Violent crimes

The violent crime rate (number of violent crimes per 1,000 residents) in Meadowbrook is 60 percent higher now than it was four years ago. The corresponding increase for Parkdale is only 10 percent. These figures support the conclusion that residents of Meadowbrook are more likely to become victims of violent crime than are residents of Parkdale. The argument above is flawed because it fails to take into account

a) Changes in population density of both Parkdale and Meadowbrook over the past four yrs.
b) how the rate of population growth in Meadowbrook over the past four yrs compares to the corresponding rate for Parkdale
c) the ratio of violent to non-violent crimes committed during the past four yrs in Meadowbrook and Parkdale
d) the violent crime rates in Meadowbrook and Parkdale four years ago
e) how Meadowbrook's expenditures for crime prevention over the past four yrs compare to Parkdale's expenditures

The summarized argument: violent crime rate from four years ago is up 60% in Meadowbrook and the crime rate from four years ago is up 10% in Parkdale. Therefore citizens in Meadowbrook are more likely to be the victims of a violent crime.

Anytime we're looking at percentages, we always want to ask ourselves about the base rates. Meadowbrook's crime rate is up 60% from what? Parkdale's crime rate is up 10% from what?

To see why this is the crucial issue, consider a simple scenario. Suppose Meadowbrook and Parkdale both had a crime rate of 100 crimes/1000 residents four years ago. In this case, Meadowbrook's new crime rate will be 160 crimes/1000 residents (60% increase) and Parkdale's will be 110 crimes/1000 residents (10% increase), and Meadowbrook is, in fact, more dangerous.

But imagine another case in which Meadowbrook had a crime rate of 10 crimes/1000 residents and Parkdale had a crime rate of 1000 crimes/1000 residents four years ago. Now, Meadowbrook's crime rate will be 16 crimes/1000 citizens (60% increase) and Parkdale's crime rate will be 1100 crimes/1000 residents (10% increase.) In this case Meadowbrook is much safer, despite the 60% increase in the crime rate. It all depends on what the initial crime rate was four years ago.

The answer choice that addresses this is D.

DavidG@VeritasPrep wrote:

The violent crime rate (number of violent crimes per 1,000 residents) in Meadowbrook is 60 percent higher now than it was four years ago. The corresponding increase for Parkdale is only 10 percent. These figures support the conclusion that residents of Meadowbrook are more likely to become victims of violent crime than are residents of Parkdale. The argument above is flawed because it fails to take into account

a) Changes in population density of both Parkdale and Meadowbrook over the past four yrs.
b) how the rate of population growth in Meadowbrook over the past four yrs compares to the corresponding rate for Parkdale
c) the ratio of violent to non-violent crimes committed during the past four yrs in Meadowbrook and Parkdale
d) the violent crime rates in Meadowbrook and Parkdale four years ago
e) how Meadowbrook's expenditures for crime prevention over the past four yrs compare to Parkdale's expenditures

The summarized argument: violent crime rate from four years ago is up 60% in Meadowbrook and the crime rate from four years ago is up 10% in Parkdale. Therefore citizens in Meadowbrook are more likely to be the victims of a violent crime.

Anytime we're looking at percentages, we always want to ask ourselves about the base rates. Meadowbrook's crime rate is up 60% from what? Parkdale's crime rate is up 10% from what?

To see why this is the crucial issue, consider a simple scenario. Suppose Meadowbrook and Parkdale both had a crime rate of 100 crimes/1000 residents four years ago. In this case, Meadowbrook's new crime rate will be 160 crimes/1000 residents (60% increase) and Parkdale's will be 110 crimes/1000 residents (10% increase), and Meadowbrook is, in fact, more dangerous.

But imagine another case in which Meadowbrook had a crime rate of 10 crimes/1000 residents and Parkdale had a crime rate of 1000 crimes/1000 residents four years ago. Now, Meadowbrook's crime rate will be 16 crimes/1000 citizens (60% increase) and Parkdale's crime rate will be 1100 crimes/1000 residents (10% increase.) In this case Meadowbrook is much safer, despite the 60% increase in the crime rate. It all depends on what the initial crime rate was four years ago.

The answer choice that addresses this is D.

Thanks Dave!! Got it!

Thanks,
Mallika

DavidG@VeritasPrep wrote:

The violent crime rate (number of violent crimes per 1,000 residents) in Meadowbrook is 60 percent higher now than it was four years ago. The corresponding increase for Parkdale is only 10 percent. These figures support the conclusion that residents of Meadowbrook are more likely to become victims of violent crime than are residents of Parkdale. The argument above is flawed because it fails to take into account

a) Changes in population density of both Parkdale and Meadowbrook over the past four yrs.
b) how the rate of population growth in Meadowbrook over the past four yrs compares to the corresponding rate for Parkdale
c) the ratio of violent to non-violent crimes committed during the past four yrs in Meadowbrook and Parkdale
d) the violent crime rates in Meadowbrook and Parkdale four years ago
e) how Meadowbrook's expenditures for crime prevention over the past four yrs compare to Parkdale's expenditures

The summarized argument: violent crime rate from four years ago is up 60% in Meadowbrook and the crime rate from four years ago is up 10% in Parkdale. Therefore citizens in Meadowbrook are more likely to be the victims of a violent crime.

Anytime we're looking at percentages, we always want to ask ourselves about the base rates. Meadowbrook's crime rate is up 60% from what? Parkdale's crime rate is up 10% from what?

To see why this is the crucial issue, consider a simple scenario. Suppose Meadowbrook and Parkdale both had a crime rate of 100 crimes/1000 residents four years ago. In this case, Meadowbrook's new crime rate will be 160 crimes/1000 residents (60% increase) and Parkdale's will be 110 crimes/1000 residents (10% increase), and Meadowbrook is, in fact, more dangerous.

But imagine another case in which Meadowbrook had a crime rate of 10 crimes/1000 residents and Parkdale had a crime rate of 1000 crimes/1000 residents four years ago. Now, Meadowbrook's crime rate will be 16 crimes/1000 citizens (60% increase) and Parkdale's crime rate will be 1100 crimes/1000 residents (10% increase.) In this case Meadowbrook is much safer, despite the 60% increase in the crime rate. It all depends on what the initial crime rate was four years ago.

The answer choice that addresses this is D.

Thanks Dave!! Got it!

Thanks,
Mallika

Thanks David. I see your reasoning behind D. Can either you or any other experts explain why neither A nor B describes that same phenomenon. I’m pretty sure I see it, but I’d really appreciate some expert analysis.

Thanks all!

Can either you or any other experts explain why neither A nor B describes that same phenomenon. I’m pretty sure I see it, but I’d really appreciate some expert analysis.

The argument is about changes in the crime rate of two cities and whether this change means one city is more dangerous than the other. 'A' is about changes in population density, and 'B' is about overall population growth. Any effect that the change in population would have on relative safety would already be incorporated in the crime rates of these two cities.

Put another way, a city that had 10 residents and 5 violent crimes would have the same per capita crime rate as one with 1000 residents and 500 violent crimes. It's not the the population size that's important. It's the crime rate within that population.