Is a T-Test Suitable to Compare Paper and Online Registry Participation Rates?

[bjrimel]

I have a small intervention study comparing a paper based registry versus an online based registry. In the 36 months of the paper based registry I had 185 participants in the 13 months of the online registry I had 301 participants.

The average participants per month in paper was 5.1 and online was 23.
I need to prove that this is "statistically significant".

I believe these are independent groups and should be compared with a t-test.
But I don't have any data on standard deviation. I have only the total number of enrolled participants and the length of time of enrollment.

Can someone help?

many thanks

 

[Yankel]

how come you have the means but not the standard deviations ?

you could try using the Poisson distribution

 

[Selis]

Keep in mind that a mean may not be a good method of summarizing your data, especially data with multiple peaks or long tails (or other deviations from an approximately normal distribution).

Are you certain you have no method at all to collect this data? Depending on how you executed both registries, I would expect a skewed distribution on the amount of registries collected per unit of time. If for instance you advertized your research after a lecture, you would expect to see a bump just after that.

Anyway, I'm not quite sure yet what you are trying to measure: you seem to be saying that you need to show with a certain level of certainty that both studies are independent on the variable 'date collected' ? (You do not prove differences in statistics, you reject the null hypothesis of no difference with a certain degree of certainty (for you probably a confidence interval).) What exactly do you want to do? Just explain it in normal language, e.g., I want to know if the groups had a different I.Q.)

PS. If this is what you want to measure, and you do have information on the dates these 'registers' where obtained: it is trivial to calculate standard deviations. It might also be possible to estimate the standard deviation, or at the very least attempt to construct a very conservative estimate.