Publications by Author Felipe Zambrini Santos
Assignment 4 - Group 9
1 Reading Data dat <- read.csv("https://raw.githubusercontent.com/tmatis12/datafiles/main/US_Japanese_Cars.csv") dat_us <- dat$USCars dat_jap <- dat$JapaneseCars 2 Question 1 According to the NPP graphics, the samples do not look like normal distributed. The scatter points do not seem to follow a linear trend for both american and japenese ...
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Assignment 6- Group6
1 Problem 1 1.1 Preperation First the given data was input into vectors pa and pb pa <- c(15,26,13,28,17,20,7,36,12,18) # sample A pb <- c(13,20,10,21,17,22,5,30,7,11) # sample B 1.2 Part A The null hypothesis is that the means are the same. The alternative is that there is a diffrence in the means. Thus this is a two sided T-test. \[H_o:...
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Flipped Assignment 9 - Group 3
1 Problem 1 1.1 Item a We need 14 observations from each group library(pwr) pwr.anova.test(k=4,n=NULL,f=1/sqrt(4.5), sig.level=0.05,power=0.8) ## ## Balanced one-way analysis of variance power calculation ## ## k = 4 ## n = 13.28401 ## f = 0.4714045 ## sig.level = 0.05 ## ...
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Homework 5
1 Question 3.7 The hypotheses testing in question to perform the ANOVA here is in the form: \[ H_0: \mu_1=\mu_2=\mu_3=\mu_4 \\ H_a : \mu_1\neq \mu_2 \neq\mu_3 \neq \mu_4\] 1.1 Item c After employing the ANOVA test, we can see that the null hypotheses is rejected and a pairwise comparison needs to be performed to check for difference in means...
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Homework6
1 Question 3.23 1.1 Item a We want to test the hypotheses: \[ H_0:\mu_1=\mu_2=\mu_3=\mu_4 \\ H_a:\mu_1 \neq \mu_2\neq\mu_3\neq\mu_4\] From the boxplot, the variance seems to be fairly equal. Therefore, we can proceed to employ the ANOVA test. F1 <- c(17.6,18.9,16.3,17.4,20.1,21.6) F2 <- c(16.9,15.3,18.6,17.1,19.5,20.3) F3 <- c(21.4,23.6,19....
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Flipped Assignment 11 - Group 8
1 Question 1 We want to test the hypotheses of equal means against the hypotheses of at least one being different: \[ H_0:\tau_i=0\\ H_a:\tau_i \neq 0 \] The linear effects model is shown by the equation below: \[ y_{ij}=\mu_i +\tau_i + \beta_j +\epsilon_{ij}\] where \(\mu_i\), \(\tau_i\), \(\beta_j\), and \(\epsilon_{ij}\) represents, respec...
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FA 12- Group 3
1 Question 1 This is a valid latin square because no repetitions of ingredients is observed in row or column, which ensure ortogonality. 2 Question 2 The model equation is given as: \[ x_{ijk}= \mu +\tau_i + \beta_j + \alpha_k + \epsilon_{ijk} \] where \(x\) is an individual observation, \(\mu\) is the mean of all observations, \(\tau\) is t...
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Homework7
1 Question 4.3 We want to test the hypotheses that chemicals have different effects on the strength of a particular type of cloth. However, bolts might have an additive effect and a block design might be required. We want to test: \[H_0:\tau_i=0 \\ H_a: \tau_i \neq 0 \space \text{for any} \space i\] From the ANOVA test, we can see that the bo...
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Question4 Test
1 Question 4 1.1 Item 1 The mean deaths between each type of zombies seems to differ but the variance seems to be pretty similar, which might be a initial guess about the validity of using ANOVA. library(readxl) dat <- read_excel(file.choose()) boxplot(dat$Basic,dat$Conehead,dat$Buckethead) 1.2 Item 2 The hypotheses testing can be written a...
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Flipped Assignment 14 - Group 3
1 Question 1 1.1 Letter A The full factorial model takes the form: \[ y_{ijk}=\mu+\tau_i+\beta_j+\gamma_k+(\tau \beta)_{ij}+(\tau \gamma)_{ik} + (\beta \gamma)_{jk} + (\tau \beta \gamma)_{ijk} + \epsilon_{ijkl}\] where i denotes the levels of ammonium, j denotes the levels for stir rate, k denotes temperature, and l denotes the number of re...
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