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sample exam i 上的各种问题

做了sample1 看了答案之后还是有很多地方希望更明白些,请指教,谢谢。

 

Quantitive:

 

An increase in which of the following items is most likely to result in an increase in the width of the confidence interval for the population means?

 

A sample size

B reliability factor

Cdegrees of freedom

 

答案是B 我没有查到reliability factor是什么东西?谁能帮我解释一下这道题,而且为什么不是degrees of freedom

 

An analyst determined that approximately 99 percent of the observations of daily sales for a company were within the interval from 23000 to 48000 and that daily sales for the company were normally distributed. the mean daily sales and standard deviation of daily sales, respectively, for the company were closest to

 

a. 41,667

b 62,500

c 83,333

 

这道题怎么也做不出,而且为什么问了两个量,结果只给了一个量的答案啊,真奇怪。

 

 

the least accurate statement about measures of dispersion for a distribution if that the:

A range provides no information about the shape of the data distribution

B arithmetic average of the deviations around the mean will always be equal to one.

C mean absolute deviation will always be less than or equal to the standard deviation

 

这道题似乎这些概念都没有见过,能不能一一给我解释一下,都是什么意思,为什么我看了两遍的notes还是没有看到这些概念呢?

merci

 

QUOTE:
以下是引用waterline在2009-11-23 9:33:00的发言:

1) Reliability factor refers to Notes LOS10 f, e)

Confidence Interval=Point Estimate +/- (reliability facor x standard error)

 

2) Given 230,000 to 480,000, 99% Confidence Interval (same as above), to get SD

-> Point estimate=(480,000-230,000)/2+230,000=355,000, 99% CI->reliability factor around 3

-> SD=(480,000-355,000)/3=41,667

 

3)

MAD - LOS 7f)

sum of diviations = 0 -> arithmetic average of the deviations = 0,  not 1

 

 

 

真厉害

 

TOP

1) Reliability factor refers to Notes LOS10 f, e)

Confidence Interval=Point Estimate +/- (reliability facor x standard error)

 

2) Given 230,000 to 480,000, 99% Confidence Interval (same as above), to get SD

-> Point estimate=(480,000-230,000)/2+230,000=355,000, 99% CI->reliability factor around 3

-> SD=(480,000-355,000)/3=41,667

 

3)

MAD - LOS 7f)

sum of diviations = 0 -> arithmetic average of the deviations = 0,  not 1

 

 

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