What "95 % confident" means

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The 95 % belongs to the method, not to μ: in repeated sampling 95 % of intervals built this way would contain μ. Wider for more confidence or a larger σ; narrower for a larger n; a claimed value outside the interval is not plausible.

Two sentences the marker accepts, one it does not

Kitchener Ridge's 95 % interval for mean days-to-pay is $(44.24, 48.16)$ . What does the 95 % refer to?

Not to $μ$ . The population mean is a fixed number — unknown, but not random. It either lies in $(44.24, 48.16)$ or it does not; there is no probability about it. What is random is the interval: a different sample gives a different $\overset{x}{ˉ}$ and a different pair of limits. The 95 % describes the method: of all the intervals that could be built this way, one sample at a time, 95 % would contain $μ$ and 5 % would miss it. We do not know which kind ours is.

So the two acceptable sentences are:

The client's sentence. We are 95 % confident that the mean days-to-pay of all current invoices is between 44.24 and 48.16 days.
The statistician's sentence. If we drew many samples of 64 and built an interval from each, about 95 % of those intervals would contain the true mean.

And the boxed wrong one: there is a 95 % probability that $μ$ lies in $(44.24, 48.16)$ . It puts the probability on $μ$ , which has none. A related wrong one: 95 % of invoices take between 44.24 and 48.16 days. The interval is about the mean, not about individual invoices — most invoices lie well outside it, since $σ = 8$ .

What moves the width

The margin is $z_{α /2} \times σ / n$ , so three things drive it:

Confidence level. At 90 % the critical value is $z_{0.05} = 1.645$ and the margin falls to $1.645 \times 1 = 1.645$ : $(44.56, 47.85)$ . Less confidence, narrower interval — you can be surer of a wider range or less sure of a narrower one, never both.
Sample size. With $n = 16$ instead of 64 the SE doubles to $8/ 16 = 2$ and the 95 % margin doubles to 3.92. Quadrupling $n$ halves the width (L8.2's square root again). This is the only lever the analyst controls without giving up confidence — hence Module 9.6.
Population spread. A larger $σ$ widens everything; nothing to be done about it.

The sample mean itself is not a driver: a larger $\overset{x}{ˉ}$ shifts the interval but leaves its width unchanged.

Using the interval to judge a claim

A consultant tells Kitchener Ridge that its mean days-to-pay is 50. Is that consistent with the data? 50 lies outside $(44.24, 48.16)$ , so at the 95 % level 50 is not a plausible value for $μ$ — the sample would be an unusually low one if the mean were really 50. Had the claim been 47, inside the interval, it would be plausible — not proven, merely not contradicted.

This is a hypothesis test in disguise: "is $μ_{0}$ inside the interval?" is exactly the two-sided test of Module 10, and L10.7 makes the equivalence explicit. For now, the paper wants the three-part answer: where the claim sits, what that says about plausibility, and the confidence level at which you say it.