. The expected counts for a one way frequency table are how many people you would expect in each category of the table if the null hypothesis was true. expected count of the category = total sample size hypothesized proportion for that category 7

8 Example: Births and the Lunar cycle Lunar phase Observed Number of births Expected Number under H0:no relationship between lunar cycle and births (n j ) New moon , = Waxing crescent , = First quarter , = Waxing gibbous , = Full moon , = Waning gibbous , = Last quarter , = Waning crescent , = n = 222,784 births 8

9 Example: Births and the Lunar cycle Lunar phase Observed Births (O) Expected Births (E) O E (O E) 2 /E New moon Waxing crescent First quarter Waxing gibbous Full moon Waning gibbous Last quarter Waning crescent

10 Goodness of fit statistic, 2 The goodness of fit statistic, 2, results from first computing the quantity 2 (observed cell count - expected cell count) expected cell count for each cell. The value of the 2 statistic is the sum of these terms. 2 (observed cell count - expected cell count) 2 expected cell count measures how good the fit of the hypothesized proportions under H 0 are to the observed proportions: large χ 2 poor fit if H 0 is true, it follows a chi square distribution, which can then be used to calculate p values. 10

11 Chi square distributions skewed non negative Chi-square Distributions df = 1 df = 2 df = 3 df = 4 df = 5 df = 8 df = 10 df = x

12 Upper tail areas for chi square distributions Right-tail area df = 1 df = 2 df = 3 df = 4 df = 5 >.100 < 2.70 < 4.60 < 6.25 < 7.77 < <.001 > > > > > Right-tail area df = 6 df = 7 df = 8 df = 9 df = 10 >.100 < < < < <

13 Upper tail areas for chi square distributions Right-tail area df = 1 df = 2 df = 3 df = 4 df = 5 >.100 < 2.70 < 4.60 < 6.25 < 7.77 < P( X > 5.05) = where X has a chi square distribution with 2 df <.001 > > > > > Right-tail area df = 6 df = 7 df = 8 df = 9 df = 10 >.100 < < < < <

14 Goodness of fit test procedure Assumptions: 1. Observed cell counts are based on a random sample. 2. The sample size is large: every expected count is at least 5. Hypotheses: H 0 : 1 = hypothesized proportion for category 1 2 = hypothesized proportion for category 2 k = hypothesized proportion for category k H a :H 0 is not true Test statistic: 2 (observed cell count - expected cell count) 2 expected cell count 2 ~ 1 k 14

15 Goodness of fit test procedure Significance level α of the test is set before the test is performed. p value: The area to the right of 2 under the df = k 1 chi square curve. p value H 0 is rejected in favor of H a if p value < α. 15

16 Example: Births and the Lunar cycle Lunar phase Observed Births (O) Expected Births (E) O E (O E) 2 /E New moon Waxing crescent First quarter Waxing gibbous Full moon Waning gibbous Last quarter Waning crescent k 1 = 8 1 = 7 df

17 Upper tail areas for chi square distributions Right-tail area df = 1 df = 2 df = 3 df = 4 df = 5 >.100 < 2.70 < 4.60 < 6.25 < 7.77 < P( ) 0.1 since Since 0.05 is statistical level of test, and the p - value, do not reject H Conclude there is no relationship between lunar cycle, births. 0. the the <.001 > > > > > Right-tail area df = 6 df = 7 df = 8 df = 9 df = 10 >.100 < < < < <

18 Two-way tables: Classroom data Code Age Weight Height Gender Vision Smoke Male Glasses No Male None No Male None No Male None No Male None No Male Contacts No Male Glasses No Male Contacts No Data from last 39 of the Male None Yes Male Contacts Yes students Male Glasses No Male Glasses No Male Glasses Yes Summarize the two Male Contacts No Female Glasses No categorical variable Female Glasses No Female Contacts No columns, gender and vision, Female Glasses No Female Glasses No Female Contacts No in a smaller two way table Female None Yes Female Glasses No and use that to test for a Female None No Female None No relationship between them Female Contacts Yes Female Contacts No Female Glasses No Female Contacts No Female Glasses Yes Female None No Female None No Female None No Female None No Female None No Female None No Female None No Female Glasses No Female None No Female Glasses No

19 Two way tables Contacts Glasses None Female Male This is a two way frequency table, or contingency table. The number entries in the table are the observed counts. Interpretation: 5 females who wore contacts 9 females who wore glasses 11 females with no vision correction 5 males who wore contacts 22 males with glasses 27 males with no correction 19

20 Code Age Weight Height Gender Vision Smoke Male None No Classroom data Male Glasses No Male None No Male Glasses No Male Glasses No Male None No Male None No Male None No Male None No Male Glasses No This table is made simply by Male None No Male Glasses Yes counting in which cell each of Male Glasses No Male Glasses No the 79 students falls in Male Glasses No Male None No Male None No Male None No Male None No Male Contacts No Male None No Male Glasses No Male None No Contacts Glasses None Male None No Male None No Female Male None No Male Glasses No Male Male Glasses No Male Glasses No Male None Yes Male Glasses No Male Glasses No Male None No Male None No Male None No Male Glasses No Male Glasses No = Male Glasses No Male None Yes Male None No

21 Two way tables Marginal totals are obtained by adding the observed cell counts in each row and also in each column. Row Marginal Total Contacts Glasses None Female Male Column Marginal Total The sum of the column marginal totals (or the row marginal totals) is called the grand total. 21

22 Two way tables Row Marginal Total Contacts Glasses None Female Male Column Marginal Total people in 31 glasses 38 no vision the class wore correction contacts 25 females in the class, 54 males 79 students total 22

23 Tests for homogeneity in a two way table Very often in a two way table the rows indicate different populations and the columns indicate different categories. Population of males and females Contacts Glasses None Female Male Categories of vision correction A test of homogeneity (sameness) is a test that the proportion of people in the different categories is the same for both populations. For example, here, whether the proportion of females wearing contacts, glasses or nothing is the same as the proportion of males. 23

24 Tests for homogeneity in a two way table H 0 : category proportions are the same for males and females (homogeneity) vs. H a : the category proportions are not all the same for males and females First step: get the expected counts under H 0 for each cell. Row Marginal Total Contacts Glasses None Female Male Column Marginal Total

25 Tests for homogeneity in a two way table Row Marginal Total Contacts Glasses None Female Male Column Marginal Total Expected count for this cell: Under H 0, the proportion of males and females wearing contacts is the same, so the expected proportion of males wearing contacts is just the overall proportion (proportion over both males and females). 10/79 = proportion of all students using contacts 54 = number of male students 10/79 54 = 6.84 = expected number of males that use contacts 25

26 Tests for homogeneity in a two way table This is equivalent to using the formula: Expected cell count = (Row total)(column total) Grand total Row Marginal Total Contacts Glasses None Female Male Column Marginal Total column total row total grand total

27 Tests for homogeneity in a two way table for each cell... Female Male Column Marginal Total Contacts Glasses None Row Marginal Total

28 Tests for homogeneity in a two way table Female Male Column Marginal Total Contacts Glasses None (3.16) (9.81) (12.03) (6.84) (21.19) (25.97) Row Marginal Total Expected number of males with no vision correction under H 0. The expected counts under the assumption that males and females have the same vision correction (H 0 ) look close to what was observed in the random sample, so maybe the data support H 0. 28

29 Comparing two or more populations using the 2 statistic Hypotheses: H 0 : The true category proportions are the same for all of the populations (homogeneity of populations). H a : The true category proportions are not all the same for all of the populations. Test statistic: 2 (observed cell count - expected cell count) 2 expected cell count The expected cell counts are estimated from the sample data (assuming that H 0 is true) using the formula Expected cell count = (Row total)(column total) Grand total 29

30 Comparing two or more populations Assumptions: 1. The data consist of independently chosen random samples. 2. The sample size is large: all expected counts are at least 5. If some expected counts are less than 5, rows or columns of the table may be combined to achieve a table with satisfactory expected counts. p value: When H 0 is true, 2 has approximately a chi square distribution with df = (number of rows 1)(number of columns 1) Calculate the area to the right of the observed 2 statistic. 30

31 Back to classroom data example Perform a test at the.05 level of significance: H 0 : category proportions are the same for males and females (homogeneity) vs. H a : the category proportions are not all the same for males and females Row Marginal Total Contacts Glasses None Female Male Column Marginal Total

32 Tests for homogeneity in a two way table Female Male Column Marginal Total Contacts Glasses None (3.16) (9.81) (12.03) (6.84) (21.19) (25.97) Row Marginal Total Assumptions Females, males randomly selected (known from before). All expected counts are at least 5. 32

33 Tests for homogeneity in a two way table Female Male Column Marginal Total Contacts Glasses None (3.16) (9.81) (12.03) (6.84) (21.19) (25.97) Row Marginal Total Solutions: Either collect a larger random sample or agree to collapse contacts and glasses into a single category, such as Any Vision Correction, and then do the test. Use Fisher s exact test (not covered in this course). For now, no test

34 Example: Shopping A student randomly observed 57 shoppers in a supermarket to see if males and females carried their items the same way. Device Row Marginal Total Gender Cart Basket Nothing Male Female Column Marginal Total Determine if the carrying device proportions are the same for both genders using a 0.05 level of significance. 34

35 Example: Shopping Hypotheses: H 0 : H a : The true proportions of the device used are the same for both genders. The true proportions of the device used are not the same for both genders. Significance level: = 0.05 Test statistic: 2 (observed cell count - expected cell count) 2 expected cell count 35

36 Example: Shopping A computer program such as minitab gives: Chi-Square Test: Basket, Cart, Nothing Expected counts are printed below observed counts (All > 5) Basket Cart Nothing Total Total Chi-Sq = = DF = 2, P-Value = > 0.05, do not reject H0. Exercise: Perform the calculations using the formulas to see if you get the same answer. 36

37 2 test for independence The exact same 2 test statistic and procedure can be used to investigate the association/independence between any two categorical variables on a single population (note the difference to the test of homogeneity, which tests for a difference between multiple populations). Hypotheses: H 0 : The two variables are independent. H a : The two variables are not independent. The expected counts under the null hypothesis of independence follow the same formula: Expected cell count = (Row total)(column total) Grand total 37

38 Example: Pain A medical paper examined the relationship between a nurse s assessment of a patient s facial expression and the patient s self reported level of pain. tennis elbow Data for 89 randomly selected patients. Patient facial expression Patient report No pain Pain 40 patients reported they had pain but their facial expression did not show any pain. No pain Pain patients reported no pain but their facial expression indicated pain. 38

39 Example: Pain The question: was there an association between whether or not the patient reported pain and the look of pain on his/her face? Or did they have nothing to do with each other? Patient facial expression Patient report No pain Pain No pain Perform a test at the.05 level of signficance. Pain

40 Example: Pain Hypotheses: H 0 : Patient self report and facial expression of pain are independent. H a : Patient self report and facial expression of pain are not independent. Significance level:.05 Test statistic: 2 (observed cell count - expected cell count) 2 expected cell count 40

41 Example: Pain Expected cell count = (Row total)(column total) Grand total Patient facial expression No pain Pain Patient report No pain 17 (12.81) 3 (7.19) Pain Row totals 40 (44.19) (24.81) 32 Assumptions Random sample All expected counts 5 Column totals

42 Example: Pain Calculation: 2 ( ) ( ) The table has 2 rows and 2 columns so df = (2 1)(2 1) = 1. The entry closest to 4.92 in the chi square table is 5.02, so the approximate p value is p value.025. Conclusion: Reject H 0 in favor of H a, that patient self report and facial expression of pain are not independent. 42

43 Tutorial 9: Smoking and Alcohol Do people who drink more tend to smoke more? A study in the New England Journal of Medicine in 1999 addressed this question by asking 21,870 male doctors ages years. They collected the following data. Smoking status Never smoked Past smoker Current smoker Alcohol: number of drinks per week < > ) To answer the question should we do a test of homogeneity or a test of independence? Why? 2.) Write the null and alternative hypotheses. 43

44 Tutorial 9: Smoking and Alcohol Smoking status Never smoked Past smoker Current smoker Alcohol: number of drinks per week < > ) Calculate row percentages by dividing each observed count by the corresponding row total. Graph your results for each smoking category separately as relative frequency barcharts (Lecture 1) and compare them. Comment on whether the graphs support the null versus alternative hypothesis? 44

45 Tutorial 9: Smoking and Alcohol Smoking status Never smoked Past smoker Current smoker Alcohol: number of drinks per week < > ) The chi square statistic for performing a test of the null hypothesis at the.05 level is calculated to be and the p value is less than State the degrees of freedom of the chi square distribution used for this test, sketch the chi square distribution itself, and indicate the statistic value and the p value on the distribution. 5.) What are the conclusions of the test? Do they agree with your graphs and what you would expect? 45