Normal approximations in hypothesis tests

A suitable normal distribution can approximate a large binomial or Poisson count. Build that distribution from the null hypothesis, then apply a continuity correction to the observed discrete boundary.

Approximate a large discrete count with continuity correction

Under $H_0$ , replace a suitable binomial or Poisson distribution by a normal distribution. All parameters must come from the null model.

Y\sim\mathrm{N}(\mu,\sigma^2)

Here the continuous variable Y approximates the discrete count X, and $p_0$ or $\lambda_0$ is the value stated by $H_0$ .

Binomial: $\mu=np_0$ and $\sigma^2=np_0(1-p_0)$ .

Poisson: $\mu=\sigma^2=\lambda_0$ .

For a binomial approximation, check that $np_0>5$ and $n(1-p_0)>5$ .
Move a discrete boundary by 0.5 before standardising: $X\le78$ becomes $Y<78.5$ .
For an upper tail $X\ge r$ , use the boundary $r-0.5$ .

Worked example

Large binomial, lower-tailed test

Test $H_0:p=0.25$ against $H_1:p<0.25$ when 78 successes are observed in 400 trials, at the 5% level.

Y\sim\mathrm{N}(100,75)

z=\frac{78.5-100}{\sqrt{75}}=-2.48

P(Y<78.5)\approx0.0065

Answer

\text{Reject }H_0.

Conclusion: There is sufficient evidence that the success probability is below 0.25.

Common mistake

Do not use the observed sample proportion to build the null mean and variance. A hypothesis test asks what would happen if the claimed parameter were true.

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