In Section 2.1, we showed how to use least squares estimators to predict the lottery sales for a zip code, outside of our sample, having a population of 10,000. Because prediction is such an important task for actuaries, we formalize the procedure so that it can be used on a regular basis. <\/p>\n
To predict an additional observation, we assume that the level of explanatory variable is known and is denoted by \\(x_{\\ast}\\). For example, in our previous lottery sales example we used \\(x_{\\ast} = 10,000\\). We also assume that the additional observation follows the same linear regression model as the observations in the sample. <\/p>\n
Using our least square estimators, our point prediction is \\(\\widehat{y}_{\\ast} = b_0 + b_1 x_{\\ast}\\), the height of the fitted regression line at \\(x_{\\ast}\\) We may decompose the prediction error into two parts: <\/p>\n
\\begin{matrix}
\n \\begin{array}{ccccc} \\underbrace{y_{\\ast} – \\widehat{y}_{\\ast}} & = & \\underbrace{\\beta_0 – b_0 + \\left( \\beta_1 – b_1 \\right) x_{\\ast}} & + & \\underbrace{\\varepsilon_{\\ast}} \\\\ \\text{prediction error} & {\\small =} & \\text{error in estimating the } & {\\small +} & \\text{deviation of the additional } \\\\ & & \\text{regression line at }x_{\\ast} & & \\text{response from its mean}
\n\\end{array} \\end{matrix} <\/p>\n
It can be shown that the standard error of the prediction is \\begin{equation*} se(pred) = s \\sqrt{1+\\frac{1}{n}+\\frac{\\left( x_{\\ast}-\\overline{x}\\right) ^2}{(n-1)s_x^2}}. \\end{equation*} As with \\(se(b_1)\\), the terms \\(n^{-1}\\) and \\(\\left( x_{\\ast}-\\overline{x} \\right) ^2\/\\left[ (n-1)s_x^2\\right] \\) become close to zero as the sample size \\(n\\) becomes large. Thus, for large \\(n\\), we have that \\(se(pred)\\approx s\\), reflecting that the error in estimating the regression line at a point becomes negligible and deviation of the additional response from its mean becomes the entire source of uncertainty.
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For example, the point prediction at \\(x_{\\ast} = 10,000\\) is \\(\\widehat{y}_{\\ast}\\)= 469.7 + 0.647 (10000) = 6,939.7. The standard error of this prediction is \\begin{equation*} se(pred) = 3,792 \\sqrt{1+\\frac{1}{50} + \\frac{\\left( 10,000-9,311\\right)^2}{(50-1)(11,098)^2}} = 3,829.6. \\end{equation*} With a \\(t\\)-value equal to 2.011, this yields an approximate 95% prediction interval \\begin{equation*} 6,939.7 \\pm (2.011)(3,829.6) = 6,939.7 \\pm 7,701.3 = (-761.6, ~14,641.0). \\end{equation*} We interpret these results by first pointing out that our best estimate of lottery sales for a zip code with a population of 10,000 is $6,939.70. Our 95% prediction interval represents a range of reliability for this prediction. If we could see many zip codes, each with a population of 10,000, on average we expect about 19 out of 20, or 95%, would have lottery sales between 0 and $14,641. It is customary to truncate the lower bound of the prediction interval to zero if negative values of the response are deemed to be inappropriate.
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