A review of performance in WQ-3 indicates that many students need more practice with statistical dis

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A review of
performance in WQ-3 indicates that many students need more practice with
statistical distributions and significance tests; this WQ-4 provides such an
opportunity. Please make sure that you review the feedback comments and a
sample solution provided for WQ-3.

The background:
In 2005,
Congress passed the Bankruptcy Abuse Prevention and Consumer Protection Act.
This Act made significant changes to the administration of bankruptcy relief.
The Government Accountability Office (GAO) was commissioned to study the
effects of the bankruptcy reform law on consumers. Attorney fees in consumer
bankruptcy cases were studied by the GAO and reported in their 2008
Bankruptcy Reform report. In the period February to March 2007 there were
71,106 Chapter 7 consumer bankruptcy cases.

To
estimate the legal fees for Chapter 7 consumer bankruptcy cases, the GAO
conducted a nationwide random sample of Chapter 7 consumer bankruptcy filings.
Business bankruptcy cases not involving attorneys were excluded. In the end,
the GAO had the sample size, n=292 consumer cases with attorney involvement.
The mean attorney fee in the sample is
.0/msohtmlclip1/01/clip_image002.png”> $1078 and the standard deviation is s=$592.

For 95% confidence interval, the z-score (from
Standard Normal Probabilities table) is 1.96
Note: For this large sample, we assume.0/msohtmlclip1/01/clip_image004.png”> and we use
Standard Norm al Distribution. For smaller sample sizes (.0/msohtmlclip1/01/clip_image006.png”> ) we use
t-distribution.

The margin
of error .0/msohtmlclip1/01/clip_image008.png”>
So, 95% confidence interval
for the mean of the whole population (.0/msohtmlclip1/01/clip_image010.png”>) is:
.0/msohtmlclip1/01/clip_image012.png”>

1

Find
the 80% confidence interval for the population mean .0/msohtmlclip1/01/clip_image014.png”>

2

Find
the 90% confidence interval for the population mean .0/msohtmlclip1/01/clip_image014.png”>

3

Find
the 99% confidence interval for the population mean .0/msohtmlclip1/01/clip_image014.png”>

4

Find
the 99.96% confidence interval for the population mean .0/msohtmlclip1/01/clip_image014.png”>

5

The margin of error for 95% confidence interval (as calculated
above):
=.0/msohtmlclip1/01/clip_image016.png”>
If
we wish to answer the question: “What should be the sample size if we want
the margin of error to be $100 (instead of $67.90)â€, then the desired sample
size can be calculated as follows:
.0/msohtmlclip1/01/clip_image018.png”>
So
a sample size of 135 should be chosen.

Calculate
the sample size for a margin of error of $100 for various confidence
intervals and complete the following table:

Confidence Interval

Sample Size

99.96%

99%

95%

135

90%

80%

6

Any
additional thoughts, about this course, you may like to share: