Assignment 5

A, The approach of statistical experiment is directed experimentation as we are causing changes in a systematic fashion and watching the results of those changes. The type of statistical experiment is a single factor experiment as we are studying the effect of one single independent variable (the Rear wing). Since the experiment has more than two levels it should be analyzed using ANOVA test. The experiment measures the lap time, a quantitative value For the rear wings labeled A, B, C We assume that - The observations in each treatment are independent and randomly selected. - All treatments (populations) are assumed to be normally distributed with means μ1, μ2, . . . , μk and variance σ2. - All populations are assumed to have the same variance σ 2 . b, H 0 : A = B = C ? ? ? H A : At least one the averages are significantly different A = (4.5 + 4.5 + 4 + 5 + 5.5 + 5) / 6 ? = 4.75 b = (5.5 + 5.5 + 6 + 6.5 + 7 +7) / 6 ? = 6.25 c = (3.5 + 4 + 3 + 4 + 5 + 4.5) / 6 ? = 4.0 S a = 0.524 S b = 0.689

S c = 0.707 overall = 5.0 ? SS treatment = nΣ( overall - n ) 2 ? ? = 6 * ((4.0 - 5.0) 2 + (4.75-5.0) 2 + (6.25 - 5.0) 2 ) = 6 * (1 2 + 0.25 2 + 1.25 2 ) = 15.75 SSE = (n-1)ΣS i 2 = 5 * ( 0.524 2 + 0.689 2 + 0.707 2 ) = 6.246 SSA = SS treatment = 15.75 MSA = SSA/k-1 = 15.75/3-1 =7.865 MSE = SSE/k(n-1) = 6.246/3*(5) = 0.41 F calc = MSA/MSE = 18.9 v 1 = k-1 = 2 v 2 = k(n-1) = 15 F table = F 0.05, 2, 15 = 3.68 Since F calc > F table we reject the null hypothesis meaning that there is a difference of speed between at least one of the rear wings at 5% significance.

C Soft (4 + 5+ 4.5)/3 ? = 4.5 Medium = 5.66 + 4.33 + 3.5 / 3 ? = 4.5 Soft = 6.83 + 5.16 + 4.5 / 3 ? = 5.5 = 5 ? A = 6.245 ? B = 4.756 ? C = 4 ? Sum Of Squares Error SSE = SS w/in = ΣΣΣ(x ijk, - iij ) 2 ? = (5.5-5.66) 2 + (5.5-5.66) 2 + …(4.5-4.5) 2 = 1.6667 Sum Of Squares Rear Wing (A) SSA = SS Wing = b * n Σ( i - … ) 2 ? ? = 2 * 3 * (6.245-5) 2 + (4.756-5) 2 + (4-5) 2 = 15.65 Sum Of Squares Tire Softness (B) SSA = SS Tire =a * n Σ( i - … ) 2 ? ? = 3 * 3 * (4.5-5) 2 + (5.5-5) 2 = 4.5 Sum of Squares Interaction SS(AB) = SS interaction = n ΣΣ( ij - i - j + ) 2 ? ? ? ? 3 * [(5.66 - 6.245-4.5+5) 2 + (4.33-4.756-4.5+5) 2 + (3.5-4-4.5+5) 2 + … (4.5-4-5.5+5) 2 ] = 0.087

Source of Variation Sum Of Squares Degrees of Freedom Mean Square Calculated F Critical F Rear Wing SSA = 15.65 A-1 = 3-1 = 2 SSA/(a-1) = 7.825 MSA/MSE = 56.3 F a, a-1, ab(n-1) =3.89 Tire Compound SSB = 4.5 B-1 = 2-1 = 1 SSB/(b-1) = 4.5 MSB/MSE = 32.4 F a, b-1, ab(n-1) = 4.75 Two-Factor Interaction SS(AB) = 0.087 (a-1)(b-1) = 2 SSAB/(a-1)(b-1) = 0.043713 = MSAB/MSE = 0.315 F a, (a-1)(b-1), ab(n-1) = 3.89 Error (within) SSE = 1.667 Ab(n-1) = 12 SSE/ab(n-1) =0.139 Since F calc > F crit for the rear wing we can reject H 0 meaning that there is a difference of speed between the rear wings at a 5% significance level Since F calc > F crit for the tire compound we can reject H 0 ’ meaning that there is a difference of speed between the tire compounds at a 5% significance level Since F calc < F crit for the interaction we can accept H 0 meaning that there is no interaction between the rear wings and tire compounds at a 5% significance level

A, In order for the data set to be random I chose random foods from all of the major food groups. I did this to try and mimic what foods the average person would eat in a day. To avoid bias towards my own dietary choices I found all of the nutrition labels online rather than use my own personal food. b, Σ sugar = 0 + 4.5 + 16.1… = 70.7 μ sugar = Sum/10 = 7.07 Σ cal = 170 + 61 + 611… = 2118 μ cal = Sum/10 = 211.8 Σ sugar^2 = 1262.07 Σ cal^2 = 685208 Σ sugar*cal = 21842.1 Sugar = x, Cal = y S xx = Σx i 2 - n 2 ? = 1262.07 - 10*7.07 2 = 762 S yy = Σy i 2 - n 2 ? = 685208 - 10*211.8 2 = 236616 = SST S xy = Σx i y i - n ? ? = 21842.1 - 10*7.07*211.8 = 6868 b= S xy /S xx = 6868/762 = 9 a= - b ? ? = 211.8 - 9*7.07 = 148.17 y(x) = 9x + 148.17 c, r = S xy /√S xx *S yy = 6868/√(762*236616) = 0.51 Since the r value is 0.51 there is a moderate correlation between the amount of sugar in a food and its total calories

D, SST = 236616 SSE = Σ(y i - y^ i ) 2 = (170-148.17) 2 + (61-188.67) 2 … = 174734 SSR = SST - SSE = 236616 - 174734 = 61882 P = 2 N = 10 df R = p - 1 = 1 df e = n-p = 8 df T = n-1 = 9 MSSR = SSR/dfr = 61882/1 = 61882 MSSE= 174734/8 = 21842 F calc = MSR/MSE = 2.83 F crit = F 0.05 1, 8 = 5.318 Since F calc < F crit we accept the null hypothesis and conclude that the model is not significant and does not explain the majority of the variability about the mean.