The normality assumption a normal p p plot of the

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The normality assumption : a normal p-p plot of the standardized resid- uals (residual divided by its standard error.) / £ ¡ ¢ EXAMPLE 11.1 A restaurant opening on a ”reservations-only” basis would like to use the number of advance reservations x to predict the number of dinners y to be prepared. Data on reservations and number of dinners served for one day chosen at random from each week in a 100-week period gave the following results. ¯ x = 150 ¯ y = 120 5
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X ( x - ¯ x ) 2 = 90 , 000 X ( y - ¯ y ) 2 = 70 , 000 . X ( x - ¯ x )( y - ¯ y ) = 60 , 000 a. Find the least squares estimates ˆ β 0 and ˆ β 1 for the linear regression line ˆ y = ˆ β 0 + ˆ β 1 x . b. Predict the number of meals to be prepared if the number of reservations is 135. c. Construct a 90% confidence interval for the slope. Does information on x (number of advance reservations) help in predicting y (number of dinners prepared)? Solution: a. The least squares estimates are given by ˆ β 1 = S XY S XX = 60 , 000 90 , 000 = . 67 and ˆ β 0 = ¯ y - ˆ β 1 ¯ x = 120 - . 67(150) = 19 . 50 . b. The predicted number of meals required for the number of advance reser- vations equal to 135 is ˆ y = 19 . 50 + . 67(135) = 109 . 95 , or 110 . c. The 90% confidence interval for β 1 uses the formula ˆ β 1 ± t (standard error) , where the standard error is s t / S XX . Although Table 4 in the Appendix does not list a t -value for α = . 05 and df = 98, we’ll use the t -value for the next higher df(df = 120); this value is 1.658. The standard deviation s ² can be computed using the summary sample data s 2 ² = SSE n - 2 , where SSE = S Y Y - ˆ β 1 S XY = 70 , 000 - 0 . 67(60 , 000) = 29 , 800 . 6
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Thus, s ² = r 29 , 800 98 = 304 . 08 = 17 . 44 and the 90% confidence interval for β 1 is 0 . 67 ± 1 . 658 (17 . 44) 90 , 000 or 0 . 67 ± . 10 . Since we are 90% confident that the true value of β 1 lies somewhere in the interval . 57 β 1 . 77, we are thus confident the increase in y ( number of dinners prepared) for every increase of one advance reservation is in the interval from .57 to .77. Also, since the interval for β 1
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