# What happens if we set the prediction interval and confidence interval around the regression line at “.9999999”

Below, I'm showing a hypothetical simple linear regression case. There are $5000$ datapoints in each of the 4 sub-populations stacked over the top of each predictor value. Each subpopulation is normally distributed. And all subpopulations have the same $\sigma^2$. However, the subpopulations' $\mu$ are different leading to an upward-sloping regression line.

# Question

When I compute the $99.999999999\%$ confidence interval for the regression line, nothing basically shows (i.e., the line is super precise due to huge number of datapoints).

However, When I compute the $99.999999999\%$ prediction interval for predicting a future observation (the light green surrounding the 4 subpopulations as well as the regression line), the prediction interval becomes hugely wide.

Why when datapoint are so huge (about the size of a population), setting the confidence interval around the regression line to about $1$, doesn't make the such confidence interval to even show BUT

setting the prediction interval for observations to about $1$, is still so huge? (Is there really any other far-away observation than the ones currently existing in each subpopulation) • It's difficult to determine what you are asking. You seem to want to set limits that are likely to contain nearly every point in a sample of size $10^{11}$. Are your "subpopulations" really that large? – whuber May 19 '17 at 17:58
• @whuber, so where does 10^11 come from? – rnorouzian May 19 '17 at 18:01
• @whuber, are you suggesting that a "subpopulation" should be extremely large to contain just about all possible observations? – rnorouzian May 19 '17 at 18:05
• $10^{11} = 1 / (1 - 99.999999999/100)$ is from your interval. It's looking like you might benefit from reviewing our threads on confidence intervals and prediction intervals: I recommend you search the site. – whuber May 19 '17 at 18:16

• Harvey, @whuber mentioned that, I need the size of my subpopulations each to be $10^{11}$ (i.e., $1 / (1 - 99.999999999/100)$ ), How this simple formula comes about? Is this formula only true in the case of normal distributions? – rnorouzian May 19 '17 at 20:48