Publications by Data Whisperer
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for maturity Reading Data ## # A tibble: 6 × 35 ## Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11_1 Q11_2 Q11_3 ## <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> ## 1 2 2 4 2 2 2 1 2 1 62 2 2 2 ## 2 2 3 3 2 2 2 2 ...
363 sym 3 img
Data Whisperer - Data Jam
Reading Data ## # A tibble: 6 × 35 ## Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11_1 Q11_2 Q11_3 ## <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> ## 1 2 2 4 2 2 2 1 2 1 62 2 2 2 ## 2 2 3 3 2 2 2 2 2 1 ...
1304 sym
Sujit_Shopify Test
Details author: Sujit Tilakraj Thakur School email : Sujthaku@ttu.edu Personal email : sjthakur.st@gmail.com contact : +1-(806)0-401-2452 Lets read the data first and convert to type in which we can wrangle the data easily data <- read_excel("C:\\Users\\sjtha\\OneDrive\\Desktop\\shopify\\Data.xlsx") str(data) ## tibble [5,000 x 7] (S3: tbl_...
1720 sym R (2350 sym/12 pcs) 4 img
FA16 Group 2
Getting the data response <- c(12,18,13,16,17,15,20,15,10,25,13,24,19,21,17,23) a <- c(-1,1) b <- c(-1,-1,1,1) c <- c(-1,-1,-1,-1,1,1,1,1) d <- c(-1,-1,-1,-1,-1,-1,-1,-1,1,1,1,1,1,1,1,1) A <- c(rep(a,8)) B <- c(rep(b,4)) C <- c(rep(c,2)) D <- c(rep(d,1)) dat1 <- cbind(A,B,C,D,response) dat <- as.data.frame(dat1) dat ## A B C ...
1561 sym R (2519 sym/18 pcs) 3 img
FA14 - Group 1
data$ï..Ammonium <- as.fixed(data$ï..Ammonium) data$StirRate<- as.fixed(data$StirRate) data$Temperature <- as.fixed(data$Temperature) Q1) a) Model Equation \(Y_{ijkl}\) = \(\mu\) + \(\alpha_i\) + \(\beta_j\) + \(\gamma_k\)+ \(\alpha\beta_{ij}\) + \(\alpha\gamma_{ik}\) + \(\beta\gamma_{jk}\) + \(\alpha\beta\gamma_{ijk}\) + \(\epsilon_{ijkl}\...
3627 sym R (8262 sym/18 pcs) 1 img
FInal FLIPPED ASSIGNMENT -12 - Group 1
Hypothesis: Null Hypothesis: \(H_o : \mu_1 = \mu_2 = \mu_3 =\mu_i\) Alternative Hypothesis: \(Ha\) : atleast one \(\mu_i\) differs As we know it has fixed effects hence we can write our hypothesis as, Null Hypothesis: \(H_o : \tau_i=0\) for all i Alternative Hypothesis: \(H_a\) : \(\tau_i \neq 0\) for some i batch <- c(rep(1,5),rep(2,5),rep(3...
1587 sym R (906 sym/3 pcs)
Flipped Assignment 11 - Group 1
Question 1 Hypothesis: Null Hypothesis: \(H_o : \mu_1 = \mu_2 = \mu_3 =\mu_i\) Alternative Hypothesis: \(Ha\) : atleast one \(\mu_i\) differs As we know it has fixed effects hence we can write our hypothesis as, Null Hypothesis: \(H_o : \tau_i=0\) for all i Alternative Hypothesis: \(H_a\) : \(\tau_i \neq 0\) for some i Linear Effects: \(y_{...
2320 sym R (992 sym/6 pcs)
Project 1.1 - Oct 5th
Title: Design of Experiment-Project Aim: Testing whether landing distance of three different balls are same or different. Instructor - Dr.Timoty Matis Author: Sujit Thakur , Tajammul Mohammed , Gowtam Sasikumar Determining the size using power calculation pwr.anova.test(k=3,n=NULL,f=0.5, sig.level = 0.05 ,power = 0.75) ## ## Balanced...
1644 sym R (2792 sym/13 pcs) 6 img
Flipped Assignment 09 - Group 1
Q1a) library(pwr) pwr.anova.test(k=4,n=NULL,f=sqrt(((1)^2)/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 ## power = 0.8 ## ## NOTE: n is number in each gr...
993 sym R (1823 sym/11 pcs) 5 img
Flip Assignment - 09 Group -- 1
Q1a) library(pwr) pwr.anova.test(k=4,n=NULL,f=sqrt(((1)^2)/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 ## power = 0.8 ## ## NOTE: n is number in each gr...
983 sym R (1824 sym/11 pcs) 5 img