In article ,
Colin Rosenstiel writes
I did a rather larger census while cycling from Putney to Westminster this
morning. I saw 154 cars with series 3 index plates; 50 had X or Z (32.5%).

Okay, we can now do some statistical analysis. A census like this is a
binomial distribution B(C,p) where C is the number of cars (you saw 154)
and p is the probability of an X or Z.

I claim that p = 3176/13824 (this is the null hypothesis). You claim it
is larger (the alternative hypothesis).

We approximate B(C,p) by the normal distribution N(Cp, Cp(1-p)), which
is N(0.23N, 0.177N). With X hits (plates with an X or Z), let:

Z = (X - mu)/sigma = (X - 0.23N)/sqrt(0.177N)

and the significance tests say that in a random sample:

Z 1.64 5% of the time
Z 2.33 1% of the time

You had N = 154, X = 50. So Z = (50 - 35.4)/5.22 = 2.79. That's highly
significant (it happens less than 1 in 378 times). It looks like you're
right.

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Clive D.W. Feather, writing for himself | Home:
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