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Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.01,0:00:04.37,Default,,0000,0000,0000,,>> In general, samples underestimate the amount of variability in a population,
Dialogue: 0,0:00:04.37,0:00:08.88,Default,,0000,0000,0000,,because samples tend to be values in the middle of the population. Especially in
Dialogue: 0,0:00:08.88,0:00:13.31,Default,,0000,0000,0000,,a normal distribution, most of the values are centered here in the middle. So
Dialogue: 0,0:00:13.31,0:00:18.65,Default,,0000,0000,0000,,when we take samples from it, most of our values are going to be around here,
Dialogue: 0,0:00:18.65,0:00:24.04,Default,,0000,0000,0000,,since most of the values are in this area. Therefore the variability in this
Dialogue: 0,0:00:24.04,0:00:29.56,Default,,0000,0000,0000,,sample will be less than the variability of the entire population. To correct
Dialogue: 0,0:00:29.56,0:00:35.22,Default,,0000,0000,0000,,for this, we use something called Bessel's correction, where instead of dividing
Dialogue: 0,0:00:35.22,0:00:40.39,Default,,0000,0000,0000,,by n, we divide by n minus 1. Same within the variance. So what will dividing by
Dialogue: 0,0:00:40.39,0:00:45.41,Default,,0000,0000,0000,,n minus 1 do to the original standard deviation and variance? Will it make them
Dialogue: 0,0:00:45.41,0:00:47.84,Default,,0000,0000,0000,,bigger, or will it make them smaller?