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Diagram showing the cumulative distribution function for the normal distribution with mean (μ) 0 and variance (σ 2) 1. These numerical values "68%, 95%, 99.7%" come from the cumulative distribution function of the normal distribution. The prediction interval for any standard score z corresponds numerically to (1 − (1 − Φ μ,σ 2 (z)) · 2).
Example: To find 0.69, one would look down the rows to find 0.6 and then across the columns to 0.09 which would yield a probability of 0.25490 for a cumulative from mean table or 0.75490 from a cumulative table. To find a negative value such as -0.83, one could use a cumulative table for negative z-values [3] which yield a probability of 0.20327.
A 95% confidence level does not mean that 95% of the sample data lie within the confidence interval. A 95% confidence level does not mean that there is a 95% probability of the parameter estimate from a repeat of the experiment falling within the confidence interval computed from a given experiment. [25]
Classically, a confidence distribution is defined by inverting the upper limits of a series of lower-sided confidence intervals. [15] [16] [page needed] In particular, For every α in (0, 1), let (−∞, ξ n (α)] be a 100α% lower-side confidence interval for θ, where ξ n (α) = ξ n (X n,α) is continuous and increasing in α for each sample X n.
For a confidence level, there is a corresponding confidence interval about the mean , that is, the interval [, +] within which values of should fall with probability . Precise values of z γ {\displaystyle z_{\gamma }} are given by the quantile function of the normal distribution (which the 68–95–99.7 rule approximates).
Confidence bands can be constructed around estimates of the empirical distribution function.Simple theory allows the construction of point-wise confidence intervals, but it is also possible to construct a simultaneous confidence band for the cumulative distribution function as a whole by inverting the Kolmogorov-Smirnov test, or by using non-parametric likelihood methods.
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For example, to calculate the 95% prediction interval for a normal distribution with a mean (μ) of 5 and a standard deviation (σ) of 1, then z is approximately 2. Therefore, the lower limit of the prediction interval is approximately 5 ‒ (2⋅1) = 3, and the upper limit is approximately 5 + (2⋅1) = 7, thus giving a prediction interval of ...