One-Way Chi-Square
Problem 1: A police chief wants to determine which month of the year is more popular for vacations: June, July or August. It is theorized that June will be more popular than the other two. Sixty officers are randomly selected and asked which of three months they prefer. Thirty prefer June, fifteen July and fifteen August. Do the results deviate from what would be expected by chance?
Theory: There is greater desire among officers to vacation in June. (O ≠ E)
Null: There is not a statistically significant difference in month preference among police officers between observed values and those expected by chance alone.
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Problem 1 answer:
Theory: There is greater desire for officers to vacation in June. (O ≠ E)
Null: There is not a statistically significant difference in vacation month preference among officers between observed values and those expected by chance alone. (O = E)
June July August total
Row 1 O 30.000 15.000 15.000 60.000
Row 2 E 20.000 20.000 20.000 60.000
Row 3 (O-E) 10.000 -5.000 -5.000
Row 4 (O-E)2 100.000 25.000 25.000
Row 5 (O-E)2/E 5.000 1.250 1.250
Row 6 X2obt = 7.500
Results: X2(2) = 7.50, p. < .05. Reject the null because the obtained X2obt of 7.50 at df = 2 exceeds X2crit of 5.99. The observed frequencies differ from the expected frequencies by an amount that would be unlikely to occur by chance if the null were true.
Analysis: Theory is supported. The probability of officers preferring vacation in June is greater than would be expected by chance. The difference in month preference is statistically significant.
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Problem 2: The FBI classifies offenders in two categories: index and minor. Sometimes these offenders fall in love; an index-crime offender will marry a minor-crime offender, and produce criminal “children”, who grow up to be offenders. If an index criminal child then marries a minor crime child, chances are 3 to 1 that the “grandchild” will grow up to be an index offenders. To test that theory, we selected a random sample of 40 “grandchildren” produced from an index-offender/minor-offender union. Out of 40 grandchildren, 23 grew up to be index-offenders, and 17 grew up to be minor-offenders. Based on the expected values of 30 index-offenders and 10 minor-offenders (3 to 1), test the theory.
Theory: The observed 23-to-17 ratio is different than the 30-to-10 ratio that is theorized by criminologists. (O ≠ E)
Null: There is not a statistically significant difference in offender-types of grandchildren between the observed values of 23 to 17, and the expected 30 to 10 ratio.
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Problem 2 answer:
Index Minor Total
Row 1 O 23.000 17.000 40.000
Row 2 E 30.000 10.000 40.000
Row 3 (O-E) -7.000 7.000
Row 4 (O-E)2 49.000 49.000
Row 5 (O-E)2/E 1.633 4.900
Row 6 X2obt = 6.533
Results: Reject, X2(1) = 6.53, p. < .05. The null hypothesis is rejected because the obtained Chi-square (X2obt = 6.53) is greater than the critical value of Chi-square (X2crit = 3.84) at df = 1.
Analysis: The null hypothesis was rejected, the theory is supported by the data. The theory predicted that there would be a difference between the expected 30-to-10 ratio and the observed 23-to-17 ratio in offender-type grandchildren. There are significantly more minor-offenders in this sample than were expected based on previous data.
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Problem 3: Among those drivers that were stopped by the police, are there more males than females? The data for police vehicle stops would seem to suggest that men are stopped in greater frequency than women. However, it is possible that the frequencies occurred by chance. A random sample of stopped drivers were identified by gender. The police had stopped 122 males and 78 females.
Theory: Males are experiencing vehicle stops in greater frequency than are females. (O ≠ E)
Null: There is not a statistically significant difference in vehicle stops among men and women between the observed count and the expected count.
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Problem 3 answer:
Theory: Men are stopped in greater frequency than females.
Null: There is not a statistically significant difference in vehicle stops among men and women between the observed values and the count by chance alone.
Male Female Total
Row 1 O 122.000 78.000 200.000
Row 2 E 100.000 100.000 200.000
Row 3 (O-E) 22.000 -22.000
Row 4 (O-E)2 484.000 484.000
Row 5 (O-E)2/E 4.840 4.840
Row 6 X2obt = 9.680
Results: X2(1) = 9.680, p. < .05. The null hypothesis is rejected because the obtained Chi-square (X2obt = 9.680) is greater than the critical value of Chi-square (X2crit = 3.84) at df = 1. Reject Null, (X2 (1) = 9.841, p. < .05)
Analysis: The null hypothesis was rejected; the theory is supported by the data. The theory predicted that there would not be a difference between men and women among stopped drivers. There were significantly more men than women among stopped drivers.
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Problem 4: In the year 2003, a number of college students were asked if it was necessary to invade Iraq. Of 305 the students surveyed, 125 said yes, 102 said no, and 78 didn’t know if it was necessary or not. Test if there is a significant difference between the observed values and what would be expected by chance.
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Problem 4 Answer:
Theory: There is a difference in student opinion, more students believe it is necessary to invade Iraq. (O ≠ E)
Null: There is not a statistically significant difference in invasion approval among students between observed values and those expected by chance alone. (O = E)
Yes No Don't Know total
Row 1 O 125.000 102.000 78.000 305.000
Row 2 E 101.667 101.667 101.667 305.000
Row 3 (O-E) 23.333 0.333 -23.667
Row 4 (O-E)2 544.444 0.111 560.111
Row 5 (O-E)2/E 5.355 0.001 5.509
Row 6 X2obt = 10.866
Results: X2(2) = 10.866, p. < .05. Reject the null because X2obt of 10.866 at df = 2 exceeds X2crit of 5.99. The observed frequencies differ from the expected frequencies by an amount that would be unlikely to occur by chance if the null were true.
Analysis: Theory is supported. The probability of students believing that the invasion is necessary is greater than would be expected by chance. The difference in opinion is statistically significant.
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Problem 5: The Gaming Commission does routine checks on slot machines in casinos to make sure that they “win” evenly. It performs the tests using the one-way Chi-square procedure. The commission tested 5 slot machines in the casino. However, using an equal number of plays the payoffs look uneven: Machine A = 17, machine B = 23, machine C = 14, machine D = 30, machine E = 16. Is there evidence to show the machines are unequal in the payoff?
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Problem 5 Answer:
Theory: There is a difference in payoffs, the wins among the machines are not equal. (O ≠ E)
Null: There is not a statistically significant difference in payoffs among machines between observed values and those expected by chance alone. (O = E)
Machine A Machine B Machine C Machine D Machine E Total
Row 1 O 17.000 23.000 14.000 30.000 16.000 100.000
Row 2 E 20.000 20.000 20.000 20.000 20.000 100.000
Row 3 (O-E) -3.000 3.000 -6.000 10.000 -4.000
Row 4 (O-E)2 9.000 9.000 36.000 100.000 16.000
Row 5 (O-E)2/E 0.450 0.450 1.800 5.000 0.800
Row 6 X2obt = 8.500
Results: Retain, X2(4) = 8.500, p. > .05. We retained the null because X2obt = 8.500 is less than X2crit of 9.49 at df = 4. The observed frequencies do not differ from the expected frequencies by enough to conclude that the null is false.
Analysis: Theory is not supported. The notion that the machines are “fixed” to win unevenly is discredited. The difference in wins between observed and expected count is not statistically significant.
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Problem 6: Is there a difference in support for gun control? Persons from the NRA forward the notion that most persons support gun ownership without first obtaining a permit. Those who support gun control see no reason why an honest person would be against registering his or her gun. A sample of 63 persons from the General Social Survey showed that 24 favored mandatory permits and 39 opposed mandatory gun permits. Is the NRA’s position supported?
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Problem 6 answer:
Theory: Most people do not support gun permits.
Null: There is not a statistically significant difference in support for gun permits among men and women between the observed values and the count by chance alone.
Favor Oppose Total
Row 1 O 24.000 39.000 63.000
Row 2 E 31.500 31.500 63.000
Row 3 (O-E) -7.500 7.500
Row 4 (O-E)2 56.250 56.250
Row 5 (O-E)2/E 1.786 1.786
Row 6 X2obt = 3.571 X2crit = 3.841
Results: X2(1) = 3.571, p. > .05. The null hypothesis is retained because Chi-square (X2obt = 3.571) is less than the critical value of Chi-square (X2crit = 3.841) at df = 1.
Analysis: The null hypothesis was retained; the NRA’s theory is not supported by the data. The theory predicted that most people would oppose gun-permits. Although more persons in the sample opposed gun permits, there was not a statistically significant difference between the observed frequencies and what would be expected by random assignment.
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Problem 7: An automobile analyst is studying car-ownership among families who have children that haven’t reached driving age. She believes that, under the definition of “family of 4 with no children drivers” there will be typically be found 2 cars. It is theorized that, although there may be some families under this definition with more or less cars, most of the defined families will have 2 cars. A random sample of married couples with two children under age 15, who are not divorced, was taken from the population. The results show that families that meet this definition were outfitted with vehicles in the following manner: One car = 15, two cars = 28, three cars = 12.
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Problem 7 answer:
Theory: Most families of four will have 2 cars. (O ≠ E)
Null: There is not a statistically significant difference car ownership among families of four between observed values and those expected by chance alone. (O = E)
One Two Three Total
Row 1 O 15.000 28.000 12.000 55.000
Row 2 E 18.333 18.333 18.333 55.000
Row 3 (O-E) -3.333 9.667 -6.333
Row 4 (O-E)2 11.111 93.444 40.111
Row 5 (O-E)2/E 0.606 5.097 2.188
Row 6 X2obt = 7.891
Results: X2(2) = 7.891, p. < .05. Reject the null because X2obt = 7.891 at df = 2 exceeds X2crit of 5.99. The observed frequencies differ from the expected frequencies by an amount that would be unlikely to occur by chance if the null were true.
Analysis: Theory is supported. More families in the sample were found to have 2 cars than would be expected by chance alone. The difference car ownership is statistically significant