From a population of trees, $n$ = 60, the sample average is 57.34 feet, and the sample variance is 4.62 feet. The question asks for a 99% confidence interval for $\mu$.
I started with 57.34 $\pm$ ...but am not sure where to go from here. I know it's 57.34 $\pm$ margin of error, but I am uncertain how to calculate margin of error. Thanks for any help, Stats geniuses.
The (approximate) interval is
$$\large{\left[\overline x-z_{(1-\frac{\alpha}{2})}\cdot \frac{s}{\sqrt n} ; \ \overline x+z_{(1-\frac{\alpha}{2})}\cdot \frac{s}{\sqrt n} \right]}$$
You have $\alpha=1-0.99=0.01$. Thus $1-\frac{\alpha}{2}=1-\frac{0.01}{2}=1-0.005=0.995$
Now you can use this calculator for a standard normal distribution and find out that $z_{0.995}=2.576$
Also given: $n=60, \overline x=57.34, s^2=4.62$
Can you proceed?