10-3 Prediction Intervals and Variation 545 FORMULA 10-5 se = BΣ1 y - yn22 n - 2 FORMULA 10-6 se = BΣy2 - b 0Σy - b1Σxy n - 2 (This is an equivalent form of Formula 10-5 that is good for manual calculations or writing computer programs.) For the paired jackpot>tickets data in Table 10-1 from the Chapter Problem, we found that there is sufficient evidence to support the claim of a linear correlation between those two variables, and we found that the regression equation is yn = -10.9 + 0.174x. We also found that if the jackpot amount is x = 625 million dollars, the predicted number of tickets sold is 97.9 million (or 98.0 million if using calculations with more decimal places). Use the jackpot amount of 625 million dollars to construct a 95% prediction interval for the number of tickets. CP EXAMPLE 1 Powerball Jackpots and Ticket Sales: Finding a Prediction Interval SOLUTION The accompanying StatCrunch and Minitab displays provide the 95% prediction interval, which is 73.7 million tickets 6 y 6 122 million tickets when rounded. StatCrunch Minitab The same 95% prediction interval could be manually calculated using these components: x0 = 625 (given) se = 4.437223 (provided by many technologies, including Statdisk, Minitab, Excel, StatCrunch, and the TI-83>84 Plus calculator) yn = 97.9 (predicted value of y found by substituting x = 625 into the regression equation) ta>2 = 2.365 (from Table A-3 with df = n - 2 = 7 and an area of 0.05 in two tails) n = 9, x = 214.8889, Σx = 1934, Σx2 = 455,364 continued
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