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STA412 Quiz2 Question: 1 2 3 4 Total Points: 30 30 24 16 100 Bonus Points: 0 0 Score: Factor B (temperature) 1. The effective life (in hours) of batteries is compared by material type (Nickel-Metal Hydride, Lithium-Ion) and operating temperature (70°F,125°F). Batteries are randomly selected from each material type and they are then randomly allocated to each temperature level. A summary of the study is given below. Factor A (material) pulation ample ¡ample Sample mean size mean variance Nickel-Metal Hydride 70°F μ11 50 80 222 Nickel-Metal Hydride 125°F [12 50 120 212 Lithium-Ion Lithium-Ion 70°F μ21 50 82 224 125°F μ22 50 117 216 For your convenience, the within-sample variation is s²w = 218.5. Consider the following contrasts: = • The average effect of factor A (material type): (A = ½μ11 + ½μ12 - μ21 - μ22 • The average effect of factor B (operating temperature): (B = 1½ μ11 - ½½μ12 + ½μ21 - μ22 • The contrast for the interaction effect: AB = μ11- μ12- μ21+ μ22 Conduct t-test to test the following hypothesis at α = 0.05. Hint: You can treat (μ11,μ12,μ21,μ22) as (µ1‚µ2‚µ³‚µ) and perform one-way ANOVA. (a) (10 points) Ho: l₁ = 0 against Ha: lд= 0. 2. (b) (10 points) Ho: B = 0 against Ha: l= 0. (c) (10 points) Ho: AB = 0 against H₁: lÃÅ= 0. A pharmaceutical company is interested in the effectiveness of a new preparation designed to relieve arthritis pain. Three variations have been prepared for investigation, which differ according to the proportion of the active ingredients. Patients enrolled in the study are randomly assigned to one of the three variations, and the time (in minutes) until pain relief is recorded on each subject. A summary of the study is given below. Proportion of the Population Sample Sample Sample active ingredients mean size mean variance 15% 40% με 30 23.45 10.75 με 35 14.08 9.62 50% Из 25 15.02 8.65 For your convenience, the within-sample variation is s²w= 9.729. Consider the contrasts below: l1=241- М2- из 12 = М2- из (a) (10 points) Construct a 95% two-sided confidence interval for 11. (b) (10 points) Construct a 95% two-sided confidence interval for l₂. (c) (10 points) Using the Bonferroni correction, construct 95% simultaneous two-sided confidence intervals for 1 and l2. Page 2 3. Aninvestigatorwishestocomparethetimetorelieffromheadachepainunderfourdistinctmedications (DrugA, B, C, and D). Twenty patients who suffer from chronic headache are randomly selected for the investigation, and five subjects are randomly assigned to each medication. The investigator use R to conduct pairwise comparisons among these drugs. The R output is given below. However, the investigator forgets to address the issue of multiple comparisons, as seen from the sentence Univariate p values reported in the R output. ## Simultaneous Tests for General Linear Hypotheses ## ## Multiple Comparisons of Means: User-defined Contrasts ## ## Fit: Im(formula = time ~ drug, data = data) ## ## Linear Hypotheses: ## Estimate Std. Error t value Pr(>|t|) ## A - B == 0 7.000 2.958 2.366 0.030913 ## A-C == 0 13.000 2.958 4.395 0.000452 ## A - D == 0 -2.000 0.676 0.508618 ## B C == 0 6.000 2.958- 2.958 2.028 0.059507 ## B D == 0 -9.000 2.958 -3.043 0.007759 ## C - D == 0 -15.000 2.958 -5.071 0.000113 ## (Univariate p values reported) (a) (6 points) Without any correction for multiple comparisons (i.e. using univariate tests), list all the pairs of drugs that show a significant difference in the time to relief from headache pain. Use an individual type I error rate of α₁ = 0.05. (b) (6 points) Using Bonferroni correction, list all the pairs of drugs that show a significant difference in the time to relief from headache pain. Use an experiment-wise type I error rate of α = 0.05. (c) (6 points) Using Scheffe's method, list all the pairs of drugs that show a significant difference in the time to relief from headache pain. Use an experiment-wise type I error rate of α = 0.05. (d) (6 points) Using Tukey's method, list all the pairs of drugs that show a significant difference in the time to relief from headache pain. Use an experiment-wise type I error rate of α = 0.05. A study is conducted to understand the relationship among the current through a conductor between two points, the voltage across the two points, and the resistance of the conductor. The response variable is the voltage (in volt). The two factors are the resistance and the current. A summary of the study is given below: Page 3 4. Resistance Current Population mean Sample size Low (12) Low (4A) μ11 Low (12) High (6A) μ12 High (202) Low (4A) μ21 22 22 High (202) High (6A) μ22 We consider three contrasts • Average effect of resistance: 1 = 1½ ½μ11 + ½ ½μ12 - μ21 μ22 • Average effect of current: 2 = 1½μ11 - 1/1μ12 + ½μ21 - μ22 • Interaction effect: l3 = μ11- μ12- μ21 + μ22 R is used to analyze the data and generate the ANOVA table below. In the output, combination refers to the combination of resistance and current. Unfortunately, the output doesn't directly provide the F test for the interaction effect between resistance and current. ## Df Sum Sq Mean Sq F value Pr(>F) ## combination 3 90 30 10 0.025 ## avg_effect_resistance 1 48 48 12 0.026 ## avg_effect_current 1 28 28 7 0.057 ## Residuals 4 16 4 (a) (4 points) Check whether l₁,l2,13 are mutually orthogonal. (b) (6 points) Based on the information given above, calculate the sum of squares for ₤3 the interaction effect between resistance and current. (c) (6 points) Conduct the F test to determine whether the interaction effect between resistance and current is significant at α = 0.05. Page 4/n1. The effective life (in hours) of batteries is compared by material type (Nickel-Metal Hydride, Lithium-Ion) and operating temperature (70°F, 125°F). Batteries are randomly selected from each material type and they are then randomly allocated to each temperature level. A summary of the study is given below. Sample Factor A (material) Factor B (temperature) pulation ample ample mean size mean variance Nickel-Metal Hydride 70-F μ11 50 80 222 Nickel-Metal Hydride 125°F [12 50 120 212 Lithium-Ion Lithium-Ion 70-F 125-F μ21 50 82 224 122 50 117 216 For your convenience, the within-sample variation is s²w = 218.5. Consider the following contrasts: • The average effect of factor A (material type): A = ½μ11 +12-21-22 •The average effect of factor B (operating temperature): B = M11 - ½μ12+21-22 • The contrast for the interaction effect: AB=μ11- μ12-H21+ μ22 Conduct t-test to test the following hypothesis at α = 0.05. Hint: You can treat (11,12,μ21,μ22) as (1234) and perform one-way ANOVA. (a) (10 points) Ho: A=0 against Ha: ld= 0.

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