Standard Deviation: Confidence Levels?

Question

I have 30 data points. The Mean is 139.2. The Standard Deviation is 1.32. How do I calculate, with 99% certainty, the range within which my next data point will fall?

If I multiply my Standard Deviation of 1.32 by 2.462 from the T-distributioin table (29 degrees of freedom) will that give me the value above and below my Mean that I can expect my next data point to fall?

Thanks.

Answer 1

Not quite: 2.462 will give you a 98% confident prediction. For 99%, you need 2.756, i.e. the 99.5 percentile (because your prediction is two-sided, having +- limits, but the percentiles in the tables are one-sided critical points). Also, you should multiply this by sqrt(1+(1/n)) to be quite correct, but for n = 30 this is very close to 1 (1.0165). So the endpoints of a 99% prediction interval for x31 are

139.2 +- 1.32*2.756*1.0165

NB: The factor 1/sqrt(n) has no place in constructing the prediction interval. It appears in estimating the true mean with the sample mean xbar because this is how fast the SD of xbar reduces as n increases. But that is another matter.

Answer 2

I think it's mean +/- (t(n-1) * std. dev) / sqrt (n)