Question

What is the relationship between Sample Size and the width of the Confidence Interval? Please explain further as to why. & Discuss the effect that the confidence level has upon the confidence interval.

Answer 1

A confidence interval with \((1-\alpha)\times100\%\) confidence level for an estimator - for instance, one estimating for the population mean - would have the following form:
\[\left(\bar{x}-Z_{\alpha/2}\frac{\sigma}{\sqrt n},~\bar{x}+Z_{\alpha/2}\frac{\sigma}{\sqrt n}\right)\]
where \(n\) is the sample size. As the sample size increases, the term after \(\bar{x}\) approaches 0, so the CI would eventually take on a point, \(\bar{x}\). This would mean \(\bar{x}\), the sample mean, perfectly estimates the population mean. Inversely, decreasing the sample size would widen the confidence interval.

The confidence level affects the value of \(Z_{\alpha/2}\). Assume you have a standard normal distribution which has the following general plot (ignore the vertical axis):

Basically, \(\alpha\) is the area under the curve outside the confidence interval. As \(\alpha\) gets smaller and smaller to approach 0 (i.e. the confidence level approaches 100%), \(Z_{\alpha/2}\) approaches \(Z_0=1\). With regard to the confidence interval, the radius of the interval would depend only on the standard error, \(\dfrac{\sigma}{\sqrt n}\).

On the other hand, if \(\alpha\) increases and approaches 1, then \(Z_{\alpha/2}\) approaches 0. Thus a 0% confidence interval would be \((\bar{x}-0,\bar{x}+0)\), or just the point \(\bar{x}\).