Ten normal observations
These numbers are a normal sample with unknown mean and variance:
2.86,\;4.16,\;6.35,\;3.92,\;5.96,\;5.12,\;6.43,\;3.03,\;4.80,\;3.79.
Give a 95% confidence interval for the mean. Use t_{9,\,0.975}=2.262.
Worked answer (click to show)
The sample mean is 4.642 and the bias-corrected sample standard deviation is 1.306. Then
4.642\pm 2.262\times\frac{1.306}{\sqrt{10}}=4.642\pm 0.935,
which is (3.71,\,5.58).