R code for estimating a Poisson parameter and its CI? What is the R code to estimate the parameter lambda of the Poisson distribution?
I tried
# of occurrence 0 1 2 3 4 5 6 7 
occ  <- 0:7 
# of sequence 150, 200, 220, 230, 240, 250, 180, 260
freq <- c(150,200,220,230,240,250,180,260)
mean_per_seq <- occ %*% freq

Is this correct?
What is the R code to find 95 % confidence interval of the parameter lambda?
 A: If your data is:
sarahs.data <- c(rep(0, 150), rep(1, 200), 
                 rep(2, 220), rep(3, 230), 
                 rep(4, 240), rep(5, 250), 
                 rep(6, 180), rep(7,260))

Then you can use fitdistr from  to find the MLE value of $\lambda$:
library(MASS)
parms <- fitdistr(sarahs.data, "poisson")
parms 
        lambda  
3.72254335 
(0.04638706)

lambda <- parms$estimate
sd_x     <- parms$sd    

To estimate the 95%CI:
ci <- c(lambda + c(-1,1) * 1.96 * sd_x)
ci
[1] 3.631625 3.813462

Or you can use the fact that $Var(x)=\lambda=E(x)$:
lambda <- mean(sarahs.data)
var    <- lambda
sd_x     <- sqrt(sd)/sqrt(length(sarahs.data))

To estimate the 95%CI:
ci <- c(lambda + c(-1,1) * 1.96 * sd_x)
ci
[1] 3.631625 3.813462

As @whuber points out, your data are not distributed as Poisson($\lambda$). Consider the following density plot of a Poisson(3.6) RV:

A: whuber is correct that your data are not Poisson distributed.  But for the record, one answer to your question is to use fit a generalized linear model.  eg
> x <- rpois(1000,5)
> model <- glm(x~1, family="poisson")
> exp(coef(model))
(Intercept) 
      5.032 # close to the correct value of 5.000
> exp(confint(model))
Waiting for profiling to be done...
   2.5 %   97.5 % 
4.894243 5.172316 

With your data, if you did this, you could tell it was not poisson distributed because the estimated dispersion parameter (residual deviance divided by degrees of freedom) is much more than the expected value of one:
> y <- rep(0:7, c(150,200,220,230,240,250,180,260))
> table(y)
y
  0   1   2   3   4   5   6   7 
150 200 220 230 240 250 180 260 
> model2 <- glm(y~1, family="poisson") 
> summary(model2)

...    
(Dispersion parameter for poisson family taken to be 1)

    Null deviance: 2835.1  on 1729  degrees of freedom
Residual deviance: 2835.1  on 1729  degrees of freedom
AIC: 7802.5

A: If you're just interested in the mean of the data (which is what lambda tells you) then with this sample size the Poisson nature of the data is not very important. Using just the central limit theorem, we can use the t.test() function to compute a 95% confidence interval;
> y <- rep(0:7, c(150,200,220,230,240,250,180,260))
> t.test(y)$conf.int
[1] 3.618070 3.827016`
attr(,"conf.level")
[1] 0.95

Compare this with the interval you get from a Poisson modeling approach that allows for variance not being equal to the mean;
 > model2q <- glm(y~1, family="quasipoisson") 
 > exp( confint.default(model2q) )
               2.5 %   97.5 %
   (Intercept) 3.619594 3.828421

...it's almost identical.
