A sample of size n = 20 is selected from a normal population to construct a 95% confidence interval estimate for a population mean. The interval was computed to be (8.20 to 9.80). Determine the sample standard deviation.

Since n<30, we use a t-distribution with d.f = 19.

Looking this up in the t table we find the rejection regions are t<-2.093 and t>2.093.

{\mu}-E=8.2

{\mu}+E=9.8

Solving the system yields \overline{x}=9, \;\ E=0.8

Or we could just find the difference in the values and divide by 2. E=(9.8-8.2)/2=0.8

Use the formula E=t\cdot \frac{s}{\sqrt{n}} to find the standard deviation, s, by subbing in your knowns and solving for s.