Two-Way Chi-Square
The problems below were bivariate frequency distribution problems from Chapter 6 that are now being tested for statistical significance.
Problem 1: The Corps of Cadets at the police academy is faced with a challenge, a pass-fail assignment. Of the 56 people in the class, (comprised of 24 females and 32 males) 11 females failed and 23 males failed. Use the bivariate tables previously created from chapter 6 to test for a statistically significant difference in pass-fail rate among male and female cadets, between the observed and the expected count. The MOE was previously computed to analyze if corps membership improved the predictability of pass fail rate among cadets.
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Problem 1 answer:
What is the theory? Females are higher achievers than are males. (O ≠ E)
What is the null hypothesis? There is not a statistically significant difference in cadets who “pass” among females and males between the observed and the expected count. (O = E)
Test the Null and Report the results: Reject, X2(1) = 3.899, P < .05
Provide the Analysis: The theory is supported Females are passing at a greater rate (54%) than are males (28%). Using the independent variable of student type (male or female) improved the predictability of pass fail by 9%.
Females Males Total
F “Pass” 13 9 (22) E1 = 22
% “Pass” 54% 28% 11
F Count “Fail” 11 23 (34) 9
% “Fail” 46% 72% E2 = 20
Total (24) (32) (56)
λ = E1 - E2 / E1
λ = 22 - 20 / 22
λ = 2 / 22
λ = .09 = 9%
Chi-square - TWO-WAY
Pass Pass Fail Fail Totals
Female Male Female Male
"O" 13 9 11 23 56
"E" 9.43 12.57 14.57 19.43 56
(O-E) 3.57 -3.57 -3.57 3.57
(O-E)2 12.76 12.76 12.76 12.76
(O-E)2/E 1.35 1.01 0.88 0.66
X2obt = 3.899
X2crit = 3.841
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Problem 2: A new infant behavior analysis device has demonstrated the ability to foretell the potential for criminality among newborns in suburban and inner city hospitals. Out of 78 suburb babies, 23 were positively identified, and 55 were negative. Inner-city positives numbered 90, with 51 negative. The sample totaled 219. Which neighborhood is more likely to produce criminals? Does knowledge about which neighborhood the baby is from improve the ability to predict whether or not a given individual will be a criminal as shown by MOE?
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Problem 2 answer:
Theory? Suburb babies are less likely to become criminals than their inner-city counterparts. (O ≠ E)
Null hypothesis? There is not a statistically significant difference in persons who become offenders among those born in the Suburbs and Inner-city, between the observed and the expected count. (O = E)
Results: Reject, X2(1) = 23.716, P. < .05
Analysis: The theory is supported. The inner-city produces a greater percentage of criminals (64% vs. 29%), there is a statistically significant difference between the observed and the expected count that would be obtained by chance if babies were not separated according to treatment. The difference could not be attributed to chance.
In City Sub Total
F “Positive” 90 23 (113) E1 = 106
% “Positive” 64% 29% 51
F “Negative” 51 55 (106) 23
% “Negative” 36% 71% E2 = 74
Total (141) (78) (219)
λ = E1 - E2 / E1
λ = 106 - 74 / 106
λ = 32 / 106
λ = .302 = 30%
Positive Positive Negative Negative Totals
Labels Innercity Suburbs Innercity Suburbs
"O" 90.000 23.000 51.000 55.000 219.000
"E" 72.753 40.247 68.247 37.753 219.000
(O-E) 17.247 -17.247 -17.247 17.247
(O-E)2 297.444 297.444 297.444 297.444
(O-E)2/E 4.088 7.391 4.358 7.879
X2obt = 23.716
X2crit = 3.841
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Problem 3: Two specialty units with the Memphis police department (the Narcotics unit and the Vice squad) are in competition to find which unit is more efficient at making arrests. Arrest efficiency is determined by whether or not a conviction is obtained (conviction or dismissed). The two divisions produced 926 arrests last year (vice = 447, Narc = 479). Vice scored 306 convictions, Narcotics scored 204).
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Problem 3 answer:
What is the theory? Narcotics officers are more likely to get a conviction than vice officers. (O ≠ E)
What is the null hypothesis? There is not a statistically significant difference in the number of convictions among Vice and Narcotics arrests, between the observed and the expected count. (O = E)
Test the Null and Report the results: Reject, X2(1) = 62.532, P. < .05.
Provide the Analysis: There is a statistically significant difference between the results we obtained and those that would be obtained by chance if no treatment were in effect. The bivariate percentages reveal that Vice had a significantly higher efficiency rating (68% vs. 43%). Using unit type as a predictor improved the predictability of convictions by 17%.
Narc Vice Total
F “Convictions” 204 306 (510) E1 = 416
% “Convictions” 43% 68% 204
F “Dismissals” 275 141 (416) 141
% “Dismissals” 57% 32% E2 = 345
Total (479) (447) (926)
λ = E1 - E2 / E1
λ = 416 - 345 / 416
λ = 71 / 416
λ = .171 = 17%
Conviction Conviction Dismissals Dismissals Totals
Labels Narc Vice Narc Vice
"O" 204.000 306.000 275.000 141.000 926.000
"E" 263.812 246.188 215.188 200.812 926.000
(O-E) -59.812 59.812 59.812 -59.812
(O-E)2 3577.487 3577.487 3577.487 3577.487
(O-E)2/E 13.561 14.532 16.625 17.815
X2obt = 62.532
X2crit = 3.841
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Problem 4: Are Asian immigrant teens more likely to join gangs in America than Latino teens? A social worker believes so, and asks Asian immigrant and Latino teens if they had been solicited to become members of an ethnic gang. Of 326 Asian immigrant teens 228 had been solicited. Latino teens numbered 78 in the affirmative category out of 185.
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Problem 4 answer:
What is the theory? Asians will be more likely to be solicited than Latinos. (O ≠ E)
What is the null hypothesis? There is not a statistically significant difference in the number of solicitations among Latino and Asian teens, between the observed and the expected count. (O = E)
Test the Null and Report the results: Reject, X2(1) = 37.904, P. < .05.
Provide the Analysis: The Chi-square test reveals that Asians were vigorously sought after for gang membership (70% vs. 42%) significantly more than if random samples of teens were drawn and the differences were calculated without respect to treatment. Using unit type as a predictor improved the predictability of convictions by 14%.
Asian Latino Total
F “Solicited” 228 78 (306) E1 = 205
% “Solicited” 70% 42% 98
F “Not Solicited” 98 107 (205) 78
% “Not Solicited” 30% 58% E2 = 176
Total (326) (185) (511)
λ = E1 - E2 / E1
λ = 205 - 176 / 205
λ = 29 / 205
λ = .141 = 14%
Gang Membership by Race
Chi-square - TWO-WAY
Solicited Solicited Not Solicited Not Solicited Totals
Asian Latino Asian Latino
"O" 228.000 78.000 98.000 107.000 511.000
"E" 195.217 110.783 130.783 74.217 511.000
(O-E) 32.783 -32.783 -32.783 32.783
(O-E)2 1074.711 1074.711 1074.711 1074.711
(O-E)2/E 5.505 9.701 8.218 14.481
X2obt = 37.904
X2crit = 3.841
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Problem 5 Of the total eligible drivers equaling 186.2 million (Male = 95.4, Female = 90.8), there were a total of 78.9 vehicle stops (Male = 50.4, Female = 28.5). You are in charge of determining if there is a disparity among males and females concerning traffic stops? Create bivariate distribution tables (freq and %) for two nominal variables and compute MOE.
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Problem 5 answer:
What is the theory? Males are stopped more than Females. (O ≠ E)
What is the null hypothesis? There is not a statistically significant difference in the number of vehicle stops among Male or Female drivers, between the observed and the expected count. (O = E)
Test the Null and Report the results: Reject, X2(1) = 8.760, P. <.05.
Provide the Analysis: There is a statistically significant difference in the count of vehicle stops among male and female drivers (male = 53%, female = 31%) between the results in the sample and what would be obtained from random sample of drivers not separated by gender. The magnitude of effect (MOE) is weak MOE = 6%.
Male Female Total
F “Stop” 50.4 28.5 (78.9) E1 = 78.9
% “Stop” 53% 31% 45.0
F “Not Stop” 45.0 62.3 (107.3) 28.5
% “Not Stop” 47% 69%
Total (95.4) (90.8) (186.2) E2 = 73.5
λ = E1 - E2 / E1
λ = 78.9 - 73.5 / 78.9
λ = 5.4 / 78.9
λ = .068 = 7%
Chi-square - TWO-WAY
Stopped Stopped Not Stopped Not Stopped Totals
Male Female Male Female
"O" 50.400 28.500 45.000 62.300 186.200
"E" 40.425 38.475 54.975 52.325 186.200
(O-E) 9.975 -9.975 -9.975 9.975
(O-E)2 99.509 99.509 99.509 99.509
(O-E)2/E 2.462 2.586 1.810 1.902
X2obt = 8.760
X2crit = 3.841
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