Two-Way Contingency Tables

14.3 Two-Way Contingency Tables

  In the scenarios of Sections 14.1 and 14.2, the observed frequencies were displayed in a single row within a rectangular table. We now study problems in which the data also consists of counts or frequencies, but the data table will now have I rows (I 2) and J columns, so IJ cells. There are two commonly encountered situations in which such data arises:

  1. There are I populations of interest, each corresponding to a different row of the table, and each population is divided into the same J categories. A sam- ple is taken from the ith population (i 5 1,…, I), and the counts are entered in the cells in the ith row of the table. For example, customers of each of

  640 Chapter 14 Goodness-of-Fit tests and Categorical Data analysis

  I5 3 department-store chains might have available the same J 5 5 payment categories: cash, check, and credit cards from American Express, Visa, and MasterCard.

  2. There is a single population of interest, with each individual in the population

  categorized with respect to two different factors. There are I categories associated with the first factor and J categories associated with the second factor. A single

  sample is taken, and the number of individuals belonging in both category i of factor 1 and category j of factor 2 is entered in the cell in row i, column

  j (i 5 1,…, I; j 5 1,…, J). As an example, customers making a purchase might be classified according to both department in which the purchase was made, with I 5 6 departments, and according to method of payment, with J 5 5 as in (1) above.

  Let n ij denote the number of individuals in the sample(s) falling in the (i, j)th cell (row i, column j) of the table—that is, the (i, j)th cell count. The table displaying the n ij ’s is called a two-way contingency table ; a prototype is shown in Table 14.9.

  Table 14.9 A Two-Way Contingency Table

  In situations of type 1, we want to investigate whether the proportions in the different categories are the same for all populations. The null hypothesis states that the populations are homogeneous with respect to these categories. In type 2 situa- tions, we investigate whether the categories of the two factors occur independently of one another in the population.

  testing for Homogeneity

  Suppose each individual in every one of the I populations belongs in exactly one of the same J categories. A sample of n i individuals is taken from the ith population; let n 5 on i and

  n ij 5 the number of individuals in the ith sample who fall into category j

  I the total number of individuals among

  n ? j 5 n ij 5

  i5 o 1 the n sample who fall into category j

  The n ij ’s are recorded in a two-way contingency table with I rows and J columns. The sum of the n ij ’s in the ith row is n i , and the sum of entries in the jth column will

  be denoted by n ?j . Let

  the proportion of the individuals in p ij 5 population i who fall into category j

  14.3 two-Way Contingency tables 641

  Thus, for population 1, the J proportions are p 11 ,p 12 ,…, p 1J (which sum to 1) and

  similarly for the other populations. The null hypothesis of homogeneity states that the proportion of individuals in category j is the same for each population and that

  this is true for every category; that is, for every j, p 1j 5 p 2j 5…5 p Ij . When H 0 is true, we can use p 1 ,p 2 ,…, p J to denote the population proportions

  in the J different categories; these proportions are common to all I populations. The expected number of individuals in the ith sample who fall in the jth category

  when H 0 is true is then E(N ij )5n i ?p j . To estimate E(N ij ), we must first estimate p j , the proportion in category j. Among the total sample of n individuals, N ?j ’s fall into

  category j, so we use pˆ j 5 N ? j yn as the estimator (this can be shown to be the maxi-

  mum likelihood estimator of p j ). Substitution of the estimate pˆ j for p j in n i p j yields a

  simple formula for estimated expected counts under H 0 :

  n ? j

  eˆ ij 5 estimated expected count in cell (i, j) 5 n i ?

  n

  (ith row total)( j th column total)

  n

  The test statistic here has the same form as in Sections 14.1 and 14.2. The num- ber of degrees of freedom comes from the general rule of thumb. In each row of Table 14.9 there are J 2 1 freely determined cell counts (each sample size n i is

  fixed), so there are a total of I sJ 2 1d freely determined cells. Parameters p 1 ,…, p J are estimated, but because op i 5 1, only J 2 1 of these are independent. Thus df 5

  I sJ 2 1d 2 sJ 2 1d 5 sJ 2 1d sI 2 1d.

  Null hypothesis: H 0 :p 1j 5 p 2j 5…5 p Ij j5 1, 2,…, J

  Alternative hypothesis: H a :H 0 is not true Test statistic value:

  all cells o

  2 (observed 2 estimated expected) 2 I J (n ij 2 eˆ ij ) x 2 5 5

  o i5 1 o j5 1 eˆ ij

  estimated expected

  When H 0 is true and eˆ ij 5 for all i, j, the test statistic has approximately a

  chi-squared distribution with (I 2 1)(J 2 1) df. The test is again upper-tailed,

  so the P-value is the area under the x 2 (I21)(J21) curve to the right of the calcu-

  lated x 2 . Table A.11 can be used to obtain P-value information as described in Section 14.1.

  example 14.13

  A company packages a particular product in cans of three different sizes, each one using a different production line. Most cans conform to specifications, but a quality control engineer has identified the following reasons for nonconformance:

  1. Blemish on can

  2. Crack in can

  3. Improper pull tab location

  4. Pull tab missing

  5. Other

  642 Chapter 14 Goodness-of-Fit tests and Categorical Data analysis

  A sample of nonconforming units is selected from each of the three lines, and each unit is categorized according to reason for nonconformity, resulting in the following contingency table data:

  Reason for Nonconformity

  Sample

  Blemish Crack

  Location Missing Other Size

  Production Line

  Does the data suggest that the proportions falling in the various nonconformance categories are not the same for the three lines? The parameters of interest are the various proportions, and the relevant hypotheses are

  H 0 : the production lines are homogeneous with respect to the five noncon ­

  formance categories; that is, p 1j 5 p 2j 5 p 3j for j 5 1, … ,5

  H a : the production lines are not homogeneous with respect to the categories The estimated expected frequencies (assuming homogeneity) must now be calcu-

  lated. Consider the first nonconformance category for the first production line. When the lines are homogeneous,

  estimated expected number among the 150 selected units that are blemished

  (first row total)(first column total)

  total of sample sizes

  The contribution of the cell in the upper-left corner to x 2 is then (observed 2 estimated expected) 2 (31 2 33.20) 2

  estimated expected

  The other contributions are calculated in a similar manner. Figure 14.5 shows Minitab output for the chi-squared test. The observed count is the top number in each cell, and directly below it is the estimated expected count. The contribution of

  Expected counts are printed below observed counts Chi–Square contributions are printed below expected counts

  Blemish Crack Location Missing Other Total

  Chi–Sq 5 21.403, DF 5 8, P–Value 5 0.006

  Figure 14.5 Minitab output for the chi-squared test of Example 14.13

  14.3 two-Way Contingency tables 643

  each cell to x 2 appears below the counts, and the test statistic value is x 2 5 21.403.

  All estimated expected counts are at least 5, so combining categories is unneces- sary. The test is based on s3 2 1ds5 2 1d 5 8 df. Appendix Table A.11 shows that the area under the 8 df chi-squared curve to the right of 20.09 is .010 and the area to the right of 21.95 is .005. Therefore we can say that .005 , P-value , .01; Minitab gives P-value 5 .006. Using a significance level of .01, the null hypoth- esis of homogeneity can be rejected in favor of the alternative that the distribution of reason for nonconformity is somehow different for the three production lines.

  At this point it is desirable to seek an explanation for why the hypothesis of homogeneity is implausible. Figure 14.6 shows a stacked comparative bar chart of the data. It appears that the three lines are relatively homogenous with respect to the Other and Missing categories but not with respect to the Location, Crack, and Blemish categories. Line 1’s incidence rate of crack nonconformities is much higher than for the other two lines, whereas location nonconformities appear to be more of

  a problem for line 2 than for the other two lines and blemish nonconformities occur much more frequently for line 3 than for the other two lines.

  Reason Other 80 Missing Location

  Crack 60 Blemish

  Figure 14.6 Stacked comparative bar chart for the data of Example 14.13 n

  testing for Independence (Lack of Association)

  We focus now on the relationship between two different factors in a single popula- tion. Each individual in the population is assumed to belong in exactly one of the I categories associated with the first factor and exactly one of the J categories associ- ated with the second factor. For example, the population of interest might consist of all individuals who regularly watch the national news on television, with the first

  factor being preferred network (ABC, CBS, NBC, or PBS, so I 5 4) and the second factor political philosophy (liberal, moderate, or conservative, giving J 5 3).

  For a sample of n individuals taken from the population, let n ij denote the number among the n who fall both in category i of the first factor and category j of the second factor. The n ij ’s can be displayed in a two-way contingency table with

  I rows and J columns. In the case of homogeneity for I populations, the row totals were fixed in advance, and only the J column totals were random. Now only the total sample size is fixed, and both the n i? ’s and n ? j ’s are observed values of random variables. To state the hypotheses of interest, let

  p ij 5 the proportion of individuals in the population who belong in category i

  of factor 1 and category j of factor 2

  5P sa randomly selected individual falls in both category i of factor 1 and

  category j of factor 2 d

  644 Chapter 14 Goodness-of-Fit tests and Categorical Data analysis

  Then

  p i? 5 p ij 5 P o (a randomly selected individual falls in category i of factor 1)

  j

  p ?j 5 p ij 5 P o (a randomly selected individual falls in category j of factor 2)

  i

  Recall that two events, A and B, are independent if P sA ù Bd 5 PsAd ? PsBd. The null hypothesis here says that an individual’s category with respect to factor 1 is independent of the category with respect to factor

  2. In symbols, this becomes

  p ij 5 p i? ?p ? j for every pair (i, j).

  The expected count in cell (i, j) is n ? p ij , so when the null hypothesis is true,

  E (N ij )5n?p i? ?p ?j . To obtain a chi-squared statistic, we must therefore estimate the

  p i? ’s (i 5 1, … , I) and p ? j ’s( j 5 1, … , J ). The (maximum likelihood) estimates are

  n i?

  pˆ i? 5 n 5 sample proportion for category i of factor 1

  and n ? j

  pˆ ? j 5 5 sample proportion for category j of factor 2

  n

  This gives estimated expected cell counts identical to those in the case of homogeneity.

  (ith row total)(jth column total)

  5 n

  The test statistic is also identical to that used in testing for homogeneity, as is the number of degrees of freedom. This is because the number of freely determined cell counts is IJ 2 1, since only the total n is fixed in advance. There are I estimated p i? ’s, but only I 2 1

  are independently estimated since op i? 5 1; and similarly J 2 1p ? j ’s are independently

  estimated, so I 1 J 2 2 parameters are independently estimated. The rule of thumb now yields df 5 IJ 2 1 2 sI 1 J 2 2d 5 IJ 2 I 2 J 1 1 5 sI 2 1d ? sJ 2 1d.

  Null hypothesis: H 0 :p ij 5 p i? ?p ? j

  i5 1,…, I; j 5 1,…, J

  Alternative hypothesis: H a :H 0 is not true Test statistic value:

  all cells o

  (observed 2 estimated expected) 2 I 2 J (n ij 2 eˆ 2 ij ) x 5 5

  o i5 1 j5 o 1 eˆ ij

  estimated expected

  When H 0 is true and eˆ ij 5 for all i, j, the test statistic has approximately a

  chi-squared distribution with (I 2 1)(J 2 1) df. The test is again upper-tailed,

  so the P-value is the area under the x 2 (I21)(J21) curve to the right of the calcu-

  lated x 2 . Table A.11 can be used to obtain P-value information as described in Section 14.1.

  14.3 two-Way Contingency tables 645

  example 14.14

  The accompanying two-way table from Minitab (Table 14.10) gives a cross- classification in which the row factor is level of paternal education (completed univer- sity, partial university, secondary, partial secondary) and the column factor represents the quartile of neonatal (i.e., newborn) weight gain (Q1 5 lowest 25, Q2 5 next

  lowest 25, Q3, Q4); the data appeared in the article “Impact of Neonatal Growth

  on IQ and Behavior at Early School Age” (Pediatrics, July 2013, e53–60) . Does it appear that educational level is independent of NWG in the sampled population?

  Table 14.10 Observed and Estimated Expected Counts for Example 14.14

  Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts

  Q4 Total

  The contribution to x 2 from the cell in the upper-left corner is (422 2 411.63) 2 y411.63 5 .261. The 15 other contributions are calculated in the same way. Then x 2 5 .261 1 … 1 .253 5 19.016. When H 0 is true, the test statistic has approximately a chi-

  squared distribution with (4 2 1)(4 2 1) 5 9 df. The expected value of a chi-squared

  rv is just its number of degrees of freedom, so E(x 2 ) 5 9 under the assumption of

  independence. Clearly the test statistic value exceeds what would be expected if the two factors were independent, but is it by enough to suggest implausibility of this null hypothesis? Table A.11 shows that .025 is the area to the right of 19.02 under the chi-squared curve with 9 df. Thus the P-value for the test is roughly .025 (which is the value calculated by Minitab; the cited article reported .03). At significance level .05, the null hypothesis of independence would be rejected since P-value < .025 .05 5 a. However, this conclusion would not be justified at a significance level of .01. The P-value is such that people might argue over what conclusion is appropriate.

  Someone persuaded by our analysis to reject the assertion of independence would want to look more closely at the data to seek an explanation for that conclu- sion. Perhaps, for example, those in a higher quartile tend to have higher educational levels. Figure 14.7 shows histograms (bar graphs) of the percentages in the various educational level categories for each of the four different quartiles. The four histo- grams appear to be very similar; the visual impression is that the distribution over the four educational levels does not depend much on the NWG quartile. This seemingly contradicts the finding of statistical significance. Now note that the sample size here is extremely large, and this inflates the value of the chi-squared statistic. With the same percentages as in Figure 14.7 but a much more moderate sample size, the value

  of x 2 would be much smaller and the P-value much larger. Our test result achieved statistical significance, but there does not seem to be any practical significance.

  646 Chapter 14 Goodness-of-Fit tests and Categorical Data analysis

  Figure 14.7 Histograms based on the data of Example 14.14 n

  Models and methods for analyzing data in which each individual is catego- rized with respect to three or more factors (multidimensional contingency tables) are discussed in several of the chapter references.

  ExErCiSES Section 14.3 (24–36)

  24. The accompanying two-way table was constructed using

  ChiSq 5 3.557 1 0.579 1

  data in the article “Television Viewing and Physical

  Fitness in Adults” (Research Quarterly for Exercise and

  Sport, 1990: 315–320) . The author hoped to determine

  whether time spent watching television is associated with

  df 5 3

  cardiovascular fitness. Subjects were asked about their

  25. In an investigation of alcohol use among college stu-

  television-viewing habits and were classified as physically

  dents, each male student in a sample was categorized

  fit if they scored in the excellent or very good category on

  both according to age group and according to the number

  a step test. We include Minitab output from a chi-squared

  of heavy drinking episodes during the previous 30 days

  analysis. The four TV groups corresponded to different

  (“Alcohol Use in Students Seeking Primary Care

  amounts of time per day spent watching TV (0, 1–2, 3–4,

  Treatment at University Health Services,” J. of Amer.

  or 5 or more hours). The 168 individuals represented in

  College Health, 2012: 217–225) .

  the first column were those judged physically fit. Expected counts appear below observed counts, and Minitab dis-

  Age Group

  plays the contribution to x from each cell. State and test

  the appropriate hypotheses using a 5 .05. 24

  18–20 21–23

  Does there appear to be an association between extent

  of binge drinking and age group in the population from

  which the sample was selected? Carry out a test of

  Total 168 1032 1200

  hypotheses at significance level .01.

  14.3 two-Way Contingency tables 647

  26. Contamination of various food products is an ongoing

  Degree of Spirituality

  problem all over the world. The article “Prevalence and Quantitative Detection of Salmonella in Retail Raw

  Very

  Moderate

  Slightly Not at all

  Chicken in Shaanxi, China” (J. of Food Production,

  N.S. 56 162

  2013) reported the following data on the occurrence of

  S.S. 56 223

  salmonella in chicken of three different types: (1) super-

  G.D. 109

  market chilled, (2) supermarket frozen, and (3) wet market fresh slaughtered.

  a. Is there substantial evidence for concluding that the three types of individuals are not homogenous with

  Salmonella

  respect to their degree of spirituality? State and test

  Sample Size

  Positive Samples

  the appropriate hypotheses. 1. 60 27 b. Considering just the natural scientists and social

  Type

  2. 60 32 scientists, is there evidence for non-homogeneity?

  45 30. Three different design configurations are being consid- ered for a particular component. There are four possible

  Does it appear that the incidence rate of salmonella

  failure modes for the component. An engineer obtained

  occurrence depends on the type of chicken? State and

  the following data on number of failures in each mode

  test the appropriate hypotheses using a significance level

  for each of the three configurations. Does the configura-

  of .05.

  tion appear to have an effect on type of failure?

  27. The article “Human Lateralization from Head to Foot:

  Sex-Related Factors” (Science, 1978: 1291–1292) Failure Mode

  reports for both a sample of right-handed men and a

  sample of right-handed women the number of individuals whose feet were the same size, had a bigger left than right

  foot (a difference of half a shoe size or more), or had a

  Configuration

  bigger right than left foot.

  Sample

  31. A random sample of smokers was obtained, and each

  individual was classified both with respect to gender and with respect to the age at which heshe first started

  Men 2 10 28 40

  smoking. The data in the accompanying table is con-

  Women 55 18

  sistent with summary results reported in the article

  14 87 “Cigarette Tar Yields in Relation to Mortality in the Cancer Prevention Study II Prospective

  Does the data indicate that gender has a strong effect on

  Cohort” (British Med. J., 2004: 72–79) .

  the development of foot asymmetry? State and test the appropriate hypotheses.

  Gender

  28. A random sample of 175 Cal Poly State University stu-

  M ale

  Female

  dents was selected, and both the email service provider and cell phone provider were determined for each one,

  resulting in the accompanying data. State and test the

  Age 16217 24 32

  appropriate hypotheses

  Cell Phone Provider

  a. Calculate the proportion of males in each age category, and then do the same for females. Based on these pro-

  ATT Verizon Other

  portions, does it appear that there might be an associa-

  gmail

  28 17 7 tion between gender and the age at which an individu-

  Email Provider Yahoo 31 26 10

  al first smokes?

  Other 26 19 11

  b. Carry out a test of hypotheses to decide whether

  there is an association between the two factors.

  29. The accompanying data on degree of spirituality for

  32. Eclosion refers to the emergence of an adult insect from

  samples of natural and social scientists at research univer-

  an egg. The following data on eclosion rates when

  sities as well as for a sample of non-academics with

  nymphs were exposed to heat for various durations was

  graduate degrees appeared in the article “Conflict

  extracted from the article “High Temperature

  Between Religion and Science Among Academic

  Determines the Ups and Downs of Small Brown

  Scientists” (J. for the Scientific Study of Religion, 2009:

  Planthopper Laodelphax Striatellus Population”

  276–292) .

  (Insect Science, 2012: 385–392) .

  648 Chapter 14 Goodness-of-Fit tests and Categorical Data analysis

  Duration (d)

  0 1 2 3 5 10 15 35. Suppose that in a particular state consisting of four distinct

  Sample size

  120 41 47 44 46 42 10 regions, a random sample of n k voters is obtained from the k th region for k 5 1, 2, 3, 4. Each voter is then classified

  Emerged:

  101 38 44 40 38 35 7 according to which candidate (1, 2, or 3) he or she prefers

  Carry out a chi-squared test to decide whether it is

  and according to voter registration s1 5 Dem., 2 5 Rep.,

  plausible that eclosion rate does not depend on exposure

  3 5 Indep. d. Let p ijk denote the proportion of voters in

  duration (the cited article included summary information

  region k who belong in candidate category i and registration

  from the test).

  category j. The null hypothesis of homogeneous regions is

  33. Show that the chi-squared statistic for the test of inde-

  H 0 :p ij 1 5 p ij 2 5 p ij 3 5 p ij 4 for all i, j (i.e., the proportion

  pendence can be written in the form

  within each candidateregistration combination is the same for all four regions). Assuming that H is true, determine pˆ ijk

  x 5 o

  and eˆ ijk as functions of the observed n ijk ’s, and use the

  1 o i5 j5 1 E ˆ ij 2 general rule of thumb to obtain the number of degrees of freedom for the chi-squared test.

  n

  Why is this formula more efficient computationally than

  36. Consider the accompanying 2 3 3 table displaying the

  the defining formula for x 2 ?

  sample proportions that fell in the various combinations

  34. Suppose that each student in a sample had been catego-

  of categories (e.g., 13 of those in the sample were in

  rized with respect to political views, marijuana usage,

  the first category of both factors).

  and religious preference, with the categories of this latter

  factor being Protestant, Catholic, and other. The data could

  be displayed in three different two-way tables, one corresponding to each category of the third factor. With

  p ijk 5 P spolitical category i, marijuana category j, and

  a. Suppose the sample consisted of n 5 100 people.

  religious category k), the null hypothesis of independence

  Use the chi-squared test for independence with sig- of all three factors states that p ijk 5 p i? ? p ? j? p ?? k . Let n ijk nificance level .10.

  denote the observed frequency in cell (i, j, k). Show how to estimate the expected cell counts assuming that H

  0 is

  b. Repeat part (a), assuming that the sample size was

  true (eˆ ijk 5 npˆ ijk , so the pˆ ijk ’s must be determined). Then

  n5 1000.

  use the general rule of thumb to determine the number of

  c. What is the smallest sample size n for which these

  degrees of freedom for the chi-squared statistic.

  observed proportions would result in rejection of the independence hypothesis?

  SuPPlEmEnTAry ExErCiSES (37–49)

  37. The article “Birth Order and Political Success” (Psych.

  Lunar Phase

  Days in Phase Births

  Reports, 1971: 1239–1242) reports that among 31 ran- domly selected candidates for political office who came

  New moon 24 7680

  from families with four children, 12 were firstborn, 11

  Waxing crescent 152

  were middle born, and 8 were last born. Use this data to

  First quarter 24 7579

  test the null hypothesis that a political candidate from

  Waxing gibbous 149 47,814

  such a family is equally likely to be in any one of the four

  Full moon

  ordinal positions.

  Waning gibbous 150

  47,595 Last quarter 24 73

  38. Does the phase of the moon have any bearing on

  Waning crescent 152

  birthrate? Each of 222,784 births that occurred during

  a period encompassing 24 full lunar cycles was clas-

  State and test the appropriate hypotheses to answer the

  sified according to lunar phase. The following data is

  question posed at the beginning of this exercise.

  consistent with summary quantities that appeared in

  39. Each individual in a sample of nursing home patients

  the article “The Effect of the Lunar Cycle on

  was cross-classified both with respect to cognitive state

  Frequency of Births and Birth Complications”

  (normal or mild impairment, moderate impairment,

  (Amer. J. of Obstetrics and Gynecology, 2005:

  severe impairment) and with respect to drug status (psy-

  1462–1464) .

  chotropic drug change, psychotropic user without a

  Supplementary exercises 649

  change, no psychotropic medication). The following

  a. State the relevant hypotheses and reach a conclusion

  Minitab output resulted from a request to perform a chi-

  using a 5 .05.

  squared analysis.

  b. Do you think that your conclusion in part (a) can be

  attributed to a single sport being an anomaly?

  Drug change No change No med

  41. The accompanying two-way frequency table appears

  90.06 64.11 34.83 in the article “Marijuana Use in College” (Youth and

  Society, 1979: 323–334) . Each of 445 college students was classified according to both frequency of mari-

  juana use and parental use of alcohol and psychoactive

  84.40 drugs. Does the data suggest that parental usage and

  student usage are independent in the population from

  Severe

  which the sample was drawn?

  Standard Level of Marijuana Use

  Never Occasional Regular

  Chi-Sq 5 11.294, DF 5 4, P-Value 5 0.023

  (“Psychotropic Drug Initiation or Increased Dosage

  Neither 141 54 40

  and the Acute Risk of Falls,” BMC Geriatrics, 2013: Parental

  a. Verify the expected frequency and contribution

  and Drugs

  to x in the normal–drug change cell of the two-way

  Both 17 11 19

  table.

  42. Much attention has recently focused on the incidence

  b. Does there appear to be an association between cog- nitive state and drug status? State and test the appro-

  of concussions among athletes. Separate samples of

  priate hypotheses using a significance level of .01.

  soccer players, non-soccer athletes, and non-athletes

  [Note: The cited article reported a P-value.]

  were selected. The accompanying table then resulted from determining the number of concussions each

  40. The authors of the article “Predicting Professional

  individual reported on a medical history questionnaire

  Sports Game Outcomes from Intermediate Game

  (“No Evidence of Impaired Neurocognitive

  Scores” (Chance, 1992: 18–22) used a chi-squared test to

  Performance in Collegiate Soccer Players,” Amer. J.

  determine whether there was any merit to the idea that

  of Sports Med., 2002: 157–162) .

  basketball games are not settled until the last quarter, whereas baseball games are over by the seventh inning.

  of Concussions

  They also considered football and hockey. Data was col-

  012 »3

  lected for 189 basketball games, 92 baseball games,

  80 hockey games, and 93 football games. The games ana-

  Soccer

  lyzed were sampled randomly from all games played dur-

  N-S Athletes

  ing the 1990 season for baseball and football and for the

  Non-athletes

  1990–1991 season for basketball and hockey. For each

  Does the distribution of of concussions appear to be

  game, the late-game leader was determined, and then it

  different for the three types of individuals? Carry out a

  was noted whether the late-game leader actually ended up

  test of hypotheses.

  winning the game. The resulting data is summarized in the accompanying table.

  43. In a study to investigate the extent to which individuals are aware of industrial odors in a certain region

  Late-Game Late-Game (“Annoyance and Health Reactions to Odor from

  Sport

  Leader Wins

  Leader Loses

  Refineries and Other Industries in Carson,

  California,” Environmental Research, 1978: 119–132) ,

  Basketball 150

  39 a sample of individuals was obtained from each of three

  Baseball

  86 6 different areas near industrial facilities. Each individual

  Hockey

  65 15 was asked whether he or she noticed odors (1) every day,

  Football

  72 21 (2) at least onceweek, (3) at least oncemonth, (4) less

  The authors state that “Late-game leader is defined as

  often than oncemonth, or (5) not at all, resulting in the

  the team that is ahead after three quarters in basketball

  data and SPSS output at the bottom of the next page.

  and football, two periods in hockey, and seven innings

  State and test the appropriate hypotheses.

  in baseball. The chi-square value on three degrees of

  44. Many shoppers have expressed unhappiness because

  freedom is 10.52 sP , .015d.”

  grocery stores have stopped putting prices on individual

  650 Chapter 14 Goodness-of-Fit tests and Categorical Data analysis

  grocery items. The article “The Impact of Item Price

  won its regional tournament 22 times, the second-ranked

  Removal on Grocery Shopping Behavior” (J. of

  team won 10 times, the third-ranked team won 5 times,

  Marketing, 1980: 73–93) reports on a study in which

  and the remaining 11 regional tournaments were won by

  each shopper in a sample was classified by age and by

  teams ranked lower than 3. Let P ij denote the probability

  whether he or she felt the need for item pricing. Based

  that the team ranked i in its region is victorious in its

  on the accompanying data, does the need for item pric-

  game against the team ranked j. Once the P ij ’s are avail-

  ing appear to be independent of age?

  able, it is possible to compute the probability that any particular seed wins its regional tournament (a compli-

  Age

  cated calculation because the number of outcomes in the

  ,30 30–39 40–49 50–59 60

  sample space is quite large). The paper “Probability Models for the NCAA Regional Basketball

  Number in

  Tournaments” (American Statistician, 1991: 35–38)

  Sample 150 141 82 63 49

  proposed several different models for the P ij ’s.

  a. One model postulated P ij 5 .5 2 l si 2 jd with

  Number

  77 61 41 l5 1 y32 (from which P 16,1 5l ,P 16,2 5 2l, etc.).

  Who Want 127 118

  Item Pricing

  Based on this, P(seed 1 wins) 5 .27477, P(seed 2 wins) 5 .20834, and P(seed 3 wins) 5 .15429. Does this model appear to provide a good fit to the

  45. Let p 1 denote the proportion of successes in a particular

  data?

  population. The test statistic value in Chapter 8 for test-

  ing H 0 :p 1 5 p 10 was z 5 spˆ

  1 2 p 10 dy Ï p 10 p 20 yn, where

  b. A more sophisticated model has game probabilities

  P ij 5 .5 1 .2813625 sz i 2 z j d, where the z’s are mea-

  p 20 5 12p 10 . Show that for the case k 5 2, the chi-

  squared test statistic value of Section 14.1 satisfies x 2 5 z 2 .

  sures of relative strengths related to standard normal

  [Hint: First show that (n 1 2 np 10 ) 2 5 (n 2 np 20 ) 2 .]

  percentiles [percentiles for successive highly seeded teams are closer together than is the case for teams

  46. The NCAA basketball tournament begins with 64 teams

  seeded lower, and .2813625 ensures that the range of

  that are apportioned into four regional tournaments,

  probabilities is the same as for the model in part (a)].

  each involving 16 teams. The 16 teams in each region

  The resulting probabilities of seeds 1, 2, or 3 winning

  are then ranked (seeded) from 1 to 16. During the

  their regional tournaments are .45883, .18813, and

  12-year period from 1991 to 2002, the top-ranked team

  .11032, respectively. Assess the fit of this model.

  SPSS output for Exercise 43 Crosstabulation: AREA BY CATEGORY

  Count

  Exp Val

  CATEGORY ¡ Row Pct Row AREA

  Col

  Pct 1.00 2.00 3.00 4.00 5.00 Total

  Column 38 74 54 48 77 291 Total

  Min E.F.

  Cells with E.F. , 5

  None

  Bibliography 651

  47. Have you ever wondered whether soccer players suffer

  a. Let p 0 denote the long run proportion of digits in

  adverse effects from hitting “headers”? The authors of

  the expansion that equal 0, and define p 1 ,…, p 9

  the article “No Evidence of Impaired Neurocognitive

  analogously. What hypotheses about these propor-

  Performance in Collegiate Soccer Players” (Amer. J.

  tions should be tested, and what is df for the chi-

  of Sports Med., 2002: 157–162) investigated this issue

  squared test?

  from several perspectives.

  b. H 0 of part (a) would not be rejected for the nonrandom

  a. The paper reported that 45 of the 91 soccer players

  sequence 012…901…901…. Consider nonoverlap-

  in their sample had suffered at least one concussion,

  ping groups of two digits, and let p ij denote the long

  28 of 96 nonsoccer athletes had suffered at least one

  run proportion of groups for which the first digit is

  concussion, and only 8 of 53 student controls had

  i and the second digit is j. What hypotheses about

  suffered at least one concussion. Analyze this data

  these proportions should be tested, and what is df for

  and draw appropriate conclusions.

  the chi-squared test?

  b. For the soccer players, the sample correlation

  c. Consider nonoverlapping groups of 5 digits. Could a

  coefficient calculated from the values of x 5 soc-

  chi-squared test of appropriate hypotheses about the

  cer exposure (total number of competitive seasons

  p ijklm ’s be based on the first 100,000 digits? Explain.

  played prior to enrollment in the study) and y 5

  d. The article “Are the Digits of p an Independent and

  score on an immediate memory recall test was

  Identically Distributed Sequence?” (The American

  r52 .220. Interpret this result.

  Statis tician, 2000: 12–16) considered the first 1,254,540

  c. Here is summary information on scores on a con-

  digits of p, and reported the following P-values for

  trolled oral word-association test for the soccer and

  group sizes of 1, … , 5: .572, .078, .529, .691, .298. What

  nonsoccer athletes:

  would you conclude? n 1 5 1 26, x 5 1 37.50, s 5 9.13 49. The Fibonacci sequence of numbers occurs in various

  n

  scientific contexts. The first two numbers in the sequence 2 5 2 56, x 5 2 39.63, s 5 10.19 are 1,1. Then every succeeding number is the sum of the

  Analyze this data and draw appropriate conclusions.

  two previous numbers: 1, 1, 1 1 1 5 2, 1 1 2 5 3, 2 1

  d. Considering the number of prior nonsoccer concus-