In statistical inference, we use sample data to make estimates about unknown population parameters. While a point estimate (such as the sample mean $\bar{x}$) provides a single value, a confidence interval provides a range of values within which the true population parameter is expected to lie, with a certain level of confidence.
KEY TAKEAWAY: A confidence interval is an interval estimate. It provides more information than a point estimate because it accounts for the variability inherent in sampling.
For a population with a known standard deviation $\sigma$, or for a large sample where $\sigma$ is approximated by $s$, the $C\%$ confidence interval for the population mean $\mu$ is given by:
$$\left( \bar{x} - z \frac{\sigma}{\sqrt{n}}, \bar{x} + z \frac{\sigma}{\sqrt{n}} \right)$$
Where:
* $\bar{x}$ is the sample mean.
* $n$ is the sample size.
* $z$ (or $z^$) is the critical value* for the desired level of confidence.
The value of $z$ is determined such that the area under the standard normal curve $Z \sim N(0,1)$ between $-z$ and $z$ is equal to the confidence level $C$.
| Confidence Level | $C$ (decimal) | Critical Value ($z$) |
|---|---|---|
| 90% | 0.90 | 1.645 |
| 95% | 0.95 | 1.960 |
| 99% | 0.99 | 2.576 |
EXAM TIP: If an unusual confidence level is requested (e.g., 92%), use the CAS command
invNormto find the $z$-score. For a $C\%$ interval, the area to the left of the upper $z$-score is $C + \frac{1-C}{2}$.
The Margin of Error ($M$) is defined as:
$$M = z \frac{\sigma}{\sqrt{n}}$$
The total Width ($W$) of the confidence interval is:
$$W = 2M = 2z \frac{\sigma}{\sqrt{n}}$$
The precision of a confidence interval is determined by its width. A narrower interval is more precise.
| Action | Effect on Width | Reason |
|---|---|---|
| Increase Confidence Level | Increases Width | Larger $z$-value required to be more certain. |
| Increase Sample Size ($n$) | Decreases Width | Standard error $\frac{\sigma}{\sqrt{n}}$ decreases. |
| Increase Standard Deviation ($\sigma$) | Increases Width | More variability in the population leads to less precision. |
The width of a confidence interval is inversely proportional to the square root of the sample size:
$$W \propto \frac{1}{\sqrt{n}}$$
To decrease the width by a factor of $k$, the sample size must be increased by a factor of $k^2$.
COMMON MISTAKE: Students often think doubling the sample size halves the width. In reality, to halve the width ($k=2$), you must quadruple the sample size ($2^2 = 4$).
The interpretation of a confidence interval is a common source of theoretical questions in VCE exams.
VCAA FOCUS: VCAA frequently tests the definition of the confidence level. Remember: the interval is the random variable, not the population parameter $\mu$. The parameter $\mu$ is a fixed (though unknown) constant.
To determine the minimum sample size required to achieve a specific margin of error $M$ at a certain confidence level:
If a researcher wants to reduce the width of a 95% confidence interval to one-third of its original size:
* New Width = $\frac{1}{3} \times$ Old Width
* Since $W \propto \frac{1}{\sqrt{n}}$, we need $\frac{1}{\sqrt{n_{new}}} = \frac{1}{3\sqrt{n_{old}}}$
* $\sqrt{n_{new}} = 3\sqrt{n_{old}}$
* $n_{new} = 9 \times n_{old}$
* The sample size must be increased by a factor of 9.
STUDY HINT: Practice rearranging the width formula $W = \frac{2z\sigma}{\sqrt{n}}$ quickly. Many multiple-choice questions require finding the ratio between two sample sizes or two widths.
