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If anyone is going to argue this way, would it not be more appropriate to
base the whole thing only on deaths? After all, we know that the children
are dead. Then the probability of a death being a cot death is 394/4959 or 1
in 12.6, that of a homicide 14/4959 or 1 in 354. 
Petra 

Dr. Petra Leimich				
Lecturer in Statistics and Information and Communication Technology
School of Computing				
University of Abertay Dundee			
Dundee 					e-mail: [log in to unmask]
DD1 1HG					phone: +44 (0)1382 308957
Scotland					fax: +44 (0)1382 308627
					


-----Original Message-----
From: [log in to unmask] [mailto:[log in to unmask]]
Sent: 12 November 1999 09:58
To: Allstat UK list
Subject: Statistics and Justice


Allan Reese has written about the case of the woman convicted of killing her
two 
children, apparently partly on the basis of a statistical argument.

I thought members might be interested in some data.  In 1996, the most
recent 
year for which data are available, there were 649,489 live births in E&W.  
There were 4,959 deaths in the first year of life, including 394 sudden
infant 
deaths (SIDS or cot deaths) and 14 homicides.  Thus the probability of a cot

death is 1 in 649489/394 = 1,648.  If deaths were independent, which they
are 
not because there are familial risk factors, the risk that a fimily of two 
babies would have two cot deaths would be 1 in 1,648*1,648 = 2,715,904.  
Presumably some other adjustment was applied for social factors as the
figure 
given in the press is said to apply to `well-to-do families'.

But the same calculation for homicide gives a risk that that two baies will
both 
be murdered as one in (649489/14)^2 =  2,152,224,291.  If the first figure
is 
relevant, so is the second.  As Allan says, neither is relevant.

Martin

.



Prof J M Bland
Dept of Public Health Sciences
St George's Hospital Medical School
London SW17 0RE, UK

Tel 0181-725 5492
email [log in to unmask]

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