The necessity of t-distributions in computing confidence intervals If the width of a 95% CI can be computed as 1.96 * SEM, where SEM is just SD/sqrt(N), then why is the t-distribution involved in computing the CIs on the means of e.g. two groups?
 A: The width of a confidence interval can be calculated using the magic number 1.96 ONLY when the SEM is calculated from the true SD of the population, $\sigma$. In almost all real-word situations we do not know $\sigma$ and so we estimate it with $s$, the SD calculated from the sample values. The use of an estimated value in place of a true value increases the degree of uncertainty, and that has to be reflected in the width of the confidence interval.
Student's $t$-distribution was devised by Gossett ('Student') to deal with the overall uncertainty that comes from estimation of the mean and the SD.  If you use the relevant value from the $t$-distribution in place of 1.96 (which is a critical value from the $z$-distribution) then you get an interval that is wide enough to cover the true population mean, $\mu$, the desired fraction of occasions (presumably 95% for you).
As the sample size increases, the calculated value of $s$ becomes more reliably close to $\sigma$, and so there is an asymptotic approach of the $t$-distribution towards the $z$-distribution. In other words, as the sample gets larger the critical value of $t$ approaches the critical value of $z$ (i.e. the 1.96 magic number for a 95% interval). If your sample is as large as, say, 50 then the difference between the critical $t$ value and the critical $z$ value is small enough to be negligible for most purposes.
