DanielTLis4273

Daniel Tafmizi Lis 4273 Dr. Ajani  March 24, 2024 Module 11 Assignment 1. ashina$subject <- factor(1:16)  attach(ashina)  act <- data.frame(vas=vas.active, subject, grp = 2)  plac <-data.frame(vas=vas.plac, subject, grp = 1)    model <- glm(unlist(act)~unlist(plac)) Coefficients: (Intercept) unlist(plac) -13.609 1.236 Degrees of Freedom: 47 Total (i.e. Null); 46 Residual Null Deviance: 95890 Residual Deviance: 61180 AIC: 485.4  plot(model) The model shows us that the active treatment patients responded better than the placebo treatment patients. I wish I could say more, but to be completely honest; I am very confused.  2. a <- gl(2, 2, 8) b <- gl(2, 4, 8) x <-- 1:8 y <- c(1:4, 8:5) z <- rnorm (8) model.matrix(z~a:b) (Intercept) a1:b1 a2:b1 a1:b2 a2:b2 1 1 1 0 0 0 2 1 1 0 0 0 3 1 0 1 0 0 4 1 0 1 ...

Daniel Tafmizi Lis 4273 Dr. Ajani  March 24, 2024 Module 10 Assignment 9.1.  library(ISwR) data <- (cystfibr) x <- lm(formula = cystfibr$pemax ~ age + weight + bmp + fev1, data=cystfibr) anova(x)          Analysis of Variance Table Response: cystfibr$pemax            Df  Sum Sq Mean Sq F value    Pr(>F)     age        1 10098.5 10098.5 18.4385 0.0003538 *** weight     1   945.2   945.2  1.7258 0.2038195     bmp        1  2379.7  2379.7  4.3450 0.0501483 .   fev1       1  2455.6  2455.6  4.4836 0.0469468 *   Residuals 20 10953.7   547.7    With the given model, we are testing the statistical significance between pemax to the other groups. Given the low P-values, we can say that age and fev1 are significa...

Daniel Tafmizi Lis 4273 Dr. Ajani  March 10, 2024 Module 9 Assignment # 1. #Your data.frame is assignment_data <- data.frame( Country = c("France","Spain","Germany","Spain","Germany", "France","Spain","France","Germany","France"), age = c(44,27,30,38,40,35,52,48,45,37), salary = c(6000,5000,7000,4000,8000), Purchased=c("No","Yes","No","No","Yes", "Yes","No","Yes","No","Yes")) #Generate simple table in R that consists of four rows: Country, age, salary and purchased. install.packages("data.table")        library(data.table) table <- setDT(assignment_data) #turns data frame into table table Country   age salary Purchased     <char> <num>  <num>    <char> 1:  France    44   6000        No 2:   Spain    27   5000       Yes 3: Germany    30   7000       ...

 Daniel Tafmizi Lis 4273 Dr. Ajani  March 3, 2024 Module 8 Assignment Question 1) We are comparing three groups by stress level, I need a qualitative(group name)and quantitative(level data) set Null Hypothesis: The means of each group are the same Alternative Hypothesis: at least one mean is different from the rest groups <- rep(c("High", "Mod", "Low"), each = 6) #Creates vector where each variable holds 6 data meansStress <- c(c(10,9,8,9,10,8),c(8,10,6,7,8,8),c(4,6,6,4,2,2)) #combines stress data datafr <- data.frame(groups, meansStress)  ggplot(datafr) + aes(x=groups, y=meansStress) + geom_jitter() #cool visual, checks that R is processing data correctly fit <- aov(meansStress ~ groups, data = datafr) #runs anova analysis summary(fit)               Df Sum Sq Mean Sq F value   Pr(>F)     groups       2  82.11   41.06   21.36 4.08e-05 *** Residuals...

Daniel Tafmizi Lis 4273 Dr. Ajani  February 23, 2024 Module 7 Assignment Question 1) x <- c(16, 17, 13, 18, 12, 14, 19, 11, 11, 10) y <- c(63, 81, 56, 91, 47, 57, 76, 72, 62, 48) Y = a + bX +e 1.1 Define the relationship model between the predictor and the response variable: model <- lm(y~x)  R equation for x,y relationship. Y is the target, x is the predictor   Relationship Model :  Y = 19.206 + 3.269X, RSE =10.48 the relationship has a positive correlation 1.2 Calculate the coefficients? summary(model)  gives information about the relationship model regression coefficient = 3.269, intercept coefficient = 19.206 question 2) discharge <- c(3.600,1.800,3.333,2.283,4.533,2.883) waiting <- c(79,54,74,62,85,55) visit <- data.frame(discharge, waiting) model2 <- lm(discharge ~ waiting, data=visit) summary(model2) Relationship Model : Y= -1.53317 + 0.06756X, RSE = 0.4724, the relationship has a positive correlation coeffs = coefficients(model2); co...

Daniel Tafmizi Lis 4273 Dr. Ajani  February 18, 2024 Module 6 Assignment #A) Consider a population consisting of the following values, which represents the number of  #ice cream purchases during the academic year for each of the five housemates.8, 14, 16, 10, 11 iceCream <- c(8,14,16,10,11) #Creates numerical vector #b. Select a random sample of size 2 out of the five members.  #See the example I used in my Power-point presentation slide # 13. randomSample <- sample(iceCream, 2) #Gets random sample of two values 10 & 8 were chosen # c. Compute the mean and standard deviation of your sample. mean(randomSample) # mean of sample = 9 sd(randomSample) # standard deviation of sample = 1.414214 mean(iceCream) # mean of population = 11.8 sd(iceCream) # standard deviation of population = 3.193744 #d. Compare the Mean and Standard deviation of your sample to the entire population of this set (8,14, 16, 10, 11). # The random sample taken chose the two lowest values of the ...

Daniel Tafmizi Lis 4273 Dr. Ajani  February 11, 2024 Module 5 Assignment 1 ) The director of manufacturing at a cookies company needs to determine whether a new machine is able to produce a particular type of cookies according to the manufacturer's specifications, which indicate that cookies should have a mean of 70 and standard deviation of 3.5 pounds. A sample of 49 cookies reveals a sample mean breaking strength of 69.1 pounds. A.   Null Hypothesis: The new machine falls within the manufacturers' specifications for cookie breaking strength. Alternative Hypothesis: The new machine does not fall within the manufacturers' specifications for cookie breaking strength. B.  Is there evidence that the machine is not meeting the manufacturer's specifications for average strength? Use a 0.05 level of significance.  No, the new machine is meeting the manufacturers' specifications. C.  Compute the p value and interpret its meaning The P-value is .0719, which is greater t...