Why is a 99% confidence interval wider than a 95% one built from the same data?

## The three moving parts Almost every classical interval has the same shape: ``` point estimate ± (critical value) × (standard error) ``` The **point estimate** is your best single guess at the unknown quantity — a sample mean, a sample proportion. The **standard error** measures how much that estimate would bounce around from sample to sample; for a mean it is `s / sqrt(n)`. The **critical value** is the only piece the confidence level touches. Choosing 90%, 95% or 99% changes that one multiplier and nothing else. ## Where the ladder comes from A two-sided interval at level C leaves `(1 - C) / 2` of the reference distribution in each tail. The critical value is the point on that reference distribution cutting off that tail. On a standard normal reference: - 90% leaves 5% in each tail, giving **1.645** - 95% leaves 2.5% in each tail, giving **1.96** - 99% leaves 0.5% in each tail, giving **2.576** Those three numbers are worth memorising; interviewers do check them. Note they are *two-sided* values. The corresponding *one-sided* bounds — 1.282, 1.645 and 2.326 — cut off the whole tail probability on one side, and mixing the two families up is one of the most common arithmetic slips in this material. The value 1.645 appearing in both lists is exactly why the mix-up happens: it is the two-sided 90% value and the one-sided 95% value. ## What the extra width actually costs Because only the multiplier changes, width ratios are just ratios of critical values: - 95% to 99%: `2.576 / 1.96 ≈ 1.31` — about 31% wider - 90% to 95%: `1.96 / 1.645 ≈ 1.19` — about 19% wider - 90% to 99%: `2.576 / 1.645 ≈ 1.57` — about 57% wider So the cost of confidence is mild in the middle of the ladder and steep at the top. Pushing to 99.9% needs about 3.29, roughly 68% wider than 95%. The reverse trade is the one that shows up in planning conversations. If you are required to report a 99% interval but you want the precision a 95% interval gave you, you must shrink the standard error to compensate. Since the standard error of a mean carries `sqrt(n)` in the denominator, you need the sample size to grow by the *square* of the critical-value ratio: `(2.576 / 1.96)^2 ≈ 1.73`, about 73% more data. Going from 95% to 90% instead buys you back a 30% reduction in required sample size for the same width. ## The thing candidates get backwards A wider interval is not a *better* result. Confidence level and precision pull against each other: at fixed data, every point of confidence you add is paid for with vagueness. In the limit, an interval spanning the entire range of the parameter is 100% confident and says nothing at all. The reason 95% is the default in most fields is convention, not optimality — it is a workable compromise, and the number itself is arbitrary. Equally important: raising the confidence level does not make the *estimate* more accurate. The centre of the interval is unchanged, and any bias in how the data were collected is unchanged too. A 99% interval built from a badly sampled population is a wider statement of the same wrong thing. ## Choosing a level in practice A higher level is defensible when being wrong is expensive and irreversible — safety limits, regulatory claims, anything you cannot walk back. A lower level is defensible when you are exploring and will follow up anyway, and the cost of an occasional miss is another round of work. What is *not* defensible is picking the level after seeing which one gives the answer you wanted; the level belongs to the analysis plan, not to the results. One last mechanical note: this all assumes a symmetric two-sided interval on a symmetric reference distribution. Once the reference is a t distribution the ladder shifts upward (t values exceed the matching normal values), and for bounded quantities like proportions the good intervals are not symmetric around the estimate at all — but the principle that confidence buys width, and width is paid for with data, survives intact.