Self-study question on Confidence Interval Good day everyone, I was making an attempt on a self-study question on Confidence Interval:

The CEO of Micosaft Inc is considering the proposal of offering a child care program for its employees. As part of the feasibility study, the CEO wishes to estimate the mean weekly child-care cost of their employees. A sample of 60 employees who use child care reveals the following statistics:



where xi is the weekly child-care cost of 1-th employees. Find a 95% confidence interval for the population mean.

My attempt as follows:

Unfortunately, my answer defers from the model answer of ***80 +_ 2.0111. Appreciate any guidance on where am I not doing right. I've looked at the question and my attempt multiple times without much success to find a way to correct it.
 A: Update
Based on whuber's pointing out that you do not have the data, I get your result for the variance (it was in an unfamiliar form), but when we use the t distribution as whuber inquires after I get exactly the text's answer:
$s_{\bar{x}} = 1.00507$
$t_{\alpha/2,\text{ }\nu=59} = 2.00100$
95% CI: $80 \pm 2.01114$ 
A: There's several minor errors here that will lead you into problems.
1) the sample standard deviation is not $\sigma$. By convention, Greek letters are for population parameters not sample quantities. You computed $s_{n-1}$, not $\sigma$. Call it $s$ and you won't be so likely to mislead yourself into:
2) You're using Z tables. You don't know $\sigma$, you estimated it, so you're dealing with a t-distribution as the basis for the interval*, not a normal distribution.
In this case (n=60) the distinction between the two isn't terribly large, but if you want to understand what you're doing, always treat $t$ as $t$ until the final step, even if $n$ is in the hundreds (that is, only bring the approximation of $t$ by a normal in at the very end, to help keep the concepts clearly distinct).
* specifically, the pivotal quantity from which the interval is derived has a t-distribution
