The width of a prediction interval is positively related to the standard error of estimate, but negatively related to the standard deviation of the independent variable and the number of observations in the sample. Why is this the case? Thanks.
False.
Where did you read that? Can you provide year, book name, chapter and page please?
CI = mean +/- SD x T-table
As you see in the above formula, there are only 2 factors that can affect the width of a CI, the SD and the specified T-table value (or Z-table value).
Thus,
(1) The higher the SD, the wider the CI and viceversa > (Positive Relation)
(2) The higher the confidence level (90%, 95%, 99%), the wider the CI > (Positive Relation)
(3) The higher the degrees of freedom, the narrower the CI (the degrees of freedom are related to the number of data observations) > (Negative Relation)
This is actually correct. If you look at the formula for a prediction interval, it’s readily apparent. Variation in X ends up in the denominator, as does the sample size. The PI endpoints are directly related to the SEE as it is in the numerator.
This makes sense because SEE represents uncertainty in the model. The larger it is, the larger your uncertainty about a prediction from the model. Larger samples size (degrees of freedom) means more precise estimation of the model, so the model predicted values should be more accurate, all else equal. More total variation in X means more information about the relationship between X and Y, which makes us more accurate in our predictions of Y (it also makes any given value of x for which we are predicting represent LESS of the total “information”/variation from X relating to Y).
Easiest way is to look at the formula and understand each piece.
A CI uses the standard error for the sample statistic utilized. The standard error is often related to the sample size (I can’t think of a case where it isn’t, in my experience). See the above post, and the PI interval formula.