HW 2_solution

HW 2_solution - Solution to Homework #2 2: a. y1· = (6.7 +...

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Unformatted text preview: Solution to Homework #2 2: a. y1· = (6.7 + 7.8 + 5.5 + + 7.0) / 10 = 7.2, y2· = (9.9 + 8.4 + 10.4 + + 10.6) / 10 = 9.33, y3· = (10.4 + 8.1 + 10.6 + + 8.0) / 10 = 9.03, y 4· = (9.3 + 9.3 + 7.2 + + 7.2) / 10 = 8.69. b. The sample means minimize the sum of square ∑ ∑ ̂ , which are called the least squares . estimates. Hence the estimates that minimize the sum of squares are 8.69. c. When plugging in ̂ 6.7 6 7.0 6, ̂ 6 9, ̂ 9.9 9 8, ̂ 7.2, . 9.33, . 9.03, . 9, the sum of squares equals 10.6 9 83.53. so we obtain the The minimum of the sum of squares is obtained by replacing ̂ by the sample mean minimum is 56.471. d. Upper confidence limit for , , 56.47100000 23.27 56.47100000 ,. 2.4268 e. Hypothesis and alternative hypothesis are: : F‐test statistic is: Method1: , , , , , ,. 0.. . . : 5.70 . , Or method2: Because the corresponding P‐value is 0.0027 which is less than 0.05, we reject H0. 3: a. ANOVA table: P‐value is <0.0001, so we reject Null hypothesis at significance level b. . ,. 0.01. . . 0.002438 The appropriate SAS program is given below: * heart-lung pump experiment; options linesize = 75; DATA HRT; INPUT ORDER RPM Y; LINES; 1 5 3.540 2 1 1.158 3 1 1.128 4 2 1.686 5 5 3.480 6 5 3.510 7 3 2.328 8 3 2.340 9 3 2.298 10 4 2.982 11 3 2.328 12 1 1.140 13 4 2.868 14 5 3.504 15 3 2.340 16 2 1.740 17 1 1.122 18 1 1.128 19 5 3.612 20 2 1.740 ; proc glm data=HRT; class rpm; model y=rpm; run; * chapter 3; ...
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