Sampling Distribution of p-hat
Введено| English | Значение |
|---|---|
| sample proportion/ˈsæmpl prəˈpɔːʃn/ | доля выборки |
| large counts condition/lɑːdʒ kaʊnts kənˈdɪʃn/ | условие больших чисел |
| 10% condition/ten pəˈsent kənˈdɪʃn/ | 10% условие |
The distribution of p-hat
- A sample proportion 样本比例 $\hat{p}$ estimates the true population proportion $p$.
- Its sampling distribution is centered at $p$: $\mu_{\hat{p}} = p$ (so $\hat{p}$ is unbiased).
- Different samples give different $\hat{p}$ — the spread is what we quantify next.
- Under the right conditions, this distribution is approximately normal.
Its standard deviation
- The standard deviation of $\hat{p}$ is:
-
$$\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$$
- Bigger $n$ → smaller spread → a more precise estimate (the $\sqrt{n}$ in the bottom).
- This measures the typical distance of $\hat{p}$ from the true $p$.

The sampling distribution of p̂ is centred at p with standard deviation √(p(1−p)/n)
The large counts condition
- To use a normal model for $\hat{p}$, check the large counts condition 大计数条件:
-
$$np \ge 10 \quad\text{and}\quad n(1-p) \ge 10$$
- In words: expect at least $10$ successes and at least $10$ failures.
- If either count is under $10$, the sampling distribution is too skewed for the normal model.
The 10% condition
- When sampling without replacement, check the 10% condition 10%条件.
- The sample must be no more than $10\%$ of the population: $n \le 0.10 N$.
- This keeps the trials "close enough" to independent for the SD formula to hold.
- A small slice of a big population barely changes as you draw.
Two different conditions do two different jobs. The large counts condition ($np\ge10$, $n(1-p)\ge10$) justifies the normal shape; the 10% condition ($n\le0.10N$) justifies the independence behind the SD formula. Check both before modeling $\hat{p}$ as normal — skipping either invalidates the later inference.
$p = 0.4$ of a population supports a plan; you sample $n = 100$.
- Center: $\mu_{\hat{p}} = 0.4$. SD: $\sqrt{\dfrac{0.4 \times 0.6}{100}} = \sqrt{0.0024} \approx 0.049$.
- Large counts: $np = 40 \ge 10$ and $n(1-p) = 60 \ge 10$. ✓ Normal is OK.
- 10%: fine as long as the population exceeds $1000$.
The sample proportion $\hat{p}$ has $\mu_{\hat{p}} = p$ and $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$. Model it as normal when the large counts condition ($np\ge10$, $n(1-p)\ge10$) holds; when sampling without replacement, also check the 10% condition ($n \le 0.10N$).
Write from a standard deviation of 0.04
- The population proportion is 0.2.
- The sample size is 100. The lesson standard deviation 0.05 is another sample size or proportion.
What is the standard deviation of the sample proportion?
The standard deviation shrinks as the sample grows.
What is the standard deviation of p-hat?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The square root of 0.0016 is 0.04. Other options — 0.05 (the other setup); 0.2 (the population proportion).
The sampling distribution of p-hat
Centered at p, spread √(p(1−p)/n); normal under the large counts condition.
p = 0.4, n = 100. Find σ of p-hat = √(p(1−p)/n). Round to two decimals.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
√(0.4·0.6/100) = √0.0024 ≈ 0.049 ≈ 0.05.
The large counts condition for a normal model of p-hat requires...
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
At least 10 expected successes AND 10 expected failures.
The mean of the sampling distribution of p-hat equals the true proportion p.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
μ(p-hat) = p, so p-hat is unbiased.
When sampling without replacement, the 10% condition requires...
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The sample must be at most 10% of the population.
Increasing n makes the standard deviation of p-hat smaller.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
n is under the square root, so larger n → smaller spread.