1. Mean of the sampling distribution: $\mu_{\hat{p}} = p$ The sample proportion is unbiased for the population proportion.

2. Standard error of $\hat{p}$: $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$ Larger $n$ and values of $p$ closer to 0 or 1 yield smaller standard errors.

3. Shape of the sampling distribution: When the Large Counts condition is met, $\hat{p}$ is approximately Normal:

   • $np \geq 10$ AND $n(1-p) \geq 10$

Large Counts Condition

The Large Counts condition ensures that the binomial distribution is sufficiently symmetric and bell-shaped to approximate normality:

• Count of successes: $np \geq 10$
• Count of failures: $n(1-p) \geq 10$

If either count falls below 10, use exact binomial probabilities instead of the normal approximation.

Standard Error and Sample Size

Just as with $\bar{x}$, the standard error of $\hat{p}$ decreases with the square root of $n$:

$\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \propto \frac{1}{\sqrt{n}}$

To cut the standard error in half, multiply $n$ by 4.

Effect of $p$ on Variability

The term $p(1-p)$ is maximized when $p = 0.5$. This means:

• Proportions near 0.5 have maximum variability.
• Proportions near 0 or 1 (rare events) have less variability.
• For $p = 0.1$, $p(1-p) = 0.09$; for $p = 0.5$, $p(1-p) = 0.25$ (2.78 times larger).

Worked Example 1: Checking Large Counts Condition

In a population, 30% of voters support a candidate. A sample of $n = 80$ is drawn. Is the Large Counts condition met?

Check:

• $np = 80 \times 0.3 = 24 \geq 10$ ✓
• $n(1-p) = 80 \times 0.7 = 56 \geq 10$ ✓

Both counts exceed 10, so the normal approximation applies.

Sampling distribution of $\hat{p}$:

• $\mu_{\hat{p}} = 0.3$
• $\sigma_{\hat{p}} = \sqrt{\frac{0.3 \times 0.7}{80}} = \sqrt{\frac{0.21}{80}} = \sqrt{0.002625} \approx 0.0512$
• $\hat{p} \sim N(0.3, 0.0512)$ approximately

Find $P(\hat{p} > 0.35)$: $Z = \frac{0.35 - 0.3}{0.0512} \approx 0.98$ $P(Z > 0.98) \approx 0.164$

About 16.4% of samples have a proportion above 0.35.

Worked Example 2: Small Proportion (Rare Event)

In a population, 5% of items are defective. A sample of $n = 150$ is drawn. Check normality and find $P(\hat{p} \leq 0.02)$.

Check Large Counts:

• $np = 150 \times 0.05 = 7.5 < 10$ ✗
• $n(1-p) = 150 \times 0.95 = 142.5 \geq 10$ ✓

The condition is NOT met (insufficient defectives). Use binomial distribution, not normal approximation.

Correct approach: Count $X =$ number of defectives $\sim \text{Binomial}(n=150, p=0.05)$.

$P(\hat{p} \leq 0.02) = P(X \leq 3)$ (since $\hat{p} = 3/150 = 0.02$).

Using binomial: $P(X \leq 3) \approx 0.094$ (exact calculation).

Conditions and Assumptions

Common Pitfalls

⚠️ Forgetting to Check Large Counts: Never use the normal approximation without verifying both $np \geq 10$ and $n(1-p) \geq 10$. If either fails, use exact binomial probabilities or report that the approximation is invalid.

⚠️ Confusing $p$ and $\hat{p}$: $p$ is the population parameter (unknown); $\hat{p}$ is the sample statistic (observed). The standard error formula uses the parameter $p$, not the sample proportion.

⚠️ Mixing Up Notation: Do not confuse $\sigma_{\hat{p}}$ with $\sigma_p$ or other notations. Always use consistent terminology.

Calculator Tip

💡 TI-84 / TI-Nspire: For small samples or when Large Counts fails, use binompdf() and binomcdf() for exact probabilities. For large samples meeting the condition, use normalcdf() with mean $p$ and SD $\sqrt{p(1-p)/n}$.

If these conditions hold, $\hat{p} \sim N(p, \sqrt{p(1-p)/n})$ approximately.

Check Large Counts:

• $np = 200 \times 0.50 = 100 \geq 10$ ✓
• $n(1-p) = 200 \times 0.50 = 100 \geq 10$ ✓

Standardize: $Z = \frac{0.52 - 0.50}{0.0354} \approx 0.565$

Find $P(Z \geq 0.565)$: $P(Z \geq 0.565) \approx 1 - 0.714 = 0.286$

About 28.6% probability of observing $\hat{p} \geq 0.52$ if $p = 0.50$ is true.

Square both sides: $0.0004 = \frac{0.0736}{n}$

Solve for $n$: $n = \frac{0.0736}{0.0004} = 184$

Verify Large Counts (with $n = 184$):

• $np = 184 \times 0.08 = 14.72 \geq 10$ ✓
• $n(1-p) = 184 \times 0.92 = 169.28 \geq 10$ ✓

Answer: $n = 184$ (or approximately 184–190, accounting for rounding in practice).