Confidence Interval's in a nested function - r

Question

New student in R, taking a very accelerated class with little/no instruction. Please be patient with me...so far y'all have been extremely helpful, and I appreciate it. I apologize in advance if this doesn't make sense.
I am trying to make a function that reads columns from an object with columns "year", "complex", "mean", "2_sd", and "n" and calculates the confidence interval, then merges the lower and upper CI's as two separate columns into a new object with the same dimensions as the products of the CI calculations. However, I keep getting an error:
code for lower CI:
x=aggregate(m.all$mean, by=list(year,complex),FUN=(m.all$mean - qnorm(0.9) * sd(m.all$mean)/sqrt(m.all$n)))
error:
'(m.all$mean - qnorm(0.9) * sd(m.all$mean)/sqrt(m.all$n))' is not a function, character or symbol
I tried to use:
x=aggregate(total_male, by=list(year,complex),FUN=t.test(total_male,conf.level=0.90))
(where "total_male", "year", "complex" variables were sourced from the BASE object) but R doesn't recognize t.test when it's inside aggregate() for some reason...
The BASE object is 3 columns of "year", "complex", "total_males". The NEW object has "year", "complex", "mean", "2_sd", and "n"
I built "mean", 2_sd" and "n" out of the BASE object, with functions, and then merged them to create the NEW object, so I understand that. But CI's is confusing me.
The BASE object has been attach()'ed so I could work with the variables more easily.
Any ideas?
NEW object:
m.all
year complex mean X2st.dev n
1 2007 3corners 26.28571 52.04760 7
2 2007 Blue 18.87500 20.15476 8
3 2007 book_cliffs 4.50000 13.19091 6
4 2007 Diamond 13.25000 48.83431 20
The OLD object is 41 observations (all in 2007) of 4 complexes, with various numeric tot_male values:
head(d4)
year complex tot_male
2 2007 Diamond 17
21 2007 3corners 19
36 2007 Blue 40
73 2007 Diamond 22
85 2007 Diamond 0
115 2007 Diamond 2

Answer 1

You need to learn how to construct an R function. All you have now is an expression but it's not capable of accepting arguments. Perhaps:
x=aggregate(m.all$mean, by=list(year,complex),
FUN=function(v){ v - qnorm(0.9) * sd(v)/sqrt(length(v)) })
Please do not attach data objects. It just ends up making things less stable and less easy to understand. If 'm.all' is actually a dataframe with named columns "mean", "year", then the first line you proposed might be:
with( m.all, aggregate(mean, by=list(year,complex),
FUN=function(v){ v - qnorm(0.9) * sd(v)/sqrt(length(v)) }))
With lets you create a small environment where the column names will get interpreted as objects. Generally it's is a bad idea to use names like mean and sd since those are functions names.

Answer 2

Following may be what you want (since mean and sd have already been generated):
> m.all$lowerci = with(m.all, mean - qnorm(0.9)*X2st.dev/sqrt(n))
> m.all
year complex mean X2st.dev n lowerci
1: 2007 3corners 26.28571 52.04760 7 1.0748434
2: 2007 Blue 18.87500 20.15476 8 9.7429407
3: 2007 book_cliffs 4.50000 13.19091 6 -2.4013685
4: 2007 Diamond 13.25000 48.83431 20 -0.7441377
Similarly for upper CI:
> m.all$upperci = with(m.all, mean + qnorm(0.9)*X2st.dev/sqrt(n))
> m.all
year complex mean X2st.dev n lowerci upperci
1: 2007 3corners 26.28571 52.04760 7 1.0748434 51.49658
2: 2007 Blue 18.87500 20.15476 8 9.7429407 28.00706
3: 2007 book_cliffs 4.50000 13.19091 6 -2.4013685 11.40137
4: 2007 Diamond 13.25000 48.83431 20 -0.7441377 27.24414