Width of a confidence interval

The width of a confidence interval is its upper bound minus its lower bound. Every interval on this course has the form statistic plus or minus a margin of error, so the width is exactly $2 \times ME = 2 \times (\text{critical value}) \times (\text{standard error})$. Because the interval is symmetric about the statistic, the width and the margin of error carry the same information in different units, and either one recovers the other.

The interval 12 to 18 has width $18 - 12 = 6$, so its margin of error is $6/2 = 3$ and its center, the point estimate, is $(12 + 18)/2 = 15$. Given only the endpoints you can always rebuild both.

Sample size enters through a square root, since every standard error on the course divides by $\sqrt{n}$. Holding the statistic and the confidence level fixed, the width is proportional to $1/\sqrt{n}$. So "I doubled my sample, so the interval should be half as wide" is wrong. Doubling $n$ multiplies the width by $1/\sqrt{2} = 0.7071$, so a width of 6 becomes 4.243, about 29% narrower, and not the 3 that guess predicts: the real width is 41% larger than the guess. Halving the width takes four times the data, since $1/\sqrt{4} = 0.5$ exactly, which does give 3.

The confidence level is the other multiplier, and it acts on the critical value directly rather than through a square root. At a fixed $n$, moving from 95% to 99% multiplies the width by $2.576/1.960 = 1.314$, about 31% wider, and moving down to 90% multiplies it by $1.645/1.960 = 0.839$. That is the whole trade: confidence is bought with width, and width is bought with data.

One caution about comparing widths across two real samples. The $1/\sqrt{n}$ rule holds with the statistic held fixed, and a fresh sample gives a fresh $\hat{p}$ or $\bar{x}$, so the observed width will only be near the predicted one. And a narrow interval is a claim about precision alone, never about whether the center is in the right place.