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Considerable controversy has arisen over the possible aftereffects of a nuclear weapons test conducted in Nevada in 1957. Included as part of the test were some three thousand military and civilian "observers". Now, more then fifty years later, eight cases of leukemia have been diagnosed among those three thousand. The expected number of cases, based on the demographic characteristics of the observers, was three. Assess the statistical significance of these findings. Calculate an approximation based on the central limit theorem.

the central limit theorem is as follows: $$P(a\le \frac{\bar w - \mu}{\sigma/\sqrt{n}} \le b)$$

the answer however is as follows: $$1-P(\frac{X-3}{\sqrt{3}} \le \frac{7.5-3 }{\sqrt{3}})$$

did i miss something why are they using the expected value instead of sigma divided by sqrt of n? should i assume that those two terms are equal? how are they equal?

notamathwiz
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1 Answers1

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I shudder at using CLT in this situation.

However, the usual model used is the Poisson model, with parameter $\lambda=3$. The variance of the Poisson with parameter $\lambda$ is $\lambda$, so the standard deviation is $\sqrt{3}$.

Another way of thinking about it is that the number of leukemia cases in a group of $3000$ has binomial distribution, $n=3000$, $p=\frac{1}{1000}$. The variance is $np(1-p)$. Since $p$ is very small, $1-p\approx 1$, and therefore the variance of the binomial is approximately $np$, which is $3$.

Remark: Using the Poisson model, it is not really too hard to compute the probability that the number of cases is $\le 7$, under the null hypothesis that $\lambda=3$. So there is no real need to use a normal approximation.

André Nicolas
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