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Course: DOCS 2735, Fall 2009
School: East Los Angeles College
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2735 MATH solutions 1 1 The required percentage points are Percentage point table value (a) F4,6 (5%) 4.534 (b) F6,4 (5%) 6.613 (c) F12,2 (5%) 19.413 (d) F3,3 (5%) 9.277 (e) F3,10 (5%) 3.708 (f) F3,20 (5%) 3.098 (g) F3,5 (5%) 5.409 (h) F12,5 (5%) 4.678 (i) F,5 (5%) 4.365 (j) F28,10 (5%) -- (k) F12,97 (5%) -- (l) F4,4 (0.25%) -- Comments on patterns: Increasing either n1 or n2 leads to a decrease in the...

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2735 MATH solutions 1 1 The required percentage points are Percentage point table value (a) F4,6 (5%) 4.534 (b) F6,4 (5%) 6.613 (c) F12,2 (5%) 19.413 (d) F3,3 (5%) 9.277 (e) F3,10 (5%) 3.708 (f) F3,20 (5%) 3.098 (g) F3,5 (5%) 5.409 (h) F12,5 (5%) 4.678 (i) F,5 (5%) 4.365 (j) F28,10 (5%) -- (k) F12,97 (5%) -- (l) F4,4 (0.25%) -- Comments on patterns: Increasing either n1 or n2 leads to a decrease in the percentage points. Note that Fn1 ,n2 (P %) = Fn2 ,n1 (P %). Keep your degrees of freedom in the right order. Advantages (+) and disadvantages (-) of R: + More precision (i.e. more decimal places) -- although this is only an advantage if the data have similar precision. + Can deal with any finite integer values for degrees of freedom -- the tables only list certain values. (How would you deal with infinite d.f. in R?) - Need to be at a computer -- not always the case. 2 The percentage points for the normal distribution are > tmp = c(0.1, 0.05, 0.025, 0.01, 0.005, 0.001, 0.0005) > tmp = qnorm(tmp, lower.tail=F) > round(tmp, 3) [1] 1.282 1.645 1.960 2.326 2.576 3.090 3.291 These are the same as those for the t distribution on infinite degrees of freedom. The tn distribution tends to the normal distribution as n . R value 4.533677 6.163132 19.41251 9.276628 3.708265 3.098391 5.409451 4.677704 -- 2.710420 1.853338 33.30274 1 MATH 2735 solutions 1 3 The data were to centre the data. Hence we do not need to consider xi - x values as x = 0. The necessary calculations are Sugar, y 8.1 7.8 8.5 9.8 9.5 8.9 8.6 10.2 9.3 9.2 10.5 100.4 Temp, x -0.5 -0.4 -0.3 -0.2 -0.1 0.0 0.2 0.1 0.3 0.4 0.5 0.0 x2 i 0.25 0.16 0.09 0.04 0.01 0.00 0.01 0.04 0.09 0.16 0.25 1.10 xi yi -4.05 -3.12 -2.55 -1.96 -0.95 0.00 0.86 2.04 2.79 3.68 5.25 1.99 Since y = 9.13 and Sxy /Sxx = 1.81 (both to 2 dp) we get the regression equation sugar = 9.13 + 1.81 coded temp where "sugar" is the sugar remaining after fermentation and "coded temp" is the fermentation temperature minus 21.5 degrees centigrade Alternatively, we could give the regression equation as sugar = -29.77 + 1.81 temp where "temp" is in degrees centigrade. Either form is correct, but you need to be clear as to which form you have used. 2 MATH 2735 solutions 1 4 Re-arranging the model gives us i = yi - - 1 xi - 2 wi (i = 1, . . . , n). Hence the sum of squared errors is S= i 2 = i i (yi - - 1 xi - 2 wi )2 . Differentiating S w.r.t. , we get S = -2 (yi - - 1 xi - 2 wi ) i = -2(ny - n) since i xi = i wi = 0 as these variables are already c...

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Rutgers - MATH - 291
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Rutgers - MATH - 291
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