AP Statistics 3.3-3.4: Confidence Intervals for Proportions

Confidence interval for a population mean with known population standard deviation

An interval estimate for an unknown population mean $\mu$ when the population standard deviation $\sigma$ is known. It is calculated as $\bar{x} \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$.

Point estimate for a population mean

The sample mean $\bar{x}$ is the point estimate of the unknown population mean $\mu$.

Error bound for a population mean (EBM)

The margin of error for estimating a population mean when $\sigma$ is known is $\text{EBM}=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$. It determines how far the confidence interval extends on either side of $\bar{x}$.

Standard error of the sample mean when $\sigma$ is known

The standard deviation of the sampling distribution of $\bar{x}$ is $\frac{\sigma}{\sqrt{n}}$, called the standard error of the mean.

Confidence interval structure for a single population mean

The interval is $(\bar{x}-\text{EBM},\,\bar{x}+\text{EBM})$, or equivalently $\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$.

How does the Central Limit Theorem support a z confidence interval for a population mean?

For a random sample, the sampling distribution of $\bar{x}$ is approximately normal, especially for sufficiently large $n$. Its mean is $\mu$ and its standard deviation is $\sigma/\sqrt{n}$.

Sampling distribution of the sample mean when the population is normal

If the population is normally distributed, then $\bar{X}$ is normally distributed for any sample size: $\bar{X}\sim N\left(\mu,\frac{\sigma}{\sqrt{n}}\right)$.

When is a normal-based confidence interval for a population mean appropriate?

The data should come from a random sample, the population standard deviation must be known for the z procedure, and the population should be normal or the sample size should be large enough for the Central Limit Theorem to apply.

Confidence level ($CL$)

The confidence level is the long-run proportion of intervals produced by the procedure that contain the true population parameter. For example, a 95% procedure captures the true mean in approximately 95% of repeated samples.

Why is it incorrect to say that a particular 95% confidence interval has a 95% probability of containing $\mu$?

In frequentist statistics, $\mu$ is fixed and the interval endpoints vary from sample to sample. The 95% describes the long-run success rate of the interval-generating procedure, not a probability assigned to one already-calculated interval.

Relationship among confidence level, $\alpha$, and the two tails

The significance level is $\alpha=1-CL$. The total probability outside the central confidence region is $\alpha$, split equally between the tails, so each tail has area $\alpha/2$.

Critical value $z_{\alpha/2}$

The value $z_{\alpha/2}$ is the positive standard-normal z-score with area $\alpha/2$ to its right and $1-\alpha/2$ to its left. It leaves the central area $CL=1-\alpha$ between $-z_{\alpha/2}$ and $z_{\alpha/2}$.

How is a z critical value found for a two-sided confidence interval?

First calculate $\alpha=1-CL$, then use the cumulative area to the left of the positive critical value: $1-\alpha/2$. For example, a 95% interval uses $\text{invNorm}(0.975,0,1)$, giving $z_{0.025}=1.96$.

Common standard-normal critical values for confidence intervals

For 90%, $z_{\alpha/2}=1.645$; for 95%, $z_{\alpha/2}=1.96$; and for 98%, $z_{\alpha/2}\approx2.326$.

How do you construct and interpret a z confidence interval for a population mean?

Calculate $\bar{x}$, determine $z_{\alpha/2}$, calculate $\text{EBM}=z_{\alpha/2}\sigma/\sqrt{n}$, and form $\bar{x}\pm\text{EBM}$. Interpret it in context by stating the confidence level, population mean, interval endpoints, and units.

A sample has $\bar{x}=7$ and $\text{EBM}=2.5$. What is the confidence interval?

The interval is $7\pm2.5$, giving $(4.5,9.5)$.

A population has known $\sigma=3$, a random sample has $n=36$ and $\bar{x}=68$. Find the 90% confidence interval.

Using $z_{0.05}=1.645$, $\text{EBM}=1.645(3/\sqrt{36})=0.8225$. The interval is $68\pm0.8225=(67.1775,68.8225)$, approximately $(67.178,68.822)$.

How should a 90% confidence interval for a population mean be stated in context?

A suitable statement is: “We estimate with 90% confidence that the true population mean is between the two reported endpoints,” including the relevant population and units.

A population has known $\sigma=0.337$, with $n=30$ and $\bar{x}=1.024$. Find the 98% confidence interval.

Using $z_{0.01}=2.326$, $\text{EBM}=2.326(0.337/\sqrt{30})\approx0.1431$. The interval is $(1.024-0.1431,1.024+0.1431)=(0.8809,1.1671)$.

What happens to the error bound and interval width when the confidence level increases?

The critical value increases, so the error bound increases and the confidence interval becomes wider. This greater width reflects the stronger requirement for long-run coverage.

What happens to the error bound and interval width when the sample size increases?

Because $\text{EBM}$ is proportional to $1/\sqrt{n}$, increasing $n$ decreases the error bound and produces a narrower interval.

How does reducing the sample size affect a known-$\sigma$ confidence interval?

Reducing $n$ increases $\sigma/\sqrt{n}$, which increases the error bound and makes the interval wider.

Why is the 95% confidence interval wider than the 90% interval when $\bar{x}$, $\sigma$, and $n$ are unchanged?

The 95% interval uses a larger critical value, $1.96$ instead of $1.645$. Therefore its margin of error is larger.

How can the error bound be recovered from a confidence interval $(L,U)$?

For a symmetric interval, $\text{EBM}=\frac{U-L}{2}$. If $\bar{x}$ is known, it can also be found as $U-\bar{x}$ or $\bar{x}-L$.

How can the sample mean be recovered from a confidence interval $(L,U)$?

The sample mean is the midpoint: $\bar{x}=\frac{L+U}{2}$. If the error bound is known, it can also be calculated as $U-\text{EBM}$ or $L+\text{EBM}$.

A confidence interval is $(42.12,47.88)$. What are the sample mean and error bound?

The sample mean is $\bar{x}=(42.12+47.88)/2=45.00$. The error bound is $(47.88-42.12)/2=2.88$.

Required sample size for estimating a population mean with known $\sigma$

To achieve a desired error bound $E$ at a specified confidence level, use $n=\left(\frac{z_{\alpha/2}\sigma}{E}\right)^2=\frac{z_{\alpha/2}^2\sigma^2}{E^2}$.

Why must a calculated sample size for a confidence interval be rounded up?

The sample size must be large enough to keep the actual error bound at or below the target. Rounding down could make the margin of error too large.

A population has $\sigma=15$. What sample size is needed for 95% confidence with an error bound of at most 2?

Using $z_{0.025}=1.96$, $n=(1.96\cdot15/2)^2\approx216.09$. Round up to require $n=217$.

How does the required sample size change when the desired margin of error is cut in half?

Because $n$ is proportional to $1/E^2$, cutting the error bound in half requires four times the sample size, assuming $\sigma$ and the confidence level stay fixed.

How do confidence intervals from different samples or confidence levels compare?

They may have different centers and widths because the samples, sample sizes, and confidence levels differ. Different intervals do not necessarily provide conflicting information about the same population parameter.

In a frequentist confidence interval, which quantity is fixed and which quantities vary from sample to sample?

The true population mean $\mu$ is treated as a fixed but unknown constant. The sample mean and the interval endpoints are random because they depend on the random sample.

What does a confidence interval estimate that a point estimate alone does not?

A point estimate such as $\bar{x}$ gives one best estimate, whereas a confidence interval also communicates sampling uncertainty by giving a plausible range generated by a specified procedure.