Confidence Interval Question - Steps Taken, no given standard deviation I just wanted to make sure I was doing this Confidence Interval problem correctly (or incorrectly).
Q: The following are the daily number of steps taken by a certain individual in 20 weekdays.
(some values here, omitted due to size and because I'm not sure if all of them are completely necessary to show since I'm more interested in explanations)
Assuming that the daily number of steps is normally distributed, construct a (a) 95% and (b) 99% two-sided confidence interval for the mean number of steps.
sample size = 20
sample mean = 2062.75
standard deviation = 104.3435
(+ or -)1.96 is from a 95% Confidence Interval
(+ or -)2.575 is from a 99% Confidence Interval
For (a), I would get: 2062.75 (+ or -) (104.3435 / sqrt(20) )
Would this be correct reasoning, or am I missing something? It was mentioned that there's a normal distribution of steps.
 A: When the population distribution is assumed normal but the population standard deviation $\sigma$ is not known, it needs to be estimated from the sample.  This estimate is $$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar x)^2}.$$  Then the standard error of the sample mean is $$SE = s/\sqrt{n}.$$  But once we use this estimate, the margin of error for the $100(1-\alpha)\%$ confidence interval is $$ME = t_{n-1, \alpha/2} SE = t_{n-1, \alpha/2} \frac{s}{\sqrt{n}},$$ where $t_{n-1,\alpha/2}$ is the upper $\alpha/2$ quantile of the Student $t$ distribution with $n-1$ degrees of freedom.  The resulting confidence interval is $$[L,U] = [\bar x - ME, \bar x + ME].$$  This will be a larger confidence interval than if you used $z_{\alpha/2}$, the upper $\alpha/2$ quantile of the normal distribution, because in estimating the unknown population standard deviation from the sample, you incur additional uncertainty about the true value of the population mean $\mu$, and the confidence interval must be larger in order to have the same nominal coverage probability.
