1985 | 7 | <1993> | 6 |
1986 | 6 | <1994> | 9 |
1987 | 9 | <1995> | 8 |
1988 | 5 | <1996> | 9 |
1989 | 12 | <1997> | 7 |
1990 | 7 | <1998> | 9 |
1991 | 9 | <1999> | 12 |
1992 | 8 | <2000> | 2 |
There is no strong skewness or extreme outliers, although two upsets in 2000 was unusually low. The mean and standard deviation of this sample are x-bar = 7.8 and s = 2.5.
(estimate) ± (multiplier)(standard error)
A confidence interval for a population mean mu takes the form
x-bar ± t* s/sqrt(n)
where x-bar is the sample mean, n is the sample size, t* is the value such that the area between -t* and t* under a t distribution with n-1 degrees of freedom is the confidence level, and s is the sample standard deviation.
Plugging in numbers, we are 95% confident that mu is in the interval:
7.8 ± (2.131)(2.5)/sqrt(16)
or
7.8 ± 1.3
In the context of the problem, I am 95% confident that the process that produces NCAA tournaments has an average number of first round upsets between 6.5 and 9.1.
Bret Larget, larget@mathcs.duq.edu