Wait, isn't the t-distribution used when the population variance is unknown? I'm going with D, the estimated mean divided by the estimated standard deviation has a Chi2-distribution.
I'm pretty confident it's C. The standard deviation is the square root of the variance, so the estimated mean divided by the estimated standard deviation has a t-distribution.
Hmm, I think it's C. The estimated mean divided by the estimated standard deviation should follow a t-distribution for i.i.d. normally distributed observations.
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