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天山7-22

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楼主
发表于 2007-1-29 23:37:00 | 只看该作者

天山7-22


    

T-4-Q20
        
天山-7-22


    

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


    

 


    
  1. Changes in the
         population density of both Meadowbrook and Parkdale over the past four
         years.
  2. How the rate of
         population growth in Meadowbrook over the past four years compares to the
         corresponding rate for Parkdale
  3. The ratio of violent to
         nonviolent crimes committed during the past four years in Meadowbrook and
         arkdale
  4. The violent crime rates
         in Meadowbrook and Parkdale four years ago

    How Meadowbrooks’
expenditures for crime prevention over the past four years compare to
Parkdale’s expenditures.

这个题目的答案是D,但是因为题目中说的是,两个城市的犯罪增长率,所以即使以前的犯罪率不一样,也不能改变犯罪增长率的差异啊?
所以明显M城市的犯罪增长率比P城市的高,
但是B选项则说M城市的人口增长更快,因为人更多了,所以犯罪率也容易变大了,
请讨论!

沙发
发表于 2007-1-30 01:00:00 | 只看该作者

应该选D,假设M城的crime rate四年前是10%,P城四年前是50%,则四年后M城crime rate为16%,P城为55%,P城的犯罪率在四年后依然比M城高,即使这四年的犯罪增长率M高于P,决定结果的除了变化的比率,还有变化的基数。

因为犯罪率的变化已经包含了人口变化的因素,所以不用考虑


[此贴子已经被作者于2007-1-30 1:01:47编辑过]
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