Vertrauenswurdige Online Casinos - ww.del

Vertrauenswurdige Online Casinos - ww.del

Appendix C
In this appendix, we examine the consequences of an assumption that the ranges of 23
to 29 MPH for the maximum attainable speed from a stop and 30 to 43 MPH for the
minimum impact speeds represent either 95% or 99% confidence intervals. Let X denote
the maximum attainable speed from a stop and Y denote the minimum impact speed. If
we assume that X and Y are normally distributed random variables, then the means of
these random variables would be the centers of the confidence intervals; that is, E[X]=26
MPH and E[Y]=36.5 MPH. Now the ends of a 95% confidence interval are 1.96 standard
deviations above and below the center of the interval. Let  X denote the standard
deviation of X and  Y denote the standard deviation of Y. Then we have
1.96 X  29  26  3
3
 1.53
1.96
1.96 Y  43  36.5  6.5
X 
6.5
 3.32
1.96
The necessary acceleration from a stop would be attained only if X  Y  0. We now
calculate the probability that X  Y  0. The expected value of X-Y is E[X-Y]=E[X]E[Y]=26-36.5=-10.5. Since X and Y are clearly independent random variables, the
standard deviation of X-Y,  X Y , is given by
Y 
 X Y   X 2   Y 2  1.532  3.322  3.66
10.5
 2.87 X Y ' s above the mean of the random variable X-Y. From the
3.66
standard normal table, the probability that X-Y assumes a value this large or higher is
only .0021. If the intervals [23,29] and [30,43] are regarded as 99% confidence intervals,
then the ends of the confidence intervals are 2.58 standard deviations above and below
the centers, so that the same sort of calculations as above give
3
X 
 1.16
2.58
6.5
Y 
 2.52
2.58
So 0 is
 X Y  1.162  2.522  2.77
10.5
 3.79 X Y ' s above the mean of X-Y. The probability that
2.77
X-Y assumes a value at least this high is only .000075, i.e. less than one in ten thousand.
But remember, these intervals are not confidence intervals – they represent absolute
limits. So the police calculations imply with mathematical certainty that Walsh ran the
stop sign. Even if Downs misinterpreted the intervals as confidence intervals, it would
still imply a vanishingly small probability not that Walsh stopped but that it was
physically possible for her to have stopped.
Now we see that 0 is