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a. there are 185 total cases
b. the two row marginal frequencies are 45 (medium security) and 140 (maximum security)
c. the two column marginal frequencies are 115 (not satisfied) and 70 (satisfied)
d. there are 15 cases in Cell A, there are 40 cases in Cell D
e. this is a 2x2 contingency table, because there are 2 columns and 2 rows
f. the number of degrees of freedom in a contingency table is determined by the following formula df = (r-1) (c-1), which is the number of rows minus one times the number of columns minus one. With 2 rows and 2 columns, the number of degrees of freedom is equal to df = (2-1)(2-1) = 1. There is one degree of freedom.
g. The cell percentages would look like this:
SATISFIED?
No Yes
Medium
TYPE Security 33% 67% 100%
OF
INSTITUTION
Maximum 71% 29% 100%
Security
Since we have assumed that institution type is the independent variable, we calculate the cell percentages with the row marginals as the denominator.
h. Yes, as you can see from the table, 67% of those officers in medium security institutions are satisfied with their job, while only 29% of those in maximum security institutions are satisfied with their job.
a. Assuming that office type is the independent variable, here are the cell percentages:
OFFICER TYPE
Conflict
Minimizer Control
No 90% 65%
USED
FORCE?
Yes 10% 35%
100% 100%
Yes, while 35% of the control officers used force in making an arrest, only 10% of the conflict minimizer officers used force.
b. The question asks if the two variables are statistically independent. Our hypotheses are:
H0: Type of Officer is statistically independent of the use of force. In other words, type of officer is not related to the use of force. c2 = 0.
H1: Type of Officer is not statistically independent of the use of force. In other words, type of officer is related to the use of force. c2 > 0.
Since the hypothesis test involves whether or not two categorical variables are statistically independent, our statistical test is the chi-square test of independence. The sampling distribution is the chi-square distribution.
We have an alpha level of .01 and 1 degree of freedom. From the chi-square distribution, we can see that the critical value of chi-square is 6.635. Our decision rule, therefore, is to reject the null hypothesis if our obtained value of chi-square is greater than or equal to 6.635. We will fail to reject the null hypothesis if our obtained chi-square is less than 6.635.
Our obtained value of chi-square can now be calculated:
fo fe (fo - fe) (fo - fe)2 (fo - fe)2/fe 90 75 15 225 3.00 98 113 -15 225 1.99 10 25 -15 225 9.00 52 37 15 225 6.08 c2 = 20.07
Since our obtained chi-square is greater than the critical value (20.07 > 6.635), our decision is to reject the null hypothesis. There is a statistically significant relationship between type of officer and use of force, conflict minimizers are significantly less likely to use force than control officers.
c. The value of Yule's Q is:
The value of Yule's Q is .65, this tells us that there is a moderately strong positive relationship between officer type and use of force. More specifically, we can reduce our errors in predicting a officer's use of force by 65% if we know if they are a conflict minimizer or not.
d. Since the value of Q above is a sample estimate of the unknown population value, we want to test the null hypothesis that the association between officer type and use of force in the population is equal to zero.
H0: Q = 0
H1: Q > 0
This hypothesis test is conducted with the z test. From the z or standard normal table, the critical value of z with an alpha of .01 is 2.33. Our decision rule will be to reject the null hypothesis that Q = 0 if zobt ³ 2.33. The obtained value of z is:
Since zobt > zcrit our decision is to reject the null hypothesis that Q = 0, we instead conclude that there is an association between officer type and use of force in the population.
a: TYPE OF COUNSEL
Public
Retained Appointed Defender
Jail
Only 23% 24% 43%
Fine &
Jail 27% 30% 16%
TYPE OF
SENTENCE
RECEIVED Less
Than 60 30% 30% 23%
Days of
Jail
60 Days 20% 16% 18%
of Jail
or More
There is only a very slight difference across type of counsel in the percent of convicted defendants who receive a jail sentence of 60 days or more. For example, 20% of those with retained counsel received that sentence, 16% of those with appointed counsel, and 18% of those with a public defender. These percent differences are not very large.
b. The question asks if the two variables are statistically independent. Our hypotheses are:
H0: Type of counsel is statistically independent of the type of sentence received. In other words, type of counsel is not related to type of sentence. c2 = 0.
H1: Type of counsel is not statistically independent of the type of sentence received. In other words, type of counsel is related to type of sentence. c2 > 0.
Since the hypothesis test involves whether or not two categorical variables are statistically independent, our statistical test is the chi-square test of independence. The sampling distribution is the chi-square distribution.
We have an alpha level of .05 and (4-1)(3-1) = 6 degrees of freedom. From the chi-square distribution, we can see that the critical value of chi-square is 12.592. Our decision rule, therefore, is to reject the null hypothesis if our obtained value of chi-square is greater than or equal to 12.592. We will fail to reject the null hypothesis if our obtained chi-square is less than 12.592.
Our obtained value of chi-square can now be calculated:
fo fe (fo - fe) (fo - fe)2 (fo - fe)2/fe 18 27 -9 81 3.00 30 42 -12 144 3.43 94 74 20 400 5.40 22 18 4 16 .89 37 28 9 81 2.89 36 49 -13 169 3.45 24 21 3 9 .43 38 33 5 25 .76 50 58 -8 64 1.10 16 14 2 4 .29 20 22 -2 4 .18 40 39 1 1 .03 c2 = 21.85
Since our obtained chi-square of 21.85 is greater than the critical value of 12.594, our decision is to reject the null hypothesis. There is a statistically significant relationship between type of counsel and type of sentence received.
This finding may startle you since the percentage differences across type of counsel were fairly small (see the percentage table in part a above). Remember, however, that the chi-square test is sensitive to sample size, and 425 observations is a fairly large sample. From our chi-square test we conclude that there is a significant relationship between type of counsel and type of sentence. We will next have to examine the strength of that relationship.
c. You are asked to calculate the strength of the relationship between type of counsel and type of sentence with lambda. This is because our two variables are measured at the nominal level. Type of counsel is clearly a nominal level variable, and type of sentence received is most likely nominal. You could perhaps call it ordinal, but we're not sure that the ordinal property of rank ordering holds. Is a fine and a jail term less severe than a jail sentence of 60 days or less? Although you could argue that the types of punishment constitute an ordinal scale, it probably safer to assume that they are both nominal level variables. To calculate lambda, remember that we first decide how many prediction errors we would make in predicting the dependent variable, type of sentence, without knowing anything about the independent variable. Our best guess would be the modal category of the dependent variable, so we would predict that all 425 cases would receive a jail sentence only. Since only 142 of the 425 cases actually received a jail sentence, we would make 425 - 142 = 283 prediction errors. Now we have to determine how many prediction errors we would make if we knew what category of the independent variable someone was in. If we knew, for example, that someone had retained counsel, we would predict that they would receive a less than 30 day jail term, because that is the modal category for those with retained counsel. If we predict that all 80 defendants with retained counsel received a jail term of less than 30 days we would be correct 24 times, and we would make 80 - 24 = 56 prediction errors. If we knew that someone had appointed counsel, we would also predict that they would receive a less than 30 day jail term, because that is the modal category for those with appointed counsel. If we predict that all 125 defendants with retained counsel received a jail term of less than 30 days we would be correct 38 times, and we would make 125 - 38 = 87 prediction errors. Finally, if we knew that someone had a public defender, we would predict that they would receive a fine only, because that is the modal category for those with a public defender. If we predict that all 220 defendants with a public defender received a fine only we would be correct 94 times, and we would make 220 - 94 = 126 prediction errors. Without knowledge of the independent variable, we would make 283 prediction errors, with knowledge of the independent variable we would make 56 + 87 + 126 = 269 prediction errors. We are now ready to calculate lambda:
Our lambda value is only .05. This tells us that by knowing the independent variable the proportionate reduction in errors in predicting the dependent variable is 5%. This is not very large. It tells us that type of counsel is not very strongly related to type of sentence received.
d. H0: l = 0
H0: l > 0
The hypothesis test that the population lambda is equal to zero is based upon the chi-square test. We have already calculated that chi-square for the table of observed and expected frequencies, and rejected the null hypothesis. We would conclude, then, that the value of lambda in the population is not zero. We would also conclude that it is not very large, so the relationship between these two variables is not particularly strong.
POST-RELEASE EMPLOYMENT
Stable Sporadic
Employment Employment Unemployed
# OF
POST- 0 67% 47% 25%
RELEASE
ARRESTS
1+ 33% 53% 75%
100% 100% 100%
Yes, those with stable employment are less likely to be arrested when released from prison. Seventy-five percent of those without jobs were arrested on release, approximately fifty percent of those only sporadically employed, and only one-third of those with stable employment were arrested.
b. Our hypotheses are:
H0: Stability of employment is statistically independent of the number of arrests after release from prison. c2 = 0.
H1: Stability of employment is not statistically independent of the number of arrests. c2 > 0.
Since the hypothesis test involves whether or not two categorical variables are statistically independent, our statistical test is the chi-square test of independence. The sampling distribution is the chi-square distribution.
We have an alpha level of .01 and (2-1)(3-1) or 2 degrees of freedom. From the chi-square distribution, we can see that the critical value of chi-square is 9.210. Our decision rule, therefore, is to reject the null hypothesis if our obtained value of chi-square is greater than or equal to 9.210. We will fail to reject the null hypothesis if our obtained chi-square is less than 9.210.
Our obtained value of chi-square can now be calculated:
fo fe (fo - fe) (fo - fe)2 (fo - fe)2/fe 30 21 9 81 3.85 14 14 0 0 0.00 10 19 -9 81 4.26 15 24 -9 81 3.37 16 16 0 0 0.00 30 21 9 81 3.85 c2 = 15.33
Using the computational formula for chi-square, our result would be the same:
Since our obtained chi-square is greater than the critical value (15.33 > 9.210), our decision is to reject the null hypothesis. There is a statistically significant relationship between stability of employment and the number of post-prison arrests. Those with more stable employment when released have fewer arrests.
c. As measures of association, gamma, Kendall's tauc and Somers' d all require the calculation of the number of concordant and discordant pairs of observations. The number of concordant pairs of observations in this table are:
30(16 + 30) = 1380
14(30) = 420
1800 concordant observations
The number of discordant pairs are:
15(14 + 10) = 360
16(10) = 160
520 discordant observations
Gamma is equal to:
The gamma value indicates that there is a moderately strong relationship between job stability and the number of arrests. The more unstable the job (from stable to sporadic to unemployed) the more likely there will be at least one post-release arrest. More specifically, the gamma value tells us we can reduce our prediction errors by 55% when we use employment stability to predict the number of arrests.
Both Kendall's tauc and Somers' d take into account tied ranks. Tauc takes into account ranks tied either on the independent or dependent variable, Somers' d takes into account ranks tied on the dependent variable. The value of Kendall's tauc for these data is:
Tauc indicates a weaker relationship than did gamma. Unfortunately, tauc does not have a proportionate reduction in error interpretation.
The value of Somers' d for these data is:
In the denominator for d above, the 2030 are the number of observations tied on the dependent variable. It is derived from the following count:
30(14 + 10) = 720
14(10) = 140
860 these are tied on the 0 arrests category
15(16 + 30) = 690
480
1170 these are tied on the 1+ arrests category
860 + 1170 = 2030 number of observations tied on dependent , variable
The value of Somers' d indicates that there is a 29% proportionate reduction in prediction errors when using employment stability to predict post-release arrest.
These measures give us different numerical estimates of the measure of association because gamma does not take tied ranks into account. It will be greater than the other two measures from the same table. Both Kendall's tauc and Somers' d take tied ranks into account, and will for that reason be lower in magnitude. Kendall's taub should not be used for this table. It is only appropriate when the number of rows are equal to the number of columns.
d. With a two tailed alpha of .05, test the null hypotheses that:
H0: g = 0
H0: tc = 0
H0: d = 0
Against the alternative hypotheses that:
H1: g ¹ 0 H1: tc ¹ 0 H1: d ¹ 0 We have stated all of these alternatives as two-tailed hypothesis tests.
All three hypotheses can be tested with the z test, with the z or standard normal distribution as the sampling distribution. With a two-tailed alpha of .05, the critical value of z will be 1.96. Our decision rule for each test, therefore, will be to reject the null hypothesis if zobt £ -1.96 or if zobt ³ 1.96.
The z test for the significance of gamma is:
Since zobt > zcrit our decision will be to reject the null hypothesis that the population value of gamma is equal to zero.
The z test for the significance of tauc is:
Since our obtained value of 3.25 is greater than the critical value of 1.96, we decide to reject the null hypothesis that the population tauc value is equal to zero.
The z test for the significance of Somers' d is:
Since our obtained value of 3.82 is greater than the critical value of 1.96, our decision is to reject the null hypothesis that the value of Somers' d in the population is equal to zero.
Our conclusion is that there is a relationship between post-release employment and number of arrests in the population.
a. phi, Yule's Q, lambda, Goodman and Kruskal's tau can all be used
b. Yule's Q, Q is the same as gamma when the table is 2x2
c. gamma, tauc, and Somers' d
d. lambda, Goodman and Kruskal's tau
e. gamma, Kendall's taub and Somers' d
f. Kendall's tauc and Somers' d
feedback form |
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