Recipe

Approximate a probability for a sample proportion

Use when: you know a population proportion p and a sample size n, and need either the shape of \hat p's sampling distribution or a probability involving \hat p

Goal: the normal approximation's mean and standard error for \hat p, and, if needed, a probability found by standardizing and reading \Phi

Taught in The sampling distribution of a proportion

1. 1

   Confirm np and n(1-p) are both comfortably large, a common rule of thumb being at least 10 each, before treating \hat p's sampling distribution as approximately normal.

   Example

   p=0.25, n=150: np=37.5 and n(1-p)=112.5, both well above 10.

   Why: the normal approximation is exactly that, an approximation, and it only holds up once the underlying binomial count is large enough to look bell-shaped.

2. 2

   The mean of the sampling distribution is p itself; its standard error is \sqrt{p(1-p)/n}. Compute both, using the same n just checked in Step 1.

   Example

   Mean: 0.25. SE = \sqrt{0.25(0.75)/150} = \sqrt{0.00125} \approx 0.03536.

   Why: this is the normal approximation this section derives, \hat p \approx \operatorname{Normal}(p,\ \sqrt{p(1-p)/n}), and every later step depends on having these two numbers right.

3. 3

   Check what the question is actually asking for before doing any more arithmetic.

   Example

   c=0.30: z = (0.30-0.25)/0.03536 \approx 1.414.

   • the question only asks for the sampling distribution itself (its shape or parameters) — stop here, Step 2's mean and SE ARE the answer.
   • the question asks for a probability involving \hat p, such as P(\hat p>c) for some cutoff c — subtract the Step 2 mean from c and divide by the Step 2 SE to get a z-score, then continue to Step 4.

   Why: this is the same standardization the course already taught for any normal distribution. Subtract the mean, then divide by the standard deviation of the CORRECT distribution: here, that's \hat p's own sampling distribution, rather than a single observation's.

4. 4

   Read \Phi(z) from the standard normal distribution, then combine it with the direction the question actually asks for: P(\hat p>c)=1-\Phi(z), or P(\hat p<c)=\Phi(z).

   Example

   \Phi(1.414) \approx 0.9214, so P(\hat p>0.30) \approx 1-0.9214 = 0.0786, about 7.9\%.

   Why: \Phi always reports the probability of falling BELOW the z-score; matching that to "above" or "below" the actual question is what a reader most often gets backward.