Estimate ± critical × SE: the z-interval

From a probability statement to an interval

Module 8 ran in one direction: knowing $\mu$, it said where $\bar{x}$ would land. Inference runs the other way. The population mean is unknown — that is why a sample was drawn — and the sample mean is what we have. L8.3's result still holds: with $\bar{X}\sim N(\mu ,{\sigma }^2/n)$, 95 % of sample means lie within $1.96\times \sigma /\sqrt{n}$ of $\mu$. Turn the sentence around: in 95 % of samples, $\mu$ lies within $1.96\times \sigma /\sqrt{n}$ of $\bar{x}$. That distance either side of $\bar{x}$ is a confidence interval for $\mu$.

The grammar is the same for every interval in the course:

$$\text{estimate};\pm ;\text{critical value}\times \text{standard error}$$

For a mean with $\sigma$ known, the estimate is $\bar{x}$, the standard error is $\sigma /\sqrt{n}$, and the critical value is $z_{\alpha /2}$ — the $z$ that leaves $\alpha /2$ in each tail, where the confidence level is $1-\alpha$:

$$\bar{x}\pm z_{\alpha /2}\frac{\sigma }{\sqrt{n}}$$

The three memorised values: 90 % uses $z_{0.05}=1.645$, 95 % uses $z_{0.025}=1.96$, 99 % uses $z_{0.005}=2.576$. The product $z_{\alpha /2}\times \sigma /\sqrt{n}$ is the margin of error — the half-width of the interval. IBS1 calls it the EBM; the paper calls it the margin.

Worked example — Kitchener Ridge's days-to-pay

Kitchener Ridge Logistics has years of receivables history, so the SD of days-to-pay is taken as known: $\sigma =8$ days. A random sample of $n=64$ recent invoices has mean $\bar{x}=46.2$ days. A 95 % interval for the mean days-to-pay of all current invoices:

$$\mathrm{SE}=\frac{8}{\sqrt{64}}=1,\text{margin}=1.96\times 1=1.96$$

$$46.2\pm 1.96⟹(44.24,48.16)$$

At 99 %, only the critical value changes:

$$\text{margin}=2.576\times 1=2.576,(43.62,48.78)$$

More confidence costs width: the 99 % interval is 5.15 days wide against 3.92 for 95 %. Nothing about the data changed — $\bar{x}$, $\sigma$, $n$ and the SE are identical — only $z$ moved from 1.96 to 2.576. That is the answer to the paper's favourite follow-up, why is the 99 % interval wider?: a larger critical value, not a larger SE.

Conditions

The interval is built on $\bar{X}$ being normal, so Module 8's table applies: the sample is a simple random sample, and either the population is normal or $n>30$ so the CLT gives the shape. Kitchener Ridge's $n=64>30$ qualifies whatever the shape of days-to-pay. Write the condition on the paper — n = 64 > 30, so by the CLT — it is a mark.

The method also needs $\sigma$ known. A firm with a long history can claim that; most cannot, and then $s$ replaces $\sigma$ and $t$ replaces $z$ — L9.3 and L9.4. Which one you are in is the first thing to read off any question.

Reporting

Report the interval as a pair, in the units of the data, to two decimals: (44.24, 48.16) days. The margin alone (46.2 ± 1.96) is acceptable as a working line but the marker wants the two limits. A common slip is to report $\bar{x}\pm \mathrm{SE}$, dropping the critical value — a 68 % interval masquerading as 95 %.