Sampling distribution of p-hat practice problems

The sampling distribution of $\hat{p}$ is centered at the population proportion, so $\mu_{\hat{p}} = p = 0.40$.

Its standard deviation is $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$, where $\sigma_{\hat{p}}$ (say "sigma p-hat") measures how much sample proportions vary from sample to sample.

Compute the product inside the square root: $p(1-p) = 0.40 \times 0.60 = 0.24$.

Divide by the sample size: $\frac{0.24}{150} = 0.0016$.

Take the square root: $\sqrt{0.0016} = 0.04$.

For part (a), the 10% condition says the population must be at least 10 times the sample when you sample without replacement: check $10n \leq N$.

Compute $10n = 10 \times 64 = 640$, and compare to $N = 900$. Since $640 \leq 900$, the 10% condition is met.

The large counts condition requires at least 10 expected successes and 10 expected failures: $np \geq 10$ and $n(1-p) \geq 10$.

Expected successes: $np = 64 \times 0.20 = 12.8$, and $12.8 \geq 10$. Expected failures: $n(1-p) = 64 \times 0.80 = 51.2$, and $51.2 \geq 10$. Both hold, so the sampling distribution of $\hat{p}$ is approximately normal.

For part (b), the center is $\mu_{\hat{p}} = p = 0.20$.

For the spread, $p(1-p) = 0.20 \times 0.80 = 0.16$, then $\frac{0.16}{64} = 0.0025$, and $\sqrt{0.0025} = 0.05$.

Compute the piece both parts share: $p(1-p) = 0.36 \times 0.64 = 0.2304$.

For part (a) with $n = 100$: $\frac{0.2304}{100} = 0.002304$, and $\sqrt{0.002304} = 0.048$.

For part (b) with $n = 400$: $\frac{0.2304}{400} = 0.000576$, and $\sqrt{0.000576} = 0.024$.

For part (c), compare: $\frac{0.024}{0.048} = 0.5$, so multiplying the sample size by 4 cut the standard deviation exactly in half.

The reason is the square root in $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$. Multiplying $n$ by 4 divides $\sigma_{\hat{p}}$ by $\sqrt{4} = 2$, so precision improves more slowly than sample size grows.

For part (d), to divide $\sigma_{\hat{p}}$ by 3 you need $\sqrt{n}$ to grow by a factor of 3, so multiply $n$ by $3^2 = 9$: $n = 100 \times 9 = 900$.

Confirm: $\frac{0.2304}{900} = 0.000256$, and $\sqrt{0.000256} = 0.016$, which is exactly $\frac{0.048}{3}$.

Check the randomness condition: the sample is a simple random sample, so it is satisfied.

Check the 10% condition: $10n = 10 \times 300 = 3000$, and $3000 \leq 9000$, so the condition holds.

Check the large counts condition: $np = 300 \times 0.25 = 75 \geq 10$ and $n(1-p) = 300 \times 0.75 = 225 \geq 10$. The sampling distribution of $\hat{p}$ (p-hat) is approximately normal.

Find the center: $\mu_{\hat{p}} = p = 0.25$.

Find the spread: $p(1-p) = 0.25 \times 0.75 = 0.1875$, then $\frac{0.1875}{300} = 0.000625$, and $\sqrt{0.000625} = 0.025$.

Standardize $\hat{p} = 0.28$ to a z-score, which counts standard deviations from the center: $z = \frac{0.28 - 0.25}{0.025} = \frac{0.03}{0.025} = 1.20$.

Take the upper-tail area: $P(Z > 1.20) = 1 - P(Z < 1.20) = 1 - 0.8849 = 0.1151$.

For part (a), the center is $\mu_{\hat{p}} = p = 0.03$.

For the spread: $p(1-p) = 0.03 \times 0.97 = 0.0291$, then $\frac{0.0291}{200} = 0.0001455$, and $\sqrt{0.0001455} \approx 0.0121$.

For part (b), the expected number of successes is $np = 200 \times 0.03 = 6$, and the expected number of failures is $n(1-p) = 200 \times 0.97 = 194$.

Since $6 < 10$, the large counts condition fails, even though the failure count of 194 is far above 10. Both parts have to pass, not just one.

For part (c), the z-score method assumes the sampling distribution of $\hat{p}$ is approximately normal, and that assumption is exactly what the failed condition denies. Any area read from a z-table here would be unreliable, so the calculation is not justified.

The real shape is strongly right-skewed. With $p = 0.03$ the distribution is pressed up against the boundary at $\hat{p} = 0$ (a sample can contain no infected pines at all), while a handful of infected pines pulls a long tail to the right.

For part (d), solve $np \geq 10$ for $n$: $n \geq \frac{10}{0.03} \approx 333.33$, so $n = 334$ is the first whole number that works. Check it: $334 \times 0.03 = 10.02 \geq 10$, while $333 \times 0.03 = 9.99$ falls short.

Confirm the other half of the condition at that size: $334 \times 0.97 = 323.98 \geq 10$, so $n = 334$ satisfies both parts.

For part (a), write the distribution of the single-visitor variable: $P(X = 1) = 0.20$ and $P(X = 0) = 0.80$. Its mean is $\mu_X = 1(0.20) + 0(0.80) = 0.20$.

Apply the discrete random variable formula $\sigma_X = \sqrt{\sum (x_i - \mu_X)^2 P(x_i)}$. The squared deviations are $(0 - 0.20)^2 = 0.04$ and $(1 - 0.20)^2 = 0.64$.

Weight and add: $0.04(0.80) = 0.032$ and $0.64(0.20) = 0.128$, so the variance is $0.032 + 0.128 = 0.16$, and $\sigma_X = \sqrt{0.16} = 0.4$.

For part (b), $p(1-p) = 0.20 \times 0.80 = 0.16$, then $\frac{0.16}{400} = 0.0004$, and $\sigma_{\hat{p}} = \sqrt{0.0004} = 0.02$.

Notice the link between them: $\sigma_{\hat{p}} = \frac{\sigma_X}{\sqrt{n}} = \frac{0.4}{\sqrt{400}} = \frac{0.4}{20} = 0.02$, which is the same answer by a different route.

For part (c), the ranger has swapped the two. The population standard deviation $0.4$ describes variation among individual visitors, who each score 0 or 1. The value $0.02$ describes variation among sample proportions: if the ranger repeated the survey of 400 visitors many times, those $\hat{p}$ values would typically sit about $0.02$ away from $0.20$.

The sampling distribution is 20 times tighter than the population here, because averaging 400 individual 0s and 1s smooths out the individual variation.

For part (a), start with randomness: the families were chosen by a simple random sample, so that condition is met.

10% condition: $10n = 10 \times 600 = 6000$, and $6000 \leq 25000$, so the sample is small enough relative to the district.

Large counts condition: $np = 600 \times 0.60 = 360 \geq 10$ and $n(1-p) = 600 \times 0.40 = 240 \geq 10$. All three conditions hold, so the sampling distribution of $\hat{p}$ (p-hat) is approximately normal.

For part (b), $\mu_{\hat{p}} = p = 0.60$. For the spread: $p(1-p) = 0.60 \times 0.40 = 0.24$, then $\frac{0.24}{600} = 0.0004$, and $\sqrt{0.0004} = 0.02$.

For part (c), standardize $0.63$: $z = \frac{0.63 - 0.60}{0.02} = \frac{0.03}{0.02} = 1.50$.

Read the upper tail: $P(Z \geq 1.50) = 1 - P(Z < 1.50) = 1 - 0.9332 = 0.0668$.

For part (d), state the result as a long-run proportion of samples: if the district's claim of $p = 0.60$ is right, about 6.68% of random samples of 600 families would produce a sample proportion of $0.63$ or higher.

For part (a), the sample is a simple random sample. The 10% condition holds because $10n = 10 \times 1200 = 12000$ and $12000 \leq 40000$.

Large counts: $np = 1200 \times 0.75 = 900 \geq 10$ and $n(1-p) = 1200 \times 0.25 = 300 \geq 10$, so a normal model is justified.

Spread: $p(1-p) = 0.75 \times 0.25 = 0.1875$, then $\frac{0.1875}{1200} = 0.00015625$, and $\sigma_{\hat{p}} = \sqrt{0.00015625} = 0.0125$.

For part (b), standardize the observed value: $z = \frac{0.72 - 0.75}{0.0125} = \frac{-0.03}{0.0125} = -2.40$.

Read the lower tail: $P(Z \leq -2.40) = 0.0082$.

For part (c), interpret that number. If the city's claim held, only about 0.82% of random samples of 1,200 bins would come in at 72% clean or lower. A result that rare is convincing evidence that the true proportion is below $0.75$.

For part (d), recompute the spread at $n = 75$: $\frac{0.1875}{75} = 0.0025$, and $\sqrt{0.0025} = 0.05$. Check the conditions still pass: $75 \times 0.75 = 56.25 \geq 10$ and $75 \times 0.25 = 18.75 \geq 10$.

Standardize again: $z = \frac{0.72 - 0.75}{0.05} = \frac{-0.03}{0.05} = -0.60$, and $P(Z \leq -0.60) = 0.2743$.

Compare the two. The sample proportion is identical, but at $n = 75$ roughly 27% of samples would fall this low by chance alone, so that audit proves nothing. Shrinking the sample from 1,200 to 75 divides $n$ by 16, which multiplies $\sigma_{\hat{p}}$ by $\sqrt{16} = 4$: $0.0125 \times 4 = 0.05$.