think about the variance of linear functions...
Gary
On Fri, 20 Dec 2002, Yevgeniy Kirpichevsky wrote:
Dear Gary, Dave, and Tao,
Below is a dataframe, which
gives a value of the incumbency coefficient for each year with the
corresponding 95 % confidence interval. the t.test on the psi values
gives the 95%CI of .026 to .063, which is pretty big since it's based on
only 7 datapoints. But it seems like it could be much smaller, since for
each year the number of datapoints is largy - about 100. this is
reflected in the confidence intervals for each individual psi. Would it
be terribly wrong then to construct a 95% interval for the mean psi by
taking the mean of CI boundaries for each year?? it would be (.033, .056)
- smaller than (.026, .063).
Thank you
-yev
psi
year psi ci.low
ci.high
1 1970 0.02075644 0.01197663 0.02953625
5 1974 0.02153942 0.01094814 0.03213070
7 1976 0.02923420 0.01901227 0.03945612
9 1978 0.05216359 0.04059904 0.06372815
11 1980 0.05336988 0.04182857 0.06491120
15 1984 0.06943829 0.05631771 0.08255886
17 1986 0.06365428 0.04991084 0.07739773
mean(psi1$ci.low)
[1] 0.03294189
mean(psi1$ci.high)
[1] 0.05567414
t.test(psi$psi)
One Sample t-test
data: psi$psi
t = 5.8013, df = 6, p-value = 0.00115
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
0.02561964 0.06299639
sample estimates:
mean of x
0.04430801