HOMEWORK FOR EXST 7036
(Chapter 3)
Student Name: Tanza Erlambang
3.2) In the 2000 US Presidential election,
Palm Beach County Florida was the
focus of unusual voting patterns caused by a confusing “butterfly bailout”. Many
voters claimed they voted mistakenly for Pat Buchanan, when they intended to
vote for Al Gore. Figure 3.8 shows the total number of votes for Buchanan plotted
against number of votes for Ross Perot in 1996, by county in Florida (See
A.Agresti and B Presnell,
Statist.Sci
., 17: 436440, 2003)
a.
In county i, let π
i
denote the proportion of vote for Buchanan and let x
i
for
proportion vote of Perot in 1996.The Linear probability model fitted to all counties
except Palm Beach, π
^
i
= 0.0003+0.0304x
i
.Give the value of p in interpretation:The
estimated proportion vote for Buchanan in 2000 was roughly p% of that for Perot in
1996.
Answer: Let say x
i
= 1000, thus:
π
^
i
= 0.0003+0.0304x
i
→ π
^
i
= 0.0003+0.0304(10000)
= 303.9997 = 304
Then, P. Buchanan = 304/10000 (100%) = 3.04% of Perot
b
. For Palm Beach County, , π
i
= 0.0079 and x
i
= 0.0774. Does this result appar to be an
outlier?. Finding π
i
/π
^
i
and π
i
 π
^
i
(Analyses conducted by statisticians predicted that
fewer than 900 votes were truly intended for Buchanan, compared with the 3407 he
received. George W Bush won the state by 537 votes and, with it, the Electoral College
and the election).
Answer:
Estimated proportion: π
^
i
= 0.0003+0.0304(0.0774)
= 0.0021
π
i
 π
^
i
= 0.0079 – 0.0021 = 0.0058
π
i
/π
^
i
= 0.0079/0.0021 = 3.76 → the actual value is 3.76 times the estimated value →
suggested that it is an outlier
1
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3.8) Refer to the previous exercise for the horseshoe crab data
a.
Report the fit for the probit model, with weight predictor
Answer: Probit (π
^
i
) = 2.238 + 1.099 (weight)
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 Spring '10
 BrianMarx
 Normal Distribution, Pat Buchanan, Palm Beach County, STANDARD NORMAL PROBABILITY, Horseshoe Crab data

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