t-test difference in size of each sample? I have a daily process that's been going on for a few months. Now I'm trying to infer if a change to the process is significantly different. I let the new process run for two weeks because I can't afford to run it too much longer without knowing if it's a better process.
Questions

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*When computing the mean statistic for the original process, how many observations should I use? Does it have to be 14 to match the new process I'm evaluating?

*If so, how could I choose what those 14 observations should be? Maybe just the most recent 14?

*Could I use the 14 observations I have for the new process and measure it against a random sample of 14 observations from the original process? What would happen if I did this multiple times?

Edit - Adding clarity
To make my use case slightly more concrete, the "process" is the amount of clicks a digital ad campaign gets in a day. The original process is the ad campaign with all default settings. The new process is some change to how the campaign is optimized to hopefully produce more clicks.
 A: Comment continued: Suppose you are doing a 2-sided, one-sample t test at the 5% level, with a random sample (independent observations) of size $n=14$ from a normal population. Also, suppose $\sigma=\delta$ in the notation of my Comment above, then you will have power (probability of detection/rejection) about 93%.
The first illustration (in R) is for testing $H_0: \mu=0$ vs $H_a: \mu \ne 0.$ The specific power is against an alternative $\delta =2$ away from $0$ and with $\sigma = 2.$ But for any $\sigma=\delta,$ the power will be the same, about 93%.
set.seed(1213)
pv = replicate(10^5, t.test(rnorm(14, 2, 2))$p.val)
mean(pv <= .05)
[1] 0.93312

Now for $\delta = \sigma = 3.$
set.seed(2020)
pv = replicate(10^5, t.test(rnorm(14, 5, 3), mu=2)$p.val)
mean(pv <= .05)
[1] 0.93303

Exact computation from a recent release of Minitab software.
Power and Sample Size 

1-Sample t Test

Testing mean = null (versus ≠ null)
Calculating power for mean = null + difference
α = 0.05  Assumed standard deviation = 2

            Sample
Difference    Size     Power
         2      14  0.932369


