Testing of significance for a single proportion

Question

Suppose a large number $n$ of independent Bernoullian trials is performed and $x$ successes are obtained. Suppose we wish to test the hypothesis that probability of success in each trial is $p$, assuming the hypothesis to be correct, the mean and variance of the sampling distribution of the number of success are $np$ and $npq$ respectively. Then for large $n$ $z=({x-np})/{\sqrt {npq}}$ is distributed as a standard normal variate. Hence we get $P(\vert z \vert >3)=0.0027$ Where did the $3$ come from?

Answer 1

It is arbitrary, illustrating the probability with a normal distribution of being more than $3$ standard deviations from the mean in the two tails.

It could have said

• "Hence we get $P(\vert z \vert >2)=0.0455$"
• "Hence we get $P(\vert z \vert >1)=0.3173$"

or various similar things. This is similar to the so-called 68–95–99.7 rule but with the probabilities of being in the tails rather than the middle of the distribution

Answer 2

This statement uses the normal approximation to a binomial probability, which works well for sufficiently large $n$ and $p$ not too far from 1/2.

The idea is that you will reject the null hypothesis that $p = p_0$ at the 0.3% level if $|Z| > 3.$

For example, if $n = 36$ and $p_0 = 1/2,$ then you would reject $H_0: p = 1/2,$ if you got only $X=1$ Success.

The probability that an observation from a standard normal distribution is in the interval $(-3,3)$ is about 0.997. In R:

diff(pnorm(c(-3,3)))
[1] 0.9973002

Alternatively, perhaps you can find a printed table of the standard normal PDF that goes that far out from $0.$

Also, perhaps google the Empirical Rule.