Calculating p-values and pnorm() in R I am trying to calculate the p-values of observations by comparing them to the normal distribution in R using pnorm().  I have constructed a random distribution as my background model on which I would like to test the significance of various tests.  I know for example, my background normal distribution has a mean of 1 and a standard deviation of 3. 
Say I have one test that I would like to test the significance of test1 <- 20. To obtain the p-value of a specific observation with a value of 20, I can use pnorm(20, mean=1, sd=3). But what if for the same test, I have 5 repeated observations (technical repeats of the same test) with the values:
test1.rep1 <- 20
test1.rep1 <- 25
test1.rep1 <- 15
test1.rep1 <- 25
test1.rep1 <- 15

These 5 numbers have a mean of 20 and a standard deviation of 5. I could simply combine all of the repeated observed by taking the mean and then comparing to the normal distribution, i.e., pnorm(20, mean=1, sd=3). But in this case, I am leaving out information, namely the standard deviation of these 5 observations. Is there an alternative way to include both the mean and standard deviation of the 5 observations when calculating the p-value?

The alternative is to calculate 5 p-values for each of the 5 observations of test1.
pnorm(20, mean=1, sd=3)
pnorm(25, mean=1, sd=3)
pnorm(15, mean=1, sd=3)
pnorm(25, mean=1, sd=3)
pnorm(15, mean=1, sd=3)

But then I have to find a way to combine these p-values, to end with one final p-value to look for significance.  I ultimately want to know if test1 is significant.
I've looked into the Fisher method and the Stouffers method, but now I am thinking it may be better to just combine the values up front, rather than combining p-values.
 A: You could use Fisher's method of combining p-values, but it wouldn't be the preferred approach.  What you want to do here is a t-test, specifically the one-sample version.  In R the function is t.test().  Here is a quick tutorial I found by Googling.  The way this works is that you are just checking to see if your sample came from a population with a given mean.  In your case you want to know what the probability of getting a sample as extreme or more extreme as yours if they were drawn from a population with a mean of 1.  It's true that you want to take into account that your sample has a SD of 5.  We estimate the standard error of the sampling distribution of the mean by dividing the SD by the square root of N.  So, in your case, that would be $2.2=5/\sqrt{5}$.  
Now, I should say here that your question taken literally (i.e., as stated) is about a $z$-test, not a $t$-test.  That's because you stipulate that you know both the population mean and SD.  I think this is the way you asked the question for the sake of understanding the issues (i.e., you wanted a well-defined problem).  In practice, this just doesn't really happen.  However, if that exact situation is really what you care about, then you would calculate the standard error of the mean by dividing the known SD by the square root of the number of data in your sample.  That is $SE = 5/\sqrt 3$, and thus your $SE = 1.3$.  Your $z$-score is $14$, and $p < .001$.  Here is the R code for that:  
2*( 1-pnorm( 20, mean=1, sd=(3/sqrt(5)) ) )

