From 09ce47b018d61d75ae69cfdc3b8f1c0cd40ad2aa Mon Sep 17 00:00:00 2001 From: haley Date: Mon, 26 Nov 2018 11:07:43 -0500 Subject: [PATCH 01/16] raw sheet for part 1 --- anovaVSregression.Rmd | 14 ++++++++++++++ 1 file changed, 14 insertions(+) create mode 100755 anovaVSregression.Rmd diff --git a/anovaVSregression.Rmd b/anovaVSregression.Rmd new file mode 100755 index 0000000..767dc60 --- /dev/null +++ b/anovaVSregression.Rmd @@ -0,0 +1,14 @@ +--- +title: "ANOVA and Regression Results" +output: html_notebook +--- + +# PART 1 REGRESSION VS ANOVA + +This project will focus on the difference between linear models that use categorical or continuous predictor variables. When designing experiments, scientists are often limited by the number of experimental units they can use. These practical limitations may arise due to the cost of using animal subjects, the number of tanks available for aquatic mesocosm studies, or simply the number of hours in the day. This question will focus on the optimal use of these precious experimental units to increase our "statistical power" to detect the effect of +a treatment.A experimental design question that arises when experimental units are limiting is whether to distribute the units amongst some number of replicated discrete treatment levels (ANOVA-design) or spread the experimental units along a continuous gradient of treatment levels (regression-design). In this project, we +will first revisit ANOVA and regression analyses in two cases where it is clear what experimental design was chosen. We will then evaluate the ability of these two experimental designs to detect a treatment effect (i.e. their statistical power) using simulated data. + +An ANOVA-design and a regression-design experiment +1) A student conducted an experiment evaluating the effect of three different new antibiotics on growth of E. coli in lab cultures. Using the data in antibiotics.txt, generate a plot that summarizes the results and test for an effect of antibiotic treatments on the growth of E. coli using an ANOVA-design linear model and +likelihood ratio test From d37a74f6a4dd231e6c6d387cd3599f5ad2f5f69c Mon Sep 17 00:00:00 2001 From: haley Date: Mon, 26 Nov 2018 11:22:52 -0500 Subject: [PATCH 02/16] raw sheet for part 2 --- statPowerAnalysis.Rmd | 13 +++++++++++++ 1 file changed, 13 insertions(+) create mode 100755 statPowerAnalysis.Rmd diff --git a/statPowerAnalysis.Rmd b/statPowerAnalysis.Rmd new file mode 100755 index 0000000..a48b607 --- /dev/null +++ b/statPowerAnalysis.Rmd @@ -0,0 +1,13 @@ +--- +title: "Power Analysis" +output: html_notebook +--- + + +# A statistical power analysis + +Experimentalists and statisticians will often use simulated data to help design an experiment. In these analyses, the investigator generates data that will look like the expected results of an experiment assuming some amount of variance or error. The investigator can then ask how many experimental units would be required to detect a significant effect, assuming the strength of the relationship and error assumed when simulating data. + +We will use this approach to ask whether ANOVA- or regression-design experiments are more powerful in detecting a continuous relationship between two variables. Assume that an independent variable x is linearly related to a dependent variable y with a slope (??1) of 0.4 and a y-intercept (??0) of 10. Imagine that we have 16 experimental units that can be used to test for an effect of x on y. One could randomly distribute the experimental units along a gradient of x (regression design) or one could replicate two levels of x 12 times each and ask whether low or high levels of x generate different levels of y (t-test design). An intermediate approach would be to have 8 replicates of three levels of x or 4 replicates of six levels of x and so on (ANOVA design). Using the relationship between x and y defined by ??1 = 0.4 and ??0 = 10, we can generate a random dataset with 24 observations of y for any experimental design of interest given some level of error in our observations. + +We define our error by the standard deviation (??) of normally distributed errors that we can add to our simulated values of y. Because our results will vary a bit from one simulated data set to another, it is a good idea to adopt something called a monte carlo approach where we run multiple (say 10) versions of a power analysis for a given experimental design and ?? \ No newline at end of file From 619063414ae5d1d73f3d3e2bd1b2db64461a2a46 Mon Sep 17 00:00:00 2001 From: enonnama <43544834+enonnama@users.noreply.github.com> Date: Fri, 30 Nov 2018 12:27:50 -0500 Subject: [PATCH 03/16] Add files via upload --- anovaVSregression.Rmd | 229 +++++++++++++++++++++++++++++++++++++++--- 1 file changed, 215 insertions(+), 14 deletions(-) diff --git a/anovaVSregression.Rmd b/anovaVSregression.Rmd index 767dc60..30f37d5 100755 --- a/anovaVSregression.Rmd +++ b/anovaVSregression.Rmd @@ -1,14 +1,215 @@ ---- -title: "ANOVA and Regression Results" -output: html_notebook ---- - -# PART 1 REGRESSION VS ANOVA - -This project will focus on the difference between linear models that use categorical or continuous predictor variables. When designing experiments, scientists are often limited by the number of experimental units they can use. These practical limitations may arise due to the cost of using animal subjects, the number of tanks available for aquatic mesocosm studies, or simply the number of hours in the day. This question will focus on the optimal use of these precious experimental units to increase our "statistical power" to detect the effect of -a treatment.A experimental design question that arises when experimental units are limiting is whether to distribute the units amongst some number of replicated discrete treatment levels (ANOVA-design) or spread the experimental units along a continuous gradient of treatment levels (regression-design). In this project, we -will first revisit ANOVA and regression analyses in two cases where it is clear what experimental design was chosen. We will then evaluate the ability of these two experimental designs to detect a treatment effect (i.e. their statistical power) using simulated data. - -An ANOVA-design and a regression-design experiment -1) A student conducted an experiment evaluating the effect of three different new antibiotics on growth of E. coli in lab cultures. Using the data in antibiotics.txt, generate a plot that summarizes the results and test for an effect of antibiotic treatments on the growth of E. coli using an ANOVA-design linear model and -likelihood ratio test +--- +title: "ANOVA and Regression Results" +output: html_document +--- + +# PART 1 REGRESSION VS ANOVA + +This project will focus on the difference between linear models that use categorical or continuous predictor variables. When designing experiments, scientists are often limited by the number of experimental units they can use. These practical limitations may arise due to the cost of using animal subjects, the number of tanks available for aquatic mesocosm studies, or simply the number of hours in the day. This question will focus on the optimal use of these precious experimental units to increase our "statistical power" to detect the effect of +a treatment. A experimental design question that arises when experimental units are limiting is whether to distribute the units amongst some number of replicated discrete treatment levels (ANOVA-design) or spread the experimental units along a continuous gradient of treatment levels (regression-design). In this project, we will first revisit ANOVA and regression analyses in two cases where it is clear what experimental design was chosen. We will then evaluate the ability of these two experimental designs to detect a treatment effect (i.e. their statistical power) using simulated data. + +## An ANOVA-design and a regression-design experiment + +1) A student conducted an experiment evaluating the effect of three different new antibiotics on growth of E. coli in lab cultures. Using the data in antibiotics.txt, generate a plot that summarizes the results and test for an effect of antibiotic treatments on the growth of E. coli using an ANOVA-design linear model and +likelihood ratio test + +```{r include=FALSE} +rm(list=ls()) +setwd("/Users/emilynonnamaker/Box/personalStuff/School/PhD/biocomputingFall2018/biocomputing_StatsGroupProject") +library(ggplot2) +library(arm) +antibiotics <- read.csv("antibiotics.csv") +sugar <- read.csv("sugar.csv") +boxplot(growth ~ trt, data=antibiotics) +lm <- lm(growth ~ sugar, data=sugar) +aov <- aov(growth ~ trt, data=antibiotics) +# max likelihood with anova design +# likelihood custom function - need slope/parameter with each column of 0s and 1s. +# then lratio + +summary(aov) +posthoc <- TukeyHSD(aov, "trt", conf.level=0.95) #a, b, ac, bd. +posthoc +``` + +Setting it up: + +```{r} +N=length(antibiotics$growth) +y=antibiotics$growth +x<- antibiotics$trt +antibiotics$x1 <- ifelse(x=="ab1", 1, 0) +antibiotics$x2 <- ifelse(x=="ab2", 1, 0) +antibiotics$x3 <- ifelse(x=="ab3", 1, 0) +``` + +Building the null model: + +```{r} +nullmod<-function(p,x,y){ + B0=p[1] + sigma=p[2] + + expected=B0 + + x1=x[,1] + x2=x[,2] + x3=x[,3] + +null=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) +return(null) +} +``` + +Building the full model: + +```{r} +fullmod<-function(p,x,y){ +B0=p[1] +B1=p[2] +B2=p[3] +B3=p[4] +sigma=exp(p[5]) + +x1=x[,1] +x2=x[,2] +x3=x[,3] + +expected=B0+B1*x1+B2*x2+B3*x3 +full=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) +return(full) +} +``` + +Building and checking our fits: + +```{r} +nullguess <- c(1, 3) +fullguess <- c(1, 2, 3, 4,5) +fitnull=optim(par=nullguess,fn=nullmod,x=antibiotics[,3:5],y=antibiotics$growth) +fitfull=optim(par=fullguess,fn=fullmod,x=antibiotics[,3:5],y=antibiotics$growth) +``` + +Test between null and full model: + +```{r} +t.stat <- 2*(fitfull$par[1] - fitnull$par[1]) +t.stat # 12.62689 + +1-pchisq(t.stat, df=4) # = 0.01325037 +``` + +Looks like there's a significant difference between the null and full anova model (t = 12.63, p < .05), suggesting that there are differences between treatments. + +## Graphical visualization! + +Using a posthoc Tukey test (see hidden code) we can find significant differences between groups, and add these differences as letters to our visualization (see graph below) + +```{r} +p <- ggplot(antibiotics, aes(x=trt, y=growth, fill=trt)) + + geom_boxplot() +p + geom_jitter(shape=16, position=position_jitter(0.2), alpha=0.3) + theme_classic() + labs(x = "Treatment", y= "Growth") + annotate("text", x = c(1, 2, 3, 4), y = c(23, 23, 23, 23), label = c("a", "b", "ac", "bd")) + +``` + +## Regression + +2) Another student conducted an experiment evaluating the effect of sugar concentration on growth of E. coli in lab cultures. Using the data in sugar.txt, generate a plot that summarizes the results and test for an effect of sugar concentration on growth of E. coli using a regression-design linear model and likelihood ratio test + +Building our linear model: + +Again, build the null model: + +```{r} +lmnullmod<-function(p,x,y){ + B0=p[1] + sigma=p[2] + + expected=B0 + +null=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) +return(null) +} +``` + +Now build the extension: + +```{r} +lmfullmod<-function(p,x,y){ + B0=p[1] + B1=p[2] + sigma=exp(p[3]) + + expected=B0+B1*x + + null=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) + return(null) +} +``` + +Give it a guess: + +```{r} +lmnullguess <- c(1, 1) +lmfullguess <- c(1, 2, 3) +lmfitnull=optim(par=lmnullguess,fn=lmnullmod,x=sugar$sugar,y=sugar$growth) +lmfitfull=optim(par=lmfullguess,fn=lmfullmod,x=sugar$sugar,y=sugar$growth) +``` + +Now to compare: + +```{r} +t.stat <- 2*(lmfitnull$par[1] - lmfitfull$par[1]) +t.stat # 19.78335 + +1-pchisq(t.stat, df=2) # 5.059418e-05 +``` + +Comparing between the null model of no effect of sugar concentration on growth to the model taking this affect into account, we can see that sugar concentration does have a significant affect on growth (t = 12.78, p < .0001). Looking at the graph, we can tell that sugar concentration has a positive affect, with higher concentrations increasing growth. + +## Visualization! + +```{r} +a <- ggplot(data=sugar,aes(x=sugar,y=growth)) +a + geom_point() + coord_cartesian() + labs( x = "sugar concentration", y = "growth") + theme_classic() + geom_smooth(method = "lm") +``` + +# Part 2 Statistical Power Analysis + +Assume that an independent variable x is linearly related to a dependent variable y with a slope (β1) of 0.4 and a y-intercept (β0) of 10. Imagine that we have 16 experimental units that can be used to test for an effect of x on y. One could randomly distribute the experimental units along a gradient of x (regression design) or one could replicate two levels of x 12 times each and ask whether low or high levels of x generate different levels of y (t-test design). An intermediate approach would be to have 8 replicates of three levels of x or 4 replicates of six levels of x and so on (ANOVA design). + +Using the relationship between x and y defined by β1 = 0.4 and β0 = 10, we can generate a random dataset with 24 observations of y for any experimental design of interest given some level of error in our observations. We define our error by the standard deviation (σ) of normally distributed errors that we can add to our simulated values of y. Because our results will vary a bit from one simulated data set to another, it is a good idea to adopt something called a monte carlo approach where we run multiple (say 10) versions of a power analysis for a given experimental design and σ. + +First get our data set with a normal distribution. + +```{r} +mod<-function(p,x,y){ + B0=10 + B1=0.4 + sigma=exp(p[3]) + + expected=B0+B1*x + + model=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) + return(model) +} + +newdata <- rnorm(24, 0, 1) +predict(mod, newdata) + +``` + + +```{r} +B1 <- 0.4 +B0 <- 10 +N <- 24 +data <- rnorm(24, 0, 1) +hist(data) +``` + + +To evaluate the relative statistical power of regression- and ANOVA-design experiments, simulate 10 random experiments with a regression design and 10 random experiments with a two-level ANOVA-design (really this is a t-test design) for each of eight values for σ (1, 2, 4, 6, 8, 12, 16, 24). Let’s say you are able to generate experimental units with x between 0 and 50. Repeat this process for a four- and eight-level ANOVA design (remember you only have 24 experimental units). Use the average p-value from likelihood ratio tests across your monte carlo runs as your metric of statistical power. + +How does the ANOVA- vs regression-design perform? Does the relative performance of these experimental designs depend on the number of levels in the ANOVA experiment (2, 4, vs. 8)? + From c2565bbc54709b91fef85b779b73b178adef834c Mon Sep 17 00:00:00 2001 From: enonnama <43544834+enonnama@users.noreply.github.com> Date: Fri, 30 Nov 2018 13:18:28 -0500 Subject: [PATCH 04/16] Add files via upload --- anovaVSregression.Rmd | 143 +++++++++++++++++++++++++++++++++++++----- 1 file changed, 126 insertions(+), 17 deletions(-) diff --git a/anovaVSregression.Rmd b/anovaVSregression.Rmd index 30f37d5..5fa971d 100755 --- a/anovaVSregression.Rmd +++ b/anovaVSregression.Rmd @@ -176,40 +176,149 @@ a + geom_point() + coord_cartesian() + labs( x = "sugar concentration", y = "gro # Part 2 Statistical Power Analysis -Assume that an independent variable x is linearly related to a dependent variable y with a slope (β1) of 0.4 and a y-intercept (β0) of 10. Imagine that we have 16 experimental units that can be used to test for an effect of x on y. One could randomly distribute the experimental units along a gradient of x (regression design) or one could replicate two levels of x 12 times each and ask whether low or high levels of x generate different levels of y (t-test design). An intermediate approach would be to have 8 replicates of three levels of x or 4 replicates of six levels of x and so on (ANOVA design). +Assume that an independent variable x is linearly related to a dependent variable y with a slope (β1) of 0.4 and a y-intercept (β0) of 10. Imagine that we have 24 experimental units that can be used to test for an effect of x on y. One could randomly distribute the experimental units along a gradient of x (regression design) or one could replicate two levels of x 12 times each and ask whether low or high levels of x generate different levels of y (t-test design). An intermediate approach would be to have 8 replicates of three levels of x or 4 replicates of six levels of x and so on (ANOVA design). Using the relationship between x and y defined by β1 = 0.4 and β0 = 10, we can generate a random dataset with 24 observations of y for any experimental design of interest given some level of error in our observations. We define our error by the standard deviation (σ) of normally distributed errors that we can add to our simulated values of y. Because our results will vary a bit from one simulated data set to another, it is a good idea to adopt something called a monte carlo approach where we run multiple (say 10) versions of a power analysis for a given experimental design and σ. -First get our data set with a normal distribution. +First get our data set with a normal distribution and set errors for given slope and intercept: ```{r} -mod<-function(p,x,y){ - B0=10 - B1=0.4 +set.seed(123) +x <- rnorm(24) +e <- rnorm(24, 0, 1) +y <- 10 + 0.4*x + e +y + +length(y) +length(x) + +df <- cbind(x, y) +df <- as.data.frame(df) +plot(df$x, df$y) +``` + +Regression - continuous + +```{r} +lm<-function(p,x,y){ + B0=p[1] + B1=p[2] sigma=exp(p[3]) - expected=B0+B1*x - model=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) - return(model) + lmmod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) + return(lmmod) } -newdata <- rnorm(24, 0, 1) -predict(mod, newdata) +lmmodguess <- c(1, 2, 3) +lmfitmod=optim(par=lmmodguess,fn=lm,x=df$x,y=df$y) +``` + +T-test - two groups + +```{r} +df$x1 <- ifelse(y>10, 1, 0) +df$x2 <- ifelse(y<10, 1, 0) + +ttest<-function(p,x,y){ + B0=p[1] + B1=p[2] + B2=p[3] + sigma=exp(p[4]) + expected=B0+B1*x1+B2*x2 + + x1=df[,3] + x2=df[,4] + + tmod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) + return(tmod) +} +tmodguess <- c(1, 2, 3, 4) +tfitmod=optim(par=tmodguess,fn=ttest,x=df[,3:4],y=df$y) ``` +Anova - four groups ```{r} -B1 <- 0.4 -B0 <- 10 -N <- 24 -data <- rnorm(24, 0, 1) -hist(data) +df$x1 <- ifelse(y< 9.353552, 1, 0) +df$x2 <- ifelse(9.353552>y & y <10.027724, 1, 0) +df$x3 <- ifelse(10.79904310.027724, 1, 0) +df$x4 <- ifelse(y>10.799043, 1, 0) + +anova<-function(p,x,y){ + B0=p[1] + B1=p[2] + B2=p[3] + B3=p[4] + B4=p[5] + sigma=exp(p[6]) + expected=B0+B1*x1+B2*x2 + + x1=df[,3] + x2=df[,4] + x3=df[,5] + x4=df[,6] + + anovamod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) + return(anovamod) +} + +anovamodguess <- c(1, 2, 3, 4, 5, 6) +anovafitmod=optim(par=anovamodguess,fn=ttest,x=df[,3:6],y=df$y) ``` +Comparing between: -To evaluate the relative statistical power of regression- and ANOVA-design experiments, simulate 10 random experiments with a regression design and 10 random experiments with a two-level ANOVA-design (really this is a t-test design) for each of eight values for σ (1, 2, 4, 6, 8, 12, 16, 24). Let’s say you are able to generate experimental units with x between 0 and 50. Repeat this process for a four- and eight-level ANOVA design (remember you only have 24 experimental units). Use the average p-value from likelihood ratio tests across your monte carlo runs as your metric of statistical power. +Regression vs. t.test -How does the ANOVA- vs regression-design perform? Does the relative performance of these experimental designs depend on the number of levels in the ANOVA experiment (2, 4, vs. 8)? +```{r} +t.stat <- 2*(lmfitmod$par[1] - tfitmod$par[1]) +t.stat # 9.767275 +1-pchisq(t.stat, df=3) # 0.02065159 +``` + +Regression vs. anova + +```{r} +t.stat <- 2*(lmfitmod$par[1] - anovafitmod$par[1]) +t.stat # 13.06448 + +1-pchisq(t.stat, df=5) # 0.02278192 +``` + + +t.test vs anova + +```{r} +t.stat <- 2*(tfitmod$par[1] - anovafitmod$par[1]) +t.stat # 3.297209 + +1-pchisq(t.stat, df=5) # 0.654269 +``` + + + + +To evaluate the relative statistical power of regression- and ANOVA-design experiments, simulate 10 random experiments with a regression design and 10 random experiments with a two-level ANOVA-design (really this is a t-test design) for each of eight values for σ (1, 2, 4, 6, 8, 12, 16, 24). + +Let’s say you are able to generate experimental units with x between 0 and 50. Repeat this process for a four- and eight-level ANOVA design (remember you only have 24 experimental units). Use the average p-value from likelihood ratio tests across your monte carlo runs as your metric of statistical power. + + +Data sets of 24 values in a normal distribution with varying sigma values: + +```{r} +x.1 <- rnorm(24, c(0:50), sd=1) +length(x.vec) +x.2 <- rnorm(24, c(0:50), sd=2) +x.4 <- rnorm(24, c(0:50), sd=4) +x.6 <- rnorm(24, c(0:50), sd=6) +x.8 <- rnorm(24, c(0:50), sd=8) +x.12 <- rnorm(24, c(0:50), sd=12) +x.16 <- rnorm(24, c(0:50), sd=16) +x.24 <- rnorm(24, c(0:50), sd=24) +``` + + +How does the ANOVA- vs regression-design perform? Does the relative performance of these experimental designs depend on the number of levels in the ANOVA experiment (2, 4, vs. 8)? From ccee8f0e805acfb9402c0b095e37a7a12392a0b2 Mon Sep 17 00:00:00 2001 From: haley Date: Mon, 3 Dec 2018 10:31:41 -0500 Subject: [PATCH 05/16] Haley's --- anovaVSregression.Rmd | 33 +++++++++++++++++++++++++++++++++ 1 file changed, 33 insertions(+) diff --git a/anovaVSregression.Rmd b/anovaVSregression.Rmd index 767dc60..62fb5bf 100755 --- a/anovaVSregression.Rmd +++ b/anovaVSregression.Rmd @@ -12,3 +12,36 @@ will first revisit ANOVA and regression analyses in two cases where it is clear An ANOVA-design and a regression-design experiment 1) A student conducted an experiment evaluating the effect of three different new antibiotics on growth of E. coli in lab cultures. Using the data in antibiotics.txt, generate a plot that summarizes the results and test for an effect of antibiotic treatments on the growth of E. coli using an ANOVA-design linear model and likelihood ratio test + +```{r} +# Load the data: +antibiotics = read.table("antibiotics.csv", sep = ',', header = T) +antibiotics$x1 = ifelse(antibiotics[,antibiotics$trt == "ab1"],1,0 ) +antibiotics$x1 = ifelse(antibiotics[antibiotics$trt == "ab1"],1,0 ) + + +View(antibiotics) +nllike2<-function(p,x,y){ #liklihood function for quadradic equation + #parameters: + B0=p[1] + B1=p[2] + B2=p[3] + sigma=exp(p[4]) + #deterministic model: + expected=B0+B1*x+B2*x^2 + #calculate negative log liklihood: + nll=-sum(dnorm(x=y,mean=expected,sd=sigma,log=TRUE)) + return(nll) +} + +initialGuess2=c(1,1,1,1) #a spot to start +fit2=optim(par=initialGuess2,fn=nllike,x=data2$x, x2=(data$x)^2, y=data2$y) + +print(fit2) + + +``` + +aov( ~ ) +lm ( ~ ) + From 761218eec88d7991f35fb6263f1b161fae2a78b9 Mon Sep 17 00:00:00 2001 From: enonnama <43544834+enonnama@users.noreply.github.com> Date: Wed, 5 Dec 2018 10:31:57 -0500 Subject: [PATCH 06/16] Add files via upload --- anovaVSregression.Rmd | 331 +++++++++++++++++++++--------------------- 1 file changed, 164 insertions(+), 167 deletions(-) diff --git a/anovaVSregression.Rmd b/anovaVSregression.Rmd index 5fa971d..0f1db45 100755 --- a/anovaVSregression.Rmd +++ b/anovaVSregression.Rmd @@ -5,14 +5,8 @@ output: html_document # PART 1 REGRESSION VS ANOVA -This project will focus on the difference between linear models that use categorical or continuous predictor variables. When designing experiments, scientists are often limited by the number of experimental units they can use. These practical limitations may arise due to the cost of using animal subjects, the number of tanks available for aquatic mesocosm studies, or simply the number of hours in the day. This question will focus on the optimal use of these precious experimental units to increase our "statistical power" to detect the effect of -a treatment. A experimental design question that arises when experimental units are limiting is whether to distribute the units amongst some number of replicated discrete treatment levels (ANOVA-design) or spread the experimental units along a continuous gradient of treatment levels (regression-design). In this project, we will first revisit ANOVA and regression analyses in two cases where it is clear what experimental design was chosen. We will then evaluate the ability of these two experimental designs to detect a treatment effect (i.e. their statistical power) using simulated data. - ## An ANOVA-design and a regression-design experiment -1) A student conducted an experiment evaluating the effect of three different new antibiotics on growth of E. coli in lab cultures. Using the data in antibiotics.txt, generate a plot that summarizes the results and test for an effect of antibiotic treatments on the growth of E. coli using an ANOVA-design linear model and -likelihood ratio test - ```{r include=FALSE} rm(list=ls()) setwd("/Users/emilynonnamaker/Box/personalStuff/School/PhD/biocomputingFall2018/biocomputing_StatsGroupProject") @@ -20,50 +14,40 @@ library(ggplot2) library(arm) antibiotics <- read.csv("antibiotics.csv") sugar <- read.csv("sugar.csv") + +#playing around - check out the data boxplot(growth ~ trt, data=antibiotics) lm <- lm(growth ~ sugar, data=sugar) aov <- aov(growth ~ trt, data=antibiotics) -# max likelihood with anova design -# likelihood custom function - need slope/parameter with each column of 0s and 1s. -# then lratio +# how a normal person would do this summary(aov) posthoc <- TukeyHSD(aov, "trt", conf.level=0.95) #a, b, ac, bd. posthoc -``` -Setting it up: +# Now to do the actual assingment: + +# set up your data: -```{r} N=length(antibiotics$growth) y=antibiotics$growth x<- antibiotics$trt -antibiotics$x1 <- ifelse(x=="ab1", 1, 0) -antibiotics$x2 <- ifelse(x=="ab2", 1, 0) -antibiotics$x3 <- ifelse(x=="ab3", 1, 0) -``` - -Building the null model: +antibiotics$x1 <- ifelse(x=="ab1", 1, 0) #dummy variables +antibiotics$x2 <- ifelse(x=="ab2", 1, 0) #dummy variables +antibiotics$x3 <- ifelse(x=="ab3", 1, 0) #dummy variables -```{r} -nullmod<-function(p,x,y){ +#Build null model +nullmod<-function(p,y){ B0=p[1] - sigma=p[2] + sigma=exp(p[2]) expected=B0 - x1=x[,1] - x2=x[,2] - x3=x[,3] - null=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) return(null) } -``` - -Building the full model: -```{r} +#Build full model fullmod<-function(p,x,y){ B0=p[1] B1=p[2] @@ -71,35 +55,27 @@ B2=p[3] B3=p[4] sigma=exp(p[5]) -x1=x[,1] -x2=x[,2] -x3=x[,3] +expected=B0+B1*x[,3]+B2*x[,4]+B3*x[,5] -expected=B0+B1*x1+B2*x2+B3*x3 -full=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) +full=-sum(dnorm(x=y,mean=expected,sd=sigma,log = TRUE)) return(full) } -``` -Building and checking our fits: +# Check fit +nullguess <- c(1, 2) +fullguess <- c(1, 2, 3, 4, 5) +fitnull=optim(par=nullguess,fn=nullmod,y=antibiotics$growth) +fitfull=optim(par=fullguess,fn=fullmod,x=antibiotics,y=antibiotics$growth) -```{r} -nullguess <- c(1, 3) -fullguess <- c(1, 2, 3, 4,5) -fitnull=optim(par=nullguess,fn=nullmod,x=antibiotics[,3:5],y=antibiotics$growth) -fitfull=optim(par=fullguess,fn=fullmod,x=antibiotics[,3:5],y=antibiotics$growth) -``` - -Test between null and full model: - -```{r} -t.stat <- 2*(fitfull$par[1] - fitnull$par[1]) -t.stat # 12.62689 +# Get t.statistic and p value. +df <- length(fitfull$par) - length(fitnull$par) +t.stat <- 2*(fitnull$value - fitfull$value) +t.stat # 25.69042 -1-pchisq(t.stat, df=4) # = 0.01325037 +1-pchisq(t.stat, df=df) ``` -Looks like there's a significant difference between the null and full anova model (t = 12.63, p < .05), suggesting that there are differences between treatments. +Looks like there's a significant difference between the null and full anova model (t = 25.69, p < .001), suggesting that there are differences between treatments. ## Graphical visualization! @@ -115,15 +91,10 @@ p + geom_jitter(shape=16, position=position_jitter(0.2), alpha=0.3) + theme_clas ## Regression 2) Another student conducted an experiment evaluating the effect of sugar concentration on growth of E. coli in lab cultures. Using the data in sugar.txt, generate a plot that summarizes the results and test for an effect of sugar concentration on growth of E. coli using a regression-design linear model and likelihood ratio test - -Building our linear model: - -Again, build the null model: - -```{r} + lmnullmod<-function(p,x,y){ B0=p[1] - sigma=p[2] + sigma=exp(p[2]) expected=B0 @@ -159,13 +130,14 @@ lmfitfull=optim(par=lmfullguess,fn=lmfullmod,x=sugar$sugar,y=sugar$growth) Now to compare: ```{r} -t.stat <- 2*(lmfitnull$par[1] - lmfitfull$par[1]) -t.stat # 19.78335 -1-pchisq(t.stat, df=2) # 5.059418e-05 +t.stat <- 2*(lmfitnull$value - lmfitfull$value) +t.stat # 39.92512 + +1-pchisq(t.stat, df=1) # 2.638878e-10 ``` -Comparing between the null model of no effect of sugar concentration on growth to the model taking this affect into account, we can see that sugar concentration does have a significant affect on growth (t = 12.78, p < .0001). Looking at the graph, we can tell that sugar concentration has a positive affect, with higher concentrations increasing growth. +Comparing between the null model of no effect of sugar concentration on growth to the model taking this affect into account, we can see that sugar concentration does have a significant affect on growth (t = 39.92512, p < .0001). Looking at the graph, we can tell that sugar concentration has a positive affect, with higher concentrations increasing growth. ## Visualization! @@ -176,31 +148,24 @@ a + geom_point() + coord_cartesian() + labs( x = "sugar concentration", y = "gro # Part 2 Statistical Power Analysis -Assume that an independent variable x is linearly related to a dependent variable y with a slope (β1) of 0.4 and a y-intercept (β0) of 10. Imagine that we have 24 experimental units that can be used to test for an effect of x on y. One could randomly distribute the experimental units along a gradient of x (regression design) or one could replicate two levels of x 12 times each and ask whether low or high levels of x generate different levels of y (t-test design). An intermediate approach would be to have 8 replicates of three levels of x or 4 replicates of six levels of x and so on (ANOVA design). +```{r} -Using the relationship between x and y defined by β1 = 0.4 and β0 = 10, we can generate a random dataset with 24 observations of y for any experimental design of interest given some level of error in our observations. We define our error by the standard deviation (σ) of normally distributed errors that we can add to our simulated values of y. Because our results will vary a bit from one simulated data set to another, it is a good idea to adopt something called a monte carlo approach where we run multiple (say 10) versions of a power analysis for a given experimental design and σ. +######################## +# Put in your models # +####################### -First get our data set with a normal distribution and set errors for given slope and intercept: +# Null model -```{r} -set.seed(123) -x <- rnorm(24) -e <- rnorm(24, 0, 1) -y <- 10 + 0.4*x + e -y - -length(y) -length(x) - -df <- cbind(x, y) -df <- as.data.frame(df) -plot(df$x, df$y) -``` - -Regression - continuous +nullmod<- function(p,x,y){ + B0=p[1] + sigma=exp(p[2]) + expected=B0 + null=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) + return(null) +} -```{r} -lm<-function(p,x,y){ +# Regression model +lin<-function(p,x,y){ B0=p[1] B1=p[2] sigma=exp(p[3]) @@ -210,115 +175,147 @@ lm<-function(p,x,y){ return(lmmod) } -lmmodguess <- c(1, 2, 3) -lmfitmod=optim(par=lmmodguess,fn=lm,x=df$x,y=df$y) -``` - -T-test - two groups - -```{r} -df$x1 <- ifelse(y>10, 1, 0) -df$x2 <- ifelse(y<10, 1, 0) - +# T-test model ttest<-function(p,x,y){ B0=p[1] B1=p[2] - B2=p[3] - sigma=exp(p[4]) - expected=B0+B1*x1+B2*x2 - - x1=df[,3] - x2=df[,4] + sigma=exp(p[3]) + x1=x + expected=B0+B1*x1 tmod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) return(tmod) } -tmodguess <- c(1, 2, 3, 4) -tfitmod=optim(par=tmodguess,fn=ttest,x=df[,3:4],y=df$y) -``` - -Anova - four groups +# Anova - four groups +anovamod<-function(p,x,y){ +B0=p[1] +B1=p[2] +B2=p[3] +B3=p[4] +sigma=exp(p[5]) -```{r} -df$x1 <- ifelse(y< 9.353552, 1, 0) -df$x2 <- ifelse(9.353552>y & y <10.027724, 1, 0) -df$x3 <- ifelse(10.79904310.027724, 1, 0) -df$x4 <- ifelse(y>10.799043, 1, 0) +expected=B0+B1*x[,1]+B2*x[,2]+B3*x[,3] +full=-sum(dnorm(x=y,mean=expected,sd=sigma,log = TRUE)) +return(full) +} #optim-call call dummy -anova<-function(p,x,y){ +# Now an anova with 8 groups +eightanovamod<-function(p,x,y){ B0=p[1] B1=p[2] B2=p[3] B3=p[4] B4=p[5] - sigma=exp(p[6]) - expected=B0+B1*x1+B2*x2 - - x1=df[,3] - x2=df[,4] - x3=df[,5] - x4=df[,6] - - anovamod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) - return(anovamod) + B5=p[6] + B6=p[7] + B7=p[8] + sigma=exp(p[9]) + expected=B0+B1*x[,1]+B2*x[,2]+B3*x[,3]+B4*x[,4]+B5*x[,5]+B6*x[,6]+B7*x[,7] + eightanovamod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) + return(eightanovamod) } -anovamodguess <- c(1, 2, 3, 4, 5, 6) -anovafitmod=optim(par=anovamodguess,fn=ttest,x=df[,3:6],y=df$y) -``` - -Comparing between: - -Regression vs. t.test - -```{r} -t.stat <- 2*(lmfitmod$par[1] - tfitmod$par[1]) -t.stat # 9.767275 - -1-pchisq(t.stat, df=3) # 0.02065159 -``` - -Regression vs. anova - -```{r} -t.stat <- 2*(lmfitmod$par[1] - anovafitmod$par[1]) -t.stat # 13.06448 +### Now do your for loops! + +# Set up an empty data frame in which to store your p-values for each sigma and model. +dfp <- setNames(data.frame(matrix(ncol = 5, nrow = 8)), c("Sigmas", "Regression", "T.test", "Anova", "Eight.Anova")) + +# Set up your vector of sigmas to loop through +sigmas <- c(1,2,4,6,8,12,16,24) + +# Write your nested for loop: +for (i in 1:length(sigmas)){ # sigma runs + # set up your vectors for your p-values from each monte-carlo run. + reg.p <- numeric(10) # changes each time you use a new sigma + t.p <- numeric(10) + anova.p <- numeric(10) + eightanova.p <- numeric(10) + for (j in 1:10){ # monte carlo runs + x <- runif(24, 0, 50) # build set of x's + e <- rnorm(24, 0, sd = sigmas[i]) # put in your error + y <- 10 + 0.4*x + e # write your linear equation + df <- cbind(x, y) # bind x and y into a data frame + df <- as.data.frame(df) + df <- df[order(df$x),] # order for later sorting. + + # dummy coding: + + # For t-test + tdummy <- matrix(0, 24, 1) + tdummy[13:24,1] <- 1 + + # For anova + dummy <- matrix(0, 24, 3) + dummy[7:12,1] <- 1 + dummy[13:18,2] <-1 + dummy[19:24,3] <-1 + + #for eight-level anova + edummy <- matrix(0, 24, 7) + edummy[4:6,1]<-1 + edummy[7:9,2]<-1 + edummy[10:12,3]<-1 + edummy[13:15,4]<-1 + edummy[16:18,5]<-1 + edummy[19:21,6]<-1 + edummy[22:24,7]<-1 + + # guesses + nullmodguess <- c(1,2) + lmmodguess <- c(1, 2, 3) + tmodguess<-c(1,2,3,4,5) + aovguess<- c(1,2,3,4,5) + eightaovguess<-c(1,2,3,4,5,6,7,8,9) + + #optims + nullfitmod=optim(par=nullmodguess,fn=nullmod,x=df$x,y=df$y) + lmfitmod=optim(par=lmmodguess,fn=lin,x=df$x,y=df$y) + tfitmod=optim(par=tmodguess,fn=ttest,x=tdummy,y=df$y) + anovafitmod=optim(par=aovguess,fn=anovamod,x=dummy,y=df$y) + eightanovafitmod=optim(par=eightaovguess,fn=eightanovamod,x=edummy,y=df$y) + + # degrees of freedom + degfreg <- length(lmfitmod$par) - length(nullfitmod$par) + degft <- length(tfitmod$par) - length(nullfitmod$par) + degfanova <- length(anovafitmod$par) - length(nullfitmod$par) + degfeightanova <- length(eightanovafitmod$par) - length(nullfitmod$par) + + # test stats + t.statreg <- 2*(nullfitmod$value - lmfitmod$value) + t.statt <- 2*(nullfitmod$value - tfitmod$value) + t.statanova <- 2*(nullfitmod$value - anovafitmod$value) + t.stateightanova <- 2*(nullfitmod$value - eightanovafitmod$value) + + # pvalues + reg.p[j] <- 1-pchisq(t.statreg, df=degfreg) + t.p[j]<- 1-pchisq(t.statt, df=degft) + anova.p[j]<- 1-pchisq(t.statanova, df=degfanova) + eightanova.p[j]<- 1-pchisq(t.stateightanova, df=degfeightanova) + } + dfp[,1]<- c(1, 2, 4, 6, 8, 12, 16, 24) + dfp[i,2] <- mean(reg.p) + dfp[i,3] <- mean(t.p) + dfp[i,4] <- mean(anova.p) + dfp[i,5] <- mean(eightanova.p) +} -1-pchisq(t.stat, df=5) # 0.02278192 ``` +Scatterplot with each model as a different color! +Linear model should be best -t.test vs anova +## Visualize!! ```{r} -t.stat <- 2*(tfitmod$par[1] - anovafitmod$par[1]) -t.stat # 3.297209 - -1-pchisq(t.stat, df=5) # 0.654269 +pvalues <- read.csv("P-values.csv") +Model <- pvalues$Model +p <- ggplot(pvalues, aes(x=pvalues$Sigmas, y=pvalues$Mean_p, colour=Model)) + geom_point (size=2) + xlab("Sigma Values") + ylab("Mean p-values") +p + ``` +The regression model performs best across the sigmas, as evidenced by it's low p-value scores. The eight level anova should perform close to the regression, but there's something wrong with our code. After that, the four level anova should perform in between the t-test and the regression, again our code is wrong. Finally the t-test, as expected, performed worse (should have). Our model is linear, so a linear regression should fit best, followed by models that break up the data into larger numbers of groups (closer to a linear model). +It's clearly best to try to fit linear data to a linear model, but when resources are constrained, it seems the best strategy is to fit as many groups as possible. In this case, you need at least four experimental units (four way anova), to detect a significant effect. - -To evaluate the relative statistical power of regression- and ANOVA-design experiments, simulate 10 random experiments with a regression design and 10 random experiments with a two-level ANOVA-design (really this is a t-test design) for each of eight values for σ (1, 2, 4, 6, 8, 12, 16, 24). - -Let’s say you are able to generate experimental units with x between 0 and 50. Repeat this process for a four- and eight-level ANOVA design (remember you only have 24 experimental units). Use the average p-value from likelihood ratio tests across your monte carlo runs as your metric of statistical power. - - -Data sets of 24 values in a normal distribution with varying sigma values: - -```{r} -x.1 <- rnorm(24, c(0:50), sd=1) -length(x.vec) -x.2 <- rnorm(24, c(0:50), sd=2) -x.4 <- rnorm(24, c(0:50), sd=4) -x.6 <- rnorm(24, c(0:50), sd=6) -x.8 <- rnorm(24, c(0:50), sd=8) -x.12 <- rnorm(24, c(0:50), sd=12) -x.16 <- rnorm(24, c(0:50), sd=16) -x.24 <- rnorm(24, c(0:50), sd=24) -``` - - -How does the ANOVA- vs regression-design perform? Does the relative performance of these experimental designs depend on the number of levels in the ANOVA experiment (2, 4, vs. 8)? From d941ba28f91c0d288a14af98230fd0b70147ac37 Mon Sep 17 00:00:00 2001 From: enonnama <43544834+enonnama@users.noreply.github.com> Date: Wed, 5 Dec 2018 11:07:20 -0500 Subject: [PATCH 07/16] Peer review for Fleishman and Mears --- Peer Review Comments.Rmd | 34 ++++++++++++++++++++++++++++++++++ 1 file changed, 34 insertions(+) create mode 100644 Peer Review Comments.Rmd diff --git a/Peer Review Comments.Rmd b/Peer Review Comments.Rmd new file mode 100644 index 0000000..8bedaab --- /dev/null +++ b/Peer Review Comments.Rmd @@ -0,0 +1,34 @@ +--- +title: "Peer Review" +output: html_document +--- + +# Part 1: + +## ANOVA + +We like that this group used the following code: + +```{r} +pchisq(teststat,df, lower=F) +``` + +Instead of doing 1-pchisq. Nifty. + + +For lines 79, 81, and 83, include the results in hashtags, for more convienent comparisons. + +Add an abline in your regression plot. + + +## Regression + +Looks great. + +# Part 2: + +Combine scripts for part 2 into one script. + +In the ANOVA Monte carlo, delete the bottom lines to reduce reader confusion. + +At the end, if time, build a graph to visually display the p-values. \ No newline at end of file From 3c0ded1fb2ed5a9a960a636d9a2612944d44d0d3 Mon Sep 17 00:00:00 2001 From: enonnama <43544834+enonnama@users.noreply.github.com> Date: Wed, 5 Dec 2018 17:25:31 -0500 Subject: [PATCH 08/16] Add files via upload --- anovaVSregression.Rmd | 124 +++++++++++++++++++++++++----------------- 1 file changed, 75 insertions(+), 49 deletions(-) diff --git a/anovaVSregression.Rmd b/anovaVSregression.Rmd index 0f1db45..f1b3236 100755 --- a/anovaVSregression.Rmd +++ b/anovaVSregression.Rmd @@ -12,6 +12,8 @@ rm(list=ls()) setwd("/Users/emilynonnamaker/Box/personalStuff/School/PhD/biocomputingFall2018/biocomputing_StatsGroupProject") library(ggplot2) library(arm) +library(tidyr) +library(reshape2) antibiotics <- read.csv("antibiotics.csv") sugar <- read.csv("sugar.csv") @@ -24,20 +26,22 @@ aov <- aov(growth ~ trt, data=antibiotics) summary(aov) posthoc <- TukeyHSD(aov, "trt", conf.level=0.95) #a, b, ac, bd. posthoc +``` -# Now to do the actual assingment: +## Now to do the actual assingment: +```{r} # set up your data: -N=length(antibiotics$growth) -y=antibiotics$growth +N <- length(antibiotics$growth) +y <- antibiotics$growth x<- antibiotics$trt antibiotics$x1 <- ifelse(x=="ab1", 1, 0) #dummy variables antibiotics$x2 <- ifelse(x=="ab2", 1, 0) #dummy variables antibiotics$x3 <- ifelse(x=="ab3", 1, 0) #dummy variables #Build null model -nullmod<-function(p,y){ +nullmod<-function(p,x,y){ B0=p[1] sigma=exp(p[2]) @@ -63,16 +67,16 @@ return(full) # Check fit nullguess <- c(1, 2) -fullguess <- c(1, 2, 3, 4, 5) -fitnull=optim(par=nullguess,fn=nullmod,y=antibiotics$growth) -fitfull=optim(par=fullguess,fn=fullmod,x=antibiotics,y=antibiotics$growth) +fullguess <- c(18, -12, -3, -4, 1) +fitnull=optim(par=nullguess,fn=nullmod,y=antibiotics$growth) # converges +fitfull=optim(par=fullguess,fn=fullmod,x=antibiotics,y=antibiotics$growth) # converges # Get t.statistic and p value. df <- length(fitfull$par) - length(fitnull$par) t.stat <- 2*(fitnull$value - fitfull$value) -t.stat # 25.69042 +t.stat # 37.90134 -1-pchisq(t.stat, df=df) +1-pchisq(t.stat, df=df) # 2.965739e-08 ``` Looks like there's a significant difference between the null and full anova model (t = 25.69, p < .001), suggesting that there are differences between treatments. @@ -85,14 +89,14 @@ Using a posthoc Tukey test (see hidden code) we can find significant differences p <- ggplot(antibiotics, aes(x=trt, y=growth, fill=trt)) + geom_boxplot() p + geom_jitter(shape=16, position=position_jitter(0.2), alpha=0.3) + theme_classic() + labs(x = "Treatment", y= "Growth") + annotate("text", x = c(1, 2, 3, 4), y = c(23, 23, 23, 23), label = c("a", "b", "ac", "bd")) - ``` ## Regression -2) Another student conducted an experiment evaluating the effect of sugar concentration on growth of E. coli in lab cultures. Using the data in sugar.txt, generate a plot that summarizes the results and test for an effect of sugar concentration on growth of E. coli using a regression-design linear model and likelihood ratio test +Build the null model: -lmnullmod<-function(p,x,y){ +```{r} +lmnullmodreg<-function(p,x,y){ B0=p[1] sigma=exp(p[2]) @@ -106,7 +110,7 @@ return(null) Now build the extension: ```{r} -lmfullmod<-function(p,x,y){ +lmfullmodreg<-function(p,x,y){ B0=p[1] B1=p[2] sigma=exp(p[3]) @@ -121,23 +125,22 @@ lmfullmod<-function(p,x,y){ Give it a guess: ```{r} -lmnullguess <- c(1, 1) -lmfullguess <- c(1, 2, 3) -lmfitnull=optim(par=lmnullguess,fn=lmnullmod,x=sugar$sugar,y=sugar$growth) -lmfitfull=optim(par=lmfullguess,fn=lmfullmod,x=sugar$sugar,y=sugar$growth) +lmnullregguess <- c(1, 1) +lmfullregguess <- c(1, 2, 3) +lmfitregnull=optim(par=lmnullregguess,fn=lmnullmodreg,x=sugar$sugar,y=sugar$growth) # converges +lmfitregfull=optim(par=lmfullregguess,fn=lmfullmodreg,x=sugar$sugar,y=sugar$growth) # converges ``` Now to compare: ```{r} - -t.stat <- 2*(lmfitnull$value - lmfitfull$value) -t.stat # 39.92512 - -1-pchisq(t.stat, df=1) # 2.638878e-10 +t.statreg <- 2*(lmfitregnull$value - lmfitregfull$value) +t.statreg # 39.92512 +dfreg <- length(lmfitregfull$par) - length(lmfitregnull$par) +1-pchisq(t.stat, df=dfreg) # 7.441429e-10, woot ``` -Comparing between the null model of no effect of sugar concentration on growth to the model taking this affect into account, we can see that sugar concentration does have a significant affect on growth (t = 39.92512, p < .0001). Looking at the graph, we can tell that sugar concentration has a positive affect, with higher concentrations increasing growth. +Comparing between the null model of no effect of sugar concentration on growth to the model taking this affect into account, we can see that sugar concentration does have a significant affect on growth (t = 39.92512, p < .0001). Looking at the graph below, we can tell that sugar concentration has a positive affect, with higher concentrations of sugar increasing growth. ## Visualization! @@ -149,17 +152,13 @@ a + geom_point() + coord_cartesian() + labs( x = "sugar concentration", y = "gro # Part 2 Statistical Power Analysis ```{r} - -######################## -# Put in your models # -####################### - # Null model - nullmod<- function(p,x,y){ B0=p[1] sigma=exp(p[2]) + expected=B0 + null=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) return(null) } @@ -169,6 +168,7 @@ lin<-function(p,x,y){ B0=p[1] B1=p[2] sigma=exp(p[3]) + expected=B0+B1*x lmmod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) @@ -187,7 +187,7 @@ ttest<-function(p,x,y){ return(tmod) } -# Anova - four groups +# ANOVA - four groups anovamod<-function(p,x,y){ B0=p[1] B1=p[2] @@ -196,11 +196,12 @@ B3=p[4] sigma=exp(p[5]) expected=B0+B1*x[,1]+B2*x[,2]+B3*x[,3] -full=-sum(dnorm(x=y,mean=expected,sd=sigma,log = TRUE)) -return(full) -} #optim-call call dummy -# Now an anova with 8 groups +anova=-sum(dnorm(x=y,mean=expected,sd=sigma,log = TRUE)) +return(anova) +} + +# Now an ANOVA with 8 groups eightanovamod<-function(p,x,y){ B0=p[1] B1=p[2] @@ -211,13 +212,18 @@ eightanovamod<-function(p,x,y){ B6=p[7] B7=p[8] sigma=exp(p[9]) + expected=B0+B1*x[,1]+B2*x[,2]+B3*x[,3]+B4*x[,4]+B5*x[,5]+B6*x[,6]+B7*x[,7] + eightanovamod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) return(eightanovamod) } -### Now do your for loops! +``` +## Now do your for loops! + +```{r} # Set up an empty data frame in which to store your p-values for each sigma and model. dfp <- setNames(data.frame(matrix(ncol = 5, nrow = 8)), c("Sigmas", "Regression", "T.test", "Anova", "Eight.Anova")) @@ -262,18 +268,18 @@ for (i in 1:length(sigmas)){ # sigma runs edummy[22:24,7]<-1 # guesses - nullmodguess <- c(1,2) - lmmodguess <- c(1, 2, 3) - tmodguess<-c(1,2,3,4,5) - aovguess<- c(1,2,3,4,5) - eightaovguess<-c(1,2,3,4,5,6,7,8,9) + nullmodguess <- c(1,2) # 2 + lmmodguess <- c(1, 2, 3) # 3 + tmodguess<-c(1,1,1,1,1) # 5 + aovguess<- c(9,10,-6,8,2) # 5 + eightaovguess<-c(1, 12.5,15,17.5,20,22.5,25,27.5,30) # 9 #optims - nullfitmod=optim(par=nullmodguess,fn=nullmod,x=df$x,y=df$y) - lmfitmod=optim(par=lmmodguess,fn=lin,x=df$x,y=df$y) - tfitmod=optim(par=tmodguess,fn=ttest,x=tdummy,y=df$y) - anovafitmod=optim(par=aovguess,fn=anovamod,x=dummy,y=df$y) - eightanovafitmod=optim(par=eightaovguess,fn=eightanovamod,x=edummy,y=df$y) + nullfitmod=optim(par=nullmodguess,fn=nullmod,x=df$x,y=df$y) # converges + lmfitmod=optim(par=lmmodguess,fn=lin,x=df$x,y=df$y) # converges + tfitmod=optim(par=tmodguess,fn=ttest,x=tdummy,y=df$y) # converges + anovafitmod=optim(par=aovguess,fn=anovamod,x=dummy,y=df$y, control=list(maxit=1e5)) # converges + eightanovafitmod=optim(par=eightaovguess,fn=eightanovamod,x=edummy,y=df$y, control=list(maxit=1e5)) # converges # degrees of freedom degfreg <- length(lmfitmod$par) - length(nullfitmod$par) @@ -299,23 +305,43 @@ for (i in 1:length(sigmas)){ # sigma runs dfp[i,4] <- mean(anova.p) dfp[i,5] <- mean(eightanova.p) } +``` +Now reshape your dataframe of p-values into something you can plot! + +```{r} +mframe <- melt(dfp) +mframe <- mframe[9:40,] +allsigmas <- c(1, 2, 4, 6, 8, 12, 16, 24, 1, 2, 4, 6, 8,12, 16, 24, 1, 2, 4, 6, 8, 12, 16, 24, 1, 2, 4, 6, 8, 12, 16, 24) +mframe$sigma <- allsigmas ``` Scatterplot with each model as a different color! Linear model should be best +NOTE 8 LEVEL ANOVA STILL NOT WORKING + ## Visualize!! ```{r} -pvalues <- read.csv("P-values.csv") -Model <- pvalues$Model -p <- ggplot(pvalues, aes(x=pvalues$Sigmas, y=pvalues$Mean_p, colour=Model)) + geom_point (size=2) + xlab("Sigma Values") + ylab("Mean p-values") +mframe <- as.data.frame(mframe) +p <- ggplot(mframe, aes(x=mframe$sigma, y=mframe$value, colour=variable)) + geom_point (size=2) + xlab("Sigma Values") + ylab("Mean p-values") p ``` -The regression model performs best across the sigmas, as evidenced by it's low p-value scores. The eight level anova should perform close to the regression, but there's something wrong with our code. After that, the four level anova should perform in between the t-test and the regression, again our code is wrong. Finally the t-test, as expected, performed worse (should have). Our model is linear, so a linear regression should fit best, followed by models that break up the data into larger numbers of groups (closer to a linear model). +The regression model performs best across the sigmas, as evidenced by it's low p-value scores. The eight level anova should perform close to the regression, but there's something wrong with our code. The four level anova performs in between the t-test and the regression. Finally the t-test, as expected, performed worse than either the regression or the four level ANOVA (and should perform worse than the eight-level as well). This all makes sense, as our data is linearly correlated, so a linear regression should fit best, followed by models that break up the data into larger numbers of groups (closer to a linear model). It's clearly best to try to fit linear data to a linear model, but when resources are constrained, it seems the best strategy is to fit as many groups as possible. In this case, you need at least four experimental units (four way anova), to detect a significant effect. + + +```{r, include=FALSE} +# get values for the eight level anova guess: +x <- c(6.25, 12.5, 18.72, 25, 31.25, 37.5, 43.75, 50) +y.vec <- rep(NA, 8) +for (i in 1:length(x)){ + y[i] <- 10 + 0.4*x[i] + y.vec <- y +} +``` From 09a165a5b35479a23cbd8b10de63a9603f73e0af Mon Sep 17 00:00:00 2001 From: haley Date: Thu, 6 Dec 2018 11:45:19 -0500 Subject: [PATCH 09/16] Final changes? --- anovaVSregression.Rmd | 362 ++---------------------------------------- statPowerAnalysis.Rmd | 2 + 2 files changed, 12 insertions(+), 352 deletions(-) diff --git a/anovaVSregression.Rmd b/anovaVSregression.Rmd index 20a23cf..be0e1ee 100755 --- a/anovaVSregression.Rmd +++ b/anovaVSregression.Rmd @@ -1,4 +1,3 @@ -<<<<<<< HEAD --- title: "ANOVA vs Regression" output: html_document @@ -72,11 +71,14 @@ fullguess <- c(18, -12, -3, -4, 1) fitnull=optim(par=nullguess,fn=nullmod,y=antibiotics$growth) # converges fitfull=optim(par=fullguess,fn=fullmod,x=antibiotics,y=antibiotics$growth) # converges -# Get t.statistic and p value. -df <- length(fitfull$par) - length(fitnull$par) +# Get t.statistic . t.stat <- 2*(fitnull$value - fitfull$value) t.stat # 37.90134 +# get degrees of freedom +df <- length(fitfull$par) - length(fitnull$par) + +# and p value 1-pchisq(t.stat, df=df) # 2.965739e-08 ``` @@ -136,9 +138,14 @@ lmfitregfull=optim(par=lmfullregguess,fn=lmfullmodreg,x=sugar$sugar,y=sugar$grow Now to compare: ```{r} +# Get t statistic t.statreg <- 2*(lmfitregnull$value - lmfitregfull$value) t.statreg # 39.92512 + +# get degrees of freedom dfreg <- length(lmfitregfull$par) - length(lmfitregnull$par) + +#and p value 1-pchisq(t.stat, df=dfreg) # 7.441429e-10, woot ``` @@ -150,352 +157,3 @@ Comparing between the null model of no effect of sugar concentration on growth t a <- ggplot(data=sugar,aes(x=sugar,y=growth)) a + geom_point() + coord_cartesian() + labs( x = "sugar concentration", y = "growth") + theme_classic() + geom_smooth(method = "lm") ``` -======= ---- -title: "ANOVA and Regression Results" -output: html_document ---- - -# PART 1 REGRESSION VS ANOVA - -## An ANOVA-design and a regression-design experiment - -```{r include=FALSE} -rm(list=ls()) -setwd("/Users/emilynonnamaker/Box/personalStuff/School/PhD/biocomputingFall2018/biocomputing_StatsGroupProject") -library(ggplot2) -library(arm) -library(tidyr) -library(reshape2) -antibiotics <- read.csv("antibiotics.csv") -sugar <- read.csv("sugar.csv") - -#playing around - check out the data -boxplot(growth ~ trt, data=antibiotics) -lm <- lm(growth ~ sugar, data=sugar) -aov <- aov(growth ~ trt, data=antibiotics) - -# how a normal person would do this -summary(aov) -posthoc <- TukeyHSD(aov, "trt", conf.level=0.95) #a, b, ac, bd. -posthoc -``` - -## Now to do the actual assingment: - -```{r} -# set up your data: - -N <- length(antibiotics$growth) -y <- antibiotics$growth -x<- antibiotics$trt -antibiotics$x1 <- ifelse(x=="ab1", 1, 0) #dummy variables -antibiotics$x2 <- ifelse(x=="ab2", 1, 0) #dummy variables -antibiotics$x3 <- ifelse(x=="ab3", 1, 0) #dummy variables - -#Build null model -nullmod<-function(p,x,y){ - B0=p[1] - sigma=exp(p[2]) - - expected=B0 - -null=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) -return(null) -} - -#Build full model -fullmod<-function(p,x,y){ -B0=p[1] -B1=p[2] -B2=p[3] -B3=p[4] -sigma=exp(p[5]) - -expected=B0+B1*x[,3]+B2*x[,4]+B3*x[,5] - -full=-sum(dnorm(x=y,mean=expected,sd=sigma,log = TRUE)) -return(full) -} - -# Check fit -nullguess <- c(1, 2) -fullguess <- c(18, -12, -3, -4, 1) -fitnull=optim(par=nullguess,fn=nullmod,y=antibiotics$growth) # converges -fitfull=optim(par=fullguess,fn=fullmod,x=antibiotics,y=antibiotics$growth) # converges - -# Get t.statistic and p value. -df <- length(fitfull$par) - length(fitnull$par) -t.stat <- 2*(fitnull$value - fitfull$value) -t.stat # 37.90134 - -1-pchisq(t.stat, df=df) # 2.965739e-08 -``` - -Looks like there's a significant difference between the null and full anova model (t = 25.69, p < .001), suggesting that there are differences between treatments. - -## Graphical visualization! - -Using a posthoc Tukey test (see hidden code) we can find significant differences between groups, and add these differences as letters to our visualization (see graph below) - -```{r} -p <- ggplot(antibiotics, aes(x=trt, y=growth, fill=trt)) + - geom_boxplot() -p + geom_jitter(shape=16, position=position_jitter(0.2), alpha=0.3) + theme_classic() + labs(x = "Treatment", y= "Growth") + annotate("text", x = c(1, 2, 3, 4), y = c(23, 23, 23, 23), label = c("a", "b", "ac", "bd")) -``` - -## Regression - -Build the null model: - -```{r} -lmnullmodreg<-function(p,x,y){ - B0=p[1] - sigma=exp(p[2]) - - expected=B0 - -null=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) -return(null) -} -``` - -Now build the extension: - -```{r} -lmfullmodreg<-function(p,x,y){ - B0=p[1] - B1=p[2] - sigma=exp(p[3]) - - expected=B0+B1*x - - null=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) - return(null) -} -``` - -Give it a guess: - -```{r} -lmnullregguess <- c(1, 1) -lmfullregguess <- c(1, 2, 3) -lmfitregnull=optim(par=lmnullregguess,fn=lmnullmodreg,x=sugar$sugar,y=sugar$growth) # converges -lmfitregfull=optim(par=lmfullregguess,fn=lmfullmodreg,x=sugar$sugar,y=sugar$growth) # converges -``` - -Now to compare: - -```{r} -t.statreg <- 2*(lmfitregnull$value - lmfitregfull$value) -t.statreg # 39.92512 -dfreg <- length(lmfitregfull$par) - length(lmfitregnull$par) -1-pchisq(t.stat, df=dfreg) # 7.441429e-10, woot -``` - -Comparing between the null model of no effect of sugar concentration on growth to the model taking this affect into account, we can see that sugar concentration does have a significant affect on growth (t = 39.92512, p < .0001). Looking at the graph below, we can tell that sugar concentration has a positive affect, with higher concentrations of sugar increasing growth. - -## Visualization! - -```{r} -a <- ggplot(data=sugar,aes(x=sugar,y=growth)) -a + geom_point() + coord_cartesian() + labs( x = "sugar concentration", y = "growth") + theme_classic() + geom_smooth(method = "lm") -``` - -# Part 2 Statistical Power Analysis - -```{r} -# Null model -nullmod<- function(p,x,y){ - B0=p[1] - sigma=exp(p[2]) - - expected=B0 - - null=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) - return(null) -} - -# Regression model -lin<-function(p,x,y){ - B0=p[1] - B1=p[2] - sigma=exp(p[3]) - - expected=B0+B1*x - - lmmod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) - return(lmmod) -} - -# T-test model -ttest<-function(p,x,y){ - B0=p[1] - B1=p[2] - sigma=exp(p[3]) - x1=x - expected=B0+B1*x1 - - tmod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) - return(tmod) -} - -# ANOVA - four groups -anovamod<-function(p,x,y){ -B0=p[1] -B1=p[2] -B2=p[3] -B3=p[4] -sigma=exp(p[5]) - -expected=B0+B1*x[,1]+B2*x[,2]+B3*x[,3] - -anova=-sum(dnorm(x=y,mean=expected,sd=sigma,log = TRUE)) -return(anova) -} - -# Now an ANOVA with 8 groups -eightanovamod<-function(p,x,y){ - B0=p[1] - B1=p[2] - B2=p[3] - B3=p[4] - B4=p[5] - B5=p[6] - B6=p[7] - B7=p[8] - sigma=exp(p[9]) - - expected=B0+B1*x[,1]+B2*x[,2]+B3*x[,3]+B4*x[,4]+B5*x[,5]+B6*x[,6]+B7*x[,7] - - eightanovamod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) - return(eightanovamod) -} - -``` - -## Now do your for loops! - -```{r} -# Set up an empty data frame in which to store your p-values for each sigma and model. -dfp <- setNames(data.frame(matrix(ncol = 5, nrow = 8)), c("Sigmas", "Regression", "T.test", "Anova", "Eight.Anova")) - -# Set up your vector of sigmas to loop through -sigmas <- c(1,2,4,6,8,12,16,24) - -# Write your nested for loop: -for (i in 1:length(sigmas)){ # sigma runs - # set up your vectors for your p-values from each monte-carlo run. - reg.p <- numeric(10) # changes each time you use a new sigma - t.p <- numeric(10) - anova.p <- numeric(10) - eightanova.p <- numeric(10) - for (j in 1:10){ # monte carlo runs - x <- runif(24, 0, 50) # build set of x's - e <- rnorm(24, 0, sd = sigmas[i]) # put in your error - y <- 10 + 0.4*x + e # write your linear equation - df <- cbind(x, y) # bind x and y into a data frame - df <- as.data.frame(df) - df <- df[order(df$x),] # order for later sorting. - - # dummy coding: - - # For t-test - tdummy <- matrix(0, 24, 1) - tdummy[13:24,1] <- 1 - - # For anova - dummy <- matrix(0, 24, 3) - dummy[7:12,1] <- 1 - dummy[13:18,2] <-1 - dummy[19:24,3] <-1 - - #for eight-level anova - edummy <- matrix(0, 24, 7) - edummy[4:6,1]<-1 - edummy[7:9,2]<-1 - edummy[10:12,3]<-1 - edummy[13:15,4]<-1 - edummy[16:18,5]<-1 - edummy[19:21,6]<-1 - edummy[22:24,7]<-1 - - # guesses - nullmodguess <- c(1,2) # 2 - lmmodguess <- c(1, 2, 3) # 3 - tmodguess<-c(1,1,1,1,1) # 5 - aovguess<- c(9,10,-6,8,2) # 5 - eightaovguess<-c(1, 12.5,15,17.5,20,22.5,25,27.5,30) # 9 - - #optims - nullfitmod=optim(par=nullmodguess,fn=nullmod,x=df$x,y=df$y) # converges - lmfitmod=optim(par=lmmodguess,fn=lin,x=df$x,y=df$y) # converges - tfitmod=optim(par=tmodguess,fn=ttest,x=tdummy,y=df$y) # converges - anovafitmod=optim(par=aovguess,fn=anovamod,x=dummy,y=df$y, control=list(maxit=1e5)) # converges - eightanovafitmod=optim(par=eightaovguess,fn=eightanovamod,x=edummy,y=df$y, control=list(maxit=1e5)) # converges - - # degrees of freedom - degfreg <- length(lmfitmod$par) - length(nullfitmod$par) - degft <- length(tfitmod$par) - length(nullfitmod$par) - degfanova <- length(anovafitmod$par) - length(nullfitmod$par) - degfeightanova <- length(eightanovafitmod$par) - length(nullfitmod$par) - - # test stats - t.statreg <- 2*(nullfitmod$value - lmfitmod$value) - t.statt <- 2*(nullfitmod$value - tfitmod$value) - t.statanova <- 2*(nullfitmod$value - anovafitmod$value) - t.stateightanova <- 2*(nullfitmod$value - eightanovafitmod$value) - - # pvalues - reg.p[j] <- 1-pchisq(t.statreg, df=degfreg) - t.p[j]<- 1-pchisq(t.statt, df=degft) - anova.p[j]<- 1-pchisq(t.statanova, df=degfanova) - eightanova.p[j]<- 1-pchisq(t.stateightanova, df=degfeightanova) - } - dfp[,1]<- c(1, 2, 4, 6, 8, 12, 16, 24) - dfp[i,2] <- mean(reg.p) - dfp[i,3] <- mean(t.p) - dfp[i,4] <- mean(anova.p) - dfp[i,5] <- mean(eightanova.p) -} -``` - -Now reshape your dataframe of p-values into something you can plot! - -```{r} -mframe <- melt(dfp) -mframe <- mframe[9:40,] -allsigmas <- c(1, 2, 4, 6, 8, 12, 16, 24, 1, 2, 4, 6, 8,12, 16, 24, 1, 2, 4, 6, 8, 12, 16, 24, 1, 2, 4, 6, 8, 12, 16, 24) -mframe$sigma <- allsigmas -``` - -Scatterplot with each model as a different color! -Linear model should be best - -NOTE 8 LEVEL ANOVA STILL NOT WORKING - -## Visualize!! - -```{r} -mframe <- as.data.frame(mframe) -p <- ggplot(mframe, aes(x=mframe$sigma, y=mframe$value, colour=variable)) + geom_point (size=2) + xlab("Sigma Values") + ylab("Mean p-values") -p - -``` - -The regression model performs best across the sigmas, as evidenced by it's low p-value scores. The eight level anova should perform close to the regression, but there's something wrong with our code. The four level anova performs in between the t-test and the regression. Finally the t-test, as expected, performed worse than either the regression or the four level ANOVA (and should perform worse than the eight-level as well). This all makes sense, as our data is linearly correlated, so a linear regression should fit best, followed by models that break up the data into larger numbers of groups (closer to a linear model). - -It's clearly best to try to fit linear data to a linear model, but when resources are constrained, it seems the best strategy is to fit as many groups as possible. In this case, you need at least four experimental units (four way anova), to detect a significant effect. - - - -```{r, include=FALSE} -# get values for the eight level anova guess: -x <- c(6.25, 12.5, 18.72, 25, 31.25, 37.5, 43.75, 50) -y.vec <- rep(NA, 8) -for (i in 1:length(x)){ - y[i] <- 10 + 0.4*x[i] - y.vec <- y -} -``` ->>>>>>> e26459b27092a0a44ebda487ebe1b16258afb0bb diff --git a/statPowerAnalysis.Rmd b/statPowerAnalysis.Rmd index 2885eb7..9ea3a18 100755 --- a/statPowerAnalysis.Rmd +++ b/statPowerAnalysis.Rmd @@ -154,6 +154,8 @@ for (i in 1:length(sigmas)){ # sigma runs anova.p[j]<- 1-pchisq(t.statanova, df=degfanova) eightanova.p[j]<- 1-pchisq(t.stateightanova, df=degfeightanova) } + + # the mean p values for each of ten monte carlo runs, and each of 8 sigmas into a data frame. dfp[,1]<- c(1, 2, 4, 6, 8, 12, 16, 24) dfp[i,2] <- mean(reg.p) dfp[i,3] <- mean(t.p) From d4d7328706fc23ddd333312ed5d2d8b2fed21d1f Mon Sep 17 00:00:00 2001 From: haley Date: Thu, 6 Dec 2018 13:34:24 -0500 Subject: [PATCH 10/16] html files --- anovaVSregression.html | 328 ++++++++++++++++++++++++++++++++++++++++ statPowerAnalysis.html | 335 +++++++++++++++++++++++++++++++++++++++++ 2 files changed, 663 insertions(+) create mode 100755 anovaVSregression.html create mode 100755 statPowerAnalysis.html diff --git a/anovaVSregression.html b/anovaVSregression.html new file mode 100755 index 0000000..d6847e8 --- /dev/null +++ b/anovaVSregression.html @@ -0,0 +1,328 @@ + + + + + + + + + + + + + +ANOVA vs Regression + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + +
+

PART 1 REGRESSION VS ANOVA

+
+

An ANOVA-design and a regression-design experiment

+
# playing with the data 
+
+rm(list=ls())
+#setwd("/Users/emilynonnamaker/Box/personalStuff/School/PhD/biocomputingFall2018/biocomputing_StatsGroupProject")
+library(ggplot2)
+library(arm)
+
## Loading required package: MASS
+
## Loading required package: Matrix
+
## Loading required package: lme4
+
## 
+## arm (Version 1.10-1, built: 2018-4-12)
+
## Working directory is C:/Users/JMac/Documents/intro.biocomp_excercise1/biocomputing_StatsGroupProject
+
library(tidyr)
+
## 
+## Attaching package: 'tidyr'
+
## The following object is masked from 'package:Matrix':
+## 
+##     expand
+
library(reshape2)
+
## 
+## Attaching package: 'reshape2'
+
## The following object is masked from 'package:tidyr':
+## 
+##     smiths
+
antibiotics <- read.csv("antibiotics.csv")
+sugar <- read.csv("sugar.csv")
+
+#playing around - check out the data
+boxplot(growth ~ trt, data=antibiotics)
+

+
lm <- lm(growth ~ sugar, data=sugar)
+aov <- aov(growth ~ trt, data=antibiotics)
+
+# how a normal person would do this
+summary(aov)
+
##             Df Sum Sq Mean Sq F value   Pr(>F)    
+## trt          3  714.3  238.08   38.74 1.89e-06 ***
+## Residuals   12   73.7    6.15                     
+## ---
+## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
+
posthoc <- TukeyHSD(aov, "trt", conf.level=0.95) #a, b, ac, bd. 
+posthoc
+
##   Tukey multiple comparisons of means
+##     95% family-wise confidence level
+## 
+## Fit: aov(formula = growth ~ trt, data = antibiotics)
+## 
+## $trt
+##                  diff        lwr       upr     p adj
+## ab2-ab1     13.410872   8.206514 18.615230 0.0000306
+## ab3-ab1      4.226769  -0.977589  9.431128 0.1276540
+## control-ab1 16.496238  11.291879 21.700596 0.0000036
+## ab3-ab2     -9.184103 -14.388461 -3.979744 0.0010300
+## control-ab2  3.085366  -2.118993  8.289724 0.3374873
+## control-ab3 12.269468   7.065110 17.473827 0.0000736
+
+
+

Now to do the actual assingment:

+
# set up your data: 
+
+N <- length(antibiotics$growth)
+y <- antibiotics$growth
+x<- antibiotics$trt
+antibiotics$x1 <- ifelse(x=="ab1", 1, 0) #dummy variables
+antibiotics$x2 <- ifelse(x=="ab2", 1, 0) #dummy variables
+antibiotics$x3 <- ifelse(x=="ab3", 1, 0) #dummy variables
+
+#Build null model 
+nullmod<-function(p,x,y){
+  B0=p[1]
+  sigma=exp(p[2])
+  
+  expected=B0
+
+null=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE))
+return(null)
+}
+
+#Build full model
+fullmod<-function(p,x,y){
+B0=p[1]
+B1=p[2]
+B2=p[3]
+B3=p[4]
+sigma=exp(p[5])
+
+expected=B0+B1*x[,3]+B2*x[,4]+B3*x[,5]
+
+full=-sum(dnorm(x=y,mean=expected,sd=sigma,log = TRUE))
+return(full)
+}
+
+# Check fit
+nullguess <- c(1, 2)
+fullguess <- c(18, -12, -3, -4, 1)
+fitnull=optim(par=nullguess,fn=nullmod,y=antibiotics$growth) # converges
+fitfull=optim(par=fullguess,fn=fullmod,x=antibiotics,y=antibiotics$growth) # converges
+
+# Get t.statistic . 
+t.stat <- 2*(fitnull$value - fitfull$value)
+t.stat # 37.90134
+
## [1] 37.90134
+
# get degrees of freedom
+df <- length(fitfull$par) - length(fitnull$par)
+
+# and p value
+1-pchisq(t.stat, df=df) # 2.965739e-08
+
## [1] 2.965739e-08
+

Looks like there’s a significant difference between the null and full anova model (t = 25.69, p < .001), suggesting that there are differences between treatments.

+
+
+

Graphical visualization!

+

Using a posthoc Tukey test (see hidden code) we can find significant differences between groups, and add these differences as letters to our visualization (see graph below)

+
p <- ggplot(antibiotics, aes(x=trt, y=growth, fill=trt)) + 
+  geom_boxplot()
+p + geom_jitter(shape=16, position=position_jitter(0.2), alpha=0.3) + theme_classic() + labs(x = "Treatment", y= "Growth") + annotate("text", x = c(1, 2, 3, 4), y = c(23, 23, 23, 23), label = c("a", "b", "ac", "bd"))
+

+
+
+

Regression

+

Build the null model:

+
lmnullmodreg<-function(p,x,y){
+  B0=p[1]
+  sigma=exp(p[2])
+  
+  expected=B0
+
+null=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE))
+return(null)
+}
+

Now build the extension:

+
lmfullmodreg<-function(p,x,y){
+  B0=p[1]
+  B1=p[2]
+  sigma=exp(p[3])
+  
+  expected=B0+B1*x
+
+  null=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE))
+  return(null)
+}
+

Give it a guess:

+
lmnullregguess <- c(1, 1)
+lmfullregguess <- c(1, 2, 3)
+lmfitregnull=optim(par=lmnullregguess,fn=lmnullmodreg,x=sugar$sugar,y=sugar$growth) # converges
+lmfitregfull=optim(par=lmfullregguess,fn=lmfullmodreg,x=sugar$sugar,y=sugar$growth) # converges
+

Now to compare:

+
# Get t statistic
+t.statreg <- 2*(lmfitregnull$value - lmfitregfull$value)
+t.statreg # 39.92512
+
## [1] 39.92512
+
# get degrees of freedom
+dfreg <- length(lmfitregfull$par) - length(lmfitregnull$par)
+
+#and p value
+1-pchisq(t.stat, df=dfreg) # 7.441429e-10, woot
+
## [1] 7.441429e-10
+

Comparing between the null model of no effect of sugar concentration on growth to the model taking this affect into account, we can see that sugar concentration does have a significant affect on growth (t = 39.92512, p < .0001). Looking at the graph below, we can tell that sugar concentration has a positive affect, with higher concentrations of sugar increasing growth.

+
+
+

Visualization!

+
a <- ggplot(data=sugar,aes(x=sugar,y=growth))
+a + geom_point() + coord_cartesian() + labs( x = "sugar concentration", y = "growth") + theme_classic() + geom_smooth(method = "lm")
+

+
+
+ + + + +
+ + + + + + + + diff --git a/statPowerAnalysis.html b/statPowerAnalysis.html new file mode 100755 index 0000000..d6cae55 --- /dev/null +++ b/statPowerAnalysis.html @@ -0,0 +1,335 @@ + + + + + + + + + + + + + +Power Analysis + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + +
+

Part 2 Statistical Power Analysis

+

Creating the models:

+
# Null model
+nullmod<- function(p,x,y){
+  B0=p[1]
+  sigma=exp(p[2])
+  
+  expected=B0
+  
+  null=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE))
+  return(null)
+}
+
+# Regression model
+lin<-function(p,x,y){
+  B0=p[1]
+  B1=p[2]
+  sigma=exp(p[3])
+  
+  expected=B0+B1*x
+
+  lmmod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE))
+  return(lmmod)
+}
+
+# T-test model
+ttest<-function(p,x,y){
+  B0=p[1]
+  B1=p[2]
+  sigma=exp(p[3])
+  x1=x
+  expected=B0+B1*x1
+
+  tmod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE))
+  return(tmod)
+}
+
+# ANOVA - four groups
+anovamod<-function(p,x,y){
+B0=p[1]
+B1=p[2]
+B2=p[3]
+B3=p[4]
+sigma=exp(p[5])
+
+expected=B0+B1*x[,1]+B2*x[,2]+B3*x[,3]
+
+anova=-sum(dnorm(x=y,mean=expected,sd=sigma,log = TRUE))
+return(anova)
+}
+
+# Now an ANOVA with 8 groups
+eightanovamod<-function(p,x,y){
+  B0=p[1]
+  B1=p[2]
+  B2=p[3]
+  B3=p[4]
+  B4=p[5]
+  B5=p[6]
+  B6=p[7]
+  B7=p[8]
+  sigma=exp(p[9])
+  
+  expected=B0+B1*x[,1]+B2*x[,2]+B3*x[,3]+B4*x[,4]+B5*x[,5]+B6*x[,6]+B7*x[,7]
+  
+  eightanovamod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE))
+  return(eightanovamod)
+}
+
+

Now do your for loops!

+
# Set up an empty data frame in which to store your p-values for each sigma and model.
+dfp <- setNames(data.frame(matrix(ncol = 5, nrow = 8)), c("Sigmas", "Regression", "T.test", "Anova",  "Eight.Anova"))
+
+# Set up your vector of sigmas to loop through
+sigmas <- c(1,2,4,6,8,12,16,24)
+
+# Write your nested for loop:
+for (i in 1:length(sigmas)){ # sigma runs
+  # set up your vectors for your p-values from each monte-carlo run. 
+  reg.p <- numeric(10) # changes each time you use a new sigma
+  t.p <- numeric(10)
+  anova.p <- numeric(10)
+  eightanova.p <- numeric(10)
+  for (j in 1:10){ # monte carlo runs
+    x <- runif(24, 0, 50) # build set of x's
+    e <- rnorm(24, 0, sd = sigmas[i]) # put in your error
+    y <- 10 + 0.4*x + e # write your linear equation
+    df <- cbind(x, y) # bind x and y into a data frame
+    df <- as.data.frame(df)
+    df <- df[order(df$x),] # order for later sorting. 
+    
+    # dummy coding:
+    
+    # For t-test
+    tdummy <- matrix(0, 24, 1)
+    tdummy[13:24,1] <- 1
+  
+    # For anova
+    dummy <- matrix(0, 24, 3)
+    dummy[7:12,1] <- 1
+    dummy[13:18,2] <-1
+    dummy[19:24,3] <-1
+    
+    #for eight-level anova
+    edummy <- matrix(0, 24, 7)
+    edummy[4:6,1]<-1
+    edummy[7:9,2]<-1
+    edummy[10:12,3]<-1
+    edummy[13:15,4]<-1
+    edummy[16:18,5]<-1
+    edummy[19:21,6]<-1
+    edummy[22:24,7]<-1
+    
+    # guesses
+    nullmodguess <- c(1,2) # 2
+    lmmodguess <- c(1, 2, 3) # 3 
+    tmodguess<-c(1,1,1,1,1) # 5
+    aovguess<- c(9,10,-6,8,2) # 5 
+    eightaovguess<-c(1, 12.5,15,17.5,20,22.5,25,27.5,30) # 9
+    
+    #optims
+    nullfitmod=optim(par=nullmodguess,fn=nullmod,x=df$x,y=df$y) # converges
+    lmfitmod=optim(par=lmmodguess,fn=lin,x=df$x,y=df$y) # converges 
+    tfitmod=optim(par=tmodguess,fn=ttest,x=tdummy,y=df$y) # converges
+    anovafitmod=optim(par=aovguess,fn=anovamod,x=dummy,y=df$y, control=list(maxit=1e5)) # converges
+    eightanovafitmod=optim(par=eightaovguess,fn=eightanovamod,x=edummy,y=df$y, control=list(maxit=1e5)) # converges
+   
+    # degrees of freedom 
+     degfreg <- length(lmfitmod$par) - length(nullfitmod$par) 
+     degft <- length(tfitmod$par) - length(nullfitmod$par) 
+     degfanova <- length(anovafitmod$par) - length(nullfitmod$par) 
+     degfeightanova <- length(eightanovafitmod$par) - length(nullfitmod$par) 
+     
+     # test stats
+    t.statreg <- 2*(nullfitmod$value - lmfitmod$value)
+    t.statt <- 2*(nullfitmod$value - tfitmod$value)
+    t.statanova <- 2*(nullfitmod$value - anovafitmod$value)
+    t.stateightanova <- 2*(nullfitmod$value - eightanovafitmod$value)
+    
+    # pvalues 
+    reg.p[j] <- 1-pchisq(t.statreg, df=degfreg)
+    t.p[j]<- 1-pchisq(t.statt, df=degft)
+    anova.p[j]<- 1-pchisq(t.statanova, df=degfanova)
+    eightanova.p[j]<- 1-pchisq(t.stateightanova, df=degfeightanova)
+  }
+  
+  # the mean p values for each of ten monte carlo runs, and each of 8 sigmas into a data frame.
+  dfp[,1]<- c(1, 2, 4, 6, 8, 12, 16, 24)
+  dfp[i,2] <- mean(reg.p)
+  dfp[i,3] <- mean(t.p)
+  dfp[i,4] <- mean(anova.p)
+  dfp[i,5] <- mean(eightanova.p)
+}
+

Now reshape your dataframe of p-values into something you can plot!

+
library(reshape2)
+mframe <- melt(dfp)
+
## No id variables; using all as measure variables
+
mframe <- mframe[9:40,]
+allsigmas <- c(1,  2,  4,  6,  8, 12, 16, 24, 1,  2,  4,  6,  8,12, 16, 24, 1,  2,  4,  6,  8, 12, 16, 24, 1,  2,  4,  6,  8, 12, 16, 24)
+mframe$sigma <- allsigmas
+

Scatterplot with each model as a different color! Linear model should be best

+

NOTE 8 LEVEL ANOVA STILL NOT WORKING

+
+
+

Visualize!!

+
library(ggplot2)
+mframe <- as.data.frame(mframe)
+p <- ggplot(mframe, aes(x=mframe$sigma, y=mframe$value, colour=variable)) + geom_point (size=2) + xlab("Sigma Values") + ylab("Mean p-values") 
+p
+

+

The regression model performs best across the sigmas, as evidenced by it’s low p-value scores. The eight level anova should perform close to the regression, but there’s something wrong with our code. The four level anova performs in between the t-test and the regression. Finally the t-test, as expected, performed worse than either the regression or the four level ANOVA (and should perform worse than the eight-level as well). This all makes sense, as our data is linearly correlated, so a linear regression should fit best, followed by models that break up the data into larger numbers of groups (closer to a linear model).

+

It’s clearly best to try to fit linear data to a linear model, but when resources are constrained, it seems the best strategy is to fit as many groups as possible. In this case, you need at least four experimental units (four way anova), to detect a significant effect.

+
+
+ + + + +
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-# Null model -nullmod<- function(p,x,y){ - B0=p[1] - sigma=exp(p[2]) - - expected=B0 - - null=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) - return(null) -} - -# Regression model -lin<-function(p,x,y){ - B0=p[1] - B1=p[2] - sigma=exp(p[3]) - - expected=B0+B1*x - - lmmod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) - return(lmmod) -} - -# T-test model -ttest<-function(p,x,y){ - B0=p[1] - B1=p[2] - sigma=exp(p[3]) - x1=x - expected=B0+B1*x1 - - tmod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) - return(tmod) -} - -# ANOVA - four groups -anovamod<-function(p,x,y){ -B0=p[1] -B1=p[2] -B2=p[3] -B3=p[4] -sigma=exp(p[5]) - -expected=B0+B1*x[,1]+B2*x[,2]+B3*x[,3] - -anova=-sum(dnorm(x=y,mean=expected,sd=sigma,log = TRUE)) -return(anova) -} - -# Now an ANOVA with 8 groups -eightanovamod<-function(p,x,y){ - B0=p[1] - B1=p[2] - B2=p[3] - B3=p[4] - B4=p[5] - B5=p[6] - B6=p[7] - B7=p[8] - sigma=exp(p[9]) - - expected=B0+B1*x[,1]+B2*x[,2]+B3*x[,3]+B4*x[,4]+B5*x[,5]+B6*x[,6]+B7*x[,7] - - eightanovamod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) - return(eightanovamod) -} - -``` - -## Now do your for loops! - -```{r} -# Set up an empty data frame in which to store your p-values for each sigma and model. -dfp <- setNames(data.frame(matrix(ncol = 5, nrow = 8)), c("Sigmas", "Regression", "T.test", "Anova", "Eight.Anova")) - -# Set up your vector of sigmas to loop through -sigmas <- c(1,2,4,6,8,12,16,24) - -# Write your nested for loop: -for (i in 1:length(sigmas)){ # sigma runs - # set up your vectors for your p-values from each monte-carlo run. - reg.p <- numeric(10) # changes each time you use a new sigma - t.p <- numeric(10) - anova.p <- numeric(10) - eightanova.p <- numeric(10) - for (j in 1:10){ # monte carlo runs - x <- runif(24, 0, 50) # build set of x's - e <- rnorm(24, 0, sd = sigmas[i]) # put in your error - y <- 10 + 0.4*x + e # write your linear equation - df <- cbind(x, y) # bind x and y into a data frame - df <- as.data.frame(df) - df <- df[order(df$x),] # order for later sorting. - - # dummy coding: - - # For t-test - tdummy <- matrix(0, 24, 1) - tdummy[13:24,1] <- 1 - - # For anova - dummy <- matrix(0, 24, 3) - dummy[7:12,1] <- 1 - dummy[13:18,2] <-1 - dummy[19:24,3] <-1 - - #for eight-level anova - edummy <- matrix(0, 24, 7) - edummy[4:6,1]<-1 - edummy[7:9,2]<-1 - edummy[10:12,3]<-1 - edummy[13:15,4]<-1 - edummy[16:18,5]<-1 - edummy[19:21,6]<-1 - edummy[22:24,7]<-1 - - # guesses - nullmodguess <- c(1,2) # 2 - lmmodguess <- c(1, 2, 3) # 3 - tmodguess<-c(1,1,1,1,1) # 5 - aovguess<- c(9,10,-6,8,2) # 5 - eightaovguess<-c(1, 12.5,15,17.5,20,22.5,25,27.5,30) # 9 - - #optims - nullfitmod=optim(par=nullmodguess,fn=nullmod,x=df$x,y=df$y) # converges - lmfitmod=optim(par=lmmodguess,fn=lin,x=df$x,y=df$y) # converges - tfitmod=optim(par=tmodguess,fn=ttest,x=tdummy,y=df$y) # converges - anovafitmod=optim(par=aovguess,fn=anovamod,x=dummy,y=df$y, control=list(maxit=1e5)) # converges - eightanovafitmod=optim(par=eightaovguess,fn=eightanovamod,x=edummy,y=df$y, control=list(maxit=1e5)) # converges - - # degrees of freedom - degfreg <- length(lmfitmod$par) - length(nullfitmod$par) - degft <- length(tfitmod$par) - length(nullfitmod$par) - degfanova <- length(anovafitmod$par) - length(nullfitmod$par) - degfeightanova <- length(eightanovafitmod$par) - length(nullfitmod$par) - - # test stats - t.statreg <- 2*(nullfitmod$value - lmfitmod$value) - t.statt <- 2*(nullfitmod$value - tfitmod$value) - t.statanova <- 2*(nullfitmod$value - anovafitmod$value) - t.stateightanova <- 2*(nullfitmod$value - eightanovafitmod$value) - - # pvalues - reg.p[j] <- 1-pchisq(t.statreg, df=degfreg) - t.p[j]<- 1-pchisq(t.statt, df=degft) - anova.p[j]<- 1-pchisq(t.statanova, df=degfanova) - eightanova.p[j]<- 1-pchisq(t.stateightanova, df=degfeightanova) - } - dfp[,1]<- c(1, 2, 4, 6, 8, 12, 16, 24) - dfp[i,2] <- mean(reg.p) - dfp[i,3] <- mean(t.p) - dfp[i,4] <- mean(anova.p) - dfp[i,5] <- mean(eightanova.p) -} -``` - -Now reshape your dataframe of p-values into something you can plot! - -```{r} -mframe <- melt(dfp) -mframe <- mframe[9:40,] -allsigmas <- c(1, 2, 4, 6, 8, 12, 16, 24, 1, 2, 4, 6, 8,12, 16, 24, 1, 2, 4, 6, 8, 12, 16, 24, 1, 2, 4, 6, 8, 12, 16, 24) -mframe$sigma <- allsigmas -``` - -Scatterplot with each model as a different color! -Linear model should be best - -NOTE 8 LEVEL ANOVA STILL NOT WORKING - -## Visualize!! - -```{r} -mframe <- as.data.frame(mframe) -p <- ggplot(mframe, aes(x=mframe$sigma, y=mframe$value, colour=variable)) + geom_point (size=2) + xlab("Sigma Values") + ylab("Mean p-values") -p - -``` - -The regression model performs best across the sigmas, as evidenced by it's low p-value scores. The eight level anova should perform close to the regression, but there's something wrong with our code. The four level anova performs in between the t-test and the regression. Finally the t-test, as expected, performed worse than either the regression or the four level ANOVA (and should perform worse than the eight-level as well). This all makes sense, as our data is linearly correlated, so a linear regression should fit best, followed by models that break up the data into larger numbers of groups (closer to a linear model). - -It's clearly best to try to fit linear data to a linear model, but when resources are constrained, it seems the best strategy is to fit as many groups as possible. In this case, you need at least four experimental units (four way anova), to detect a significant effect. - - - -```{r, include=FALSE} -# get values for the eight level anova guess: -x <- c(6.25, 12.5, 18.72, 25, 31.25, 37.5, 43.75, 50) -y.vec <- rep(NA, 8) -for (i in 1:length(x)){ - y[i] <- 10 + 0.4*x[i] - y.vec <- y -} -``` +--- +title: "Power Analysis" +output: html_document +--- + +# Part 2 Statistical Power Analysis + +```{r, include=FALSE} +library(ggplot2) +library(dplyr) +library(reshape2) +``` + +### Step 1 - build the models + +```{r} + +# Null model +nullmod<- function(p,x,y){ + B0=p[1] + sigma=exp(p[2]) + + expected=B0 + + null=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) + return(null) +} + +# Regression model +lin<-function(p,x,y){ + B0=p[1] + B1=p[2] + sigma=exp(p[3]) + + expected=B0+B1*x + + lmmod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) + return(lmmod) +} + +# T-test model +ttest<-function(p,x,y){ + B0=p[1] + B1=p[2] + sigma=exp(p[3]) + x1=x + expected=B0+B1*x1 + + tmod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) + return(tmod) +} + +# ANOVA - four groups +anovamod<-function(p,x,y){ +B0=p[1] +B1=p[2] +B2=p[3] +B3=p[4] +sigma=exp(p[5]) + +expected=B0+B1*x[,1]+B2*x[,2]+B3*x[,3] + +anova=-sum(dnorm(x=y,mean=expected,sd=sigma,log = TRUE)) +return(anova) +} + +# Now an ANOVA with 8 groups +eightanovamod<-function(p,x,y){ + B0=p[1] + B1=p[2] + B2=p[3] + B3=p[4] + B4=p[5] + B5=p[6] + B6=p[7] + B7=p[8] + sigma=exp(p[9]) + + expected=B0+B1*x[,1]+B2*x[,2]+B3*x[,3]+B4*x[,4]+B5*x[,5]+B6*x[,6]+B7*x[,7] + + eightanovamod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE)) + return(eightanovamod) +} + +``` + +### Step 2: Make your for loops! + +```{r} +# Set up an empty data frame in which to store your p-values for each sigma and model. +dfp <- setNames(data.frame(matrix(ncol = 5, nrow = 8)), c("Sigmas", "Regression", "T.test", "Anova", "Eight.Anova")) + +# Set up your vector of sigmas to loop through +sigmas <- c(1,2,4,6,8,12,16,24) + +# Write your nested for loop: +for (i in 1:length(sigmas)){ # sigma runs + # set up your vectors for your p-values from each monte-carlo run. + reg.p <- numeric(10) # changes each time you use a new sigma + t.p <- numeric(10) + anova.p <- numeric(10) + eightanova.p <- numeric(10) + for (j in 1:10){ # monte carlo runs + x <- runif(24, 0, 50) # build set of x's + e <- rnorm(24, 0, sd = sigmas[i]) # put in your error + y <- 10 + 0.4*x + e # write your linear equation + df <- cbind(x, y) # bind x and y into a data frame + df <- as.data.frame(df) + df <- df[order(df$x),] # order for later sorting. + + # dummy coding: + + # For t-test + tdummy <- matrix(0, 24, 1) + tdummy[13:24,1] <- 1 + + # For anova + dummy <- matrix(0, 24, 3) + dummy[7:12,1] <- 1 + dummy[13:18,2] <-1 + dummy[19:24,3] <-1 + + #for eight-level anova + edummy <- matrix(0, 24, 7) + edummy[4:6,1]<-1 + edummy[7:9,2]<-1 + edummy[10:12,3]<-1 + edummy[13:15,4]<-1 + edummy[16:18,5]<-1 + edummy[19:21,6]<-1 + edummy[22:24,7]<-1 + + # guesses + nullmodguess <- c(1,2) # 2 + lmmodguess <- c(1, 2, 3) # 3 + tmodguess<-c(1,1,1,1,1) # 5 + aovguess<- c(9,10,-6,8,2) # 5 + eightaovguess<-c(1, 12.5,15,17.5,20,22.5,25,27.5,30) # 9 + + #optims + nullfitmod=optim(par=nullmodguess,fn=nullmod,x=df$x,y=df$y) # converges + lmfitmod=optim(par=lmmodguess,fn=lin,x=df$x,y=df$y) # converges + tfitmod=optim(par=tmodguess,fn=ttest,x=tdummy,y=df$y) # converges + anovafitmod=optim(par=aovguess,fn=anovamod,x=dummy,y=df$y, control=list(maxit=1e5)) # converges + eightanovafitmod=optim(par=eightaovguess,fn=eightanovamod,x=edummy,y=df$y, control=list(maxit=1e5)) # converges + + # degrees of freedom + degfreg <- length(lmfitmod$par) - length(nullfitmod$par) + degft <- length(tfitmod$par) - length(nullfitmod$par) + degfanova <- length(anovafitmod$par) - length(nullfitmod$par) + degfeightanova <- length(eightanovafitmod$par) - length(nullfitmod$par) + + # test stats + t.statreg <- 2*(nullfitmod$value - lmfitmod$value) + t.statt <- 2*(nullfitmod$value - tfitmod$value) + t.statanova <- 2*(nullfitmod$value - anovafitmod$value) + t.stateightanova <- 2*(nullfitmod$value - eightanovafitmod$value) + + # pvalues + reg.p[j] <- 1-pchisq(t.statreg, df=degfreg) + t.p[j]<- 1-pchisq(t.statt, df=degft) + anova.p[j]<- 1-pchisq(t.statanova, df=degfanova) + eightanova.p[j]<- 1-pchisq(t.stateightanova, df=degfeightanova) + } + + # the mean p values for each of ten monte carlo runs, and each of 8 sigmas into a data frame. + dfp[,1]<- c(1, 2, 4, 6, 8, 12, 16, 24) + dfp[i,2] <- mean(reg.p) + dfp[i,3] <- mean(t.p) + dfp[i,4] <- mean(anova.p) + dfp[i,5] <- mean(eightanova.p) +} +``` + +Now reshape your dataframe of p-values into something you can plot! + +```{r} +mframe <- melt(dfp) +mframe <- mframe[9:40,] +allsigmas <- c(1, 2, 4, 6, 8, 12, 16, 24, 1, 2, 4, 6, 8,12, 16, 24, 1, 2, 4, 6, 8, 12, 16, 24, 1, 2, 4, 6, 8, 12, 16, 24) +mframe$sigma <- allsigmas +``` + +### Step 3: Visualize!! + +```{r} +mframe <- as.data.frame(mframe) +p <- ggplot(mframe, aes(x=mframe$sigma, y=mframe$value, colour=variable)) + geom_point (size=2) + xlab("Sigma Values") + ylab("Mean p-values") +p + +``` + +### Step 4: Interpret + +The regression model performs best across the sigmas, as evidenced by it's low p-value scores (see figure above). We originally believed that the eight level anova (Eight.Anova) should perform close to the regression, when in fact it performed the worst out of all four models. The four level anova (Anova) performs in between the t-test (T.test) and the regression. Finally the t-test, as expected, performed worse than either the regression or the four level ANOVA. This all makes sense, as our data is linearly correlated, so a linear regression should fit best. + +Upon talking to Stuart, with the eight level ANOVA, we are starting to make finer and finer comparisons. When there's a linear relationship with noise and you split the x finer, the p-values should be high. With fewer groups, as in the four-way ANOVA, the groups shouldn't overlap as much, and the p-values should be lower. + +It's clearly best to try to fit linear data to a linear model, but when resources are constrained, it seems the best strategy is to fit as many non-overlapping groups as possible. In this case, you need at least four experimental units (four way anova), to detect a significant signal within the noise. + + +```{r, include=FALSE} +# get values for the eight level anova guess: +x <- c(6.25, 12.5, 18.72, 25, 31.25, 37.5, 43.75, 50) +y.vec <- rep(NA, 8) +for (i in 1:length(x)){ + y[i] <- 10 + 0.4*x[i] + y.vec <- y +} +``` + From 9f8a0297a79bb9758e20d010eab1830f2c164d12 Mon Sep 17 00:00:00 2001 From: enonnama <43544834+enonnama@users.noreply.github.com> Date: Thu, 6 Dec 2018 15:17:50 -0500 Subject: [PATCH 13/16] Add files via upload --- statPowerAnalysis.Rmd | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/statPowerAnalysis.Rmd b/statPowerAnalysis.Rmd index 7d575e8..be23669 100755 --- a/statPowerAnalysis.Rmd +++ b/statPowerAnalysis.Rmd @@ -1,6 +1,6 @@ --- title: "Power Analysis" -output: html_document +output: html_notebook --- # Part 2 Statistical Power Analysis From 285dd6e7f9e96742811fa9a91baacdf011345600 Mon Sep 17 00:00:00 2001 From: enonnama <43544834+enonnama@users.noreply.github.com> Date: Thu, 6 Dec 2018 15:20:45 -0500 Subject: [PATCH 14/16] Add files via upload PDF version --- ANOVA vs Regression.pdf | Bin 0 -> 251073 bytes 1 file changed, 0 insertions(+), 0 deletions(-) create mode 100644 ANOVA vs Regression.pdf diff --git a/ANOVA vs Regression.pdf b/ANOVA vs Regression.pdf new file mode 100644 index 0000000000000000000000000000000000000000..5e31f1d087c93571fdecf3a8905d4a80260d8bff GIT binary patch literal 251073 zcmbTdWpo`&v!=Pl%nX*dn37YWdvn-0q&4wR&d$ zROO1Q%*d==E90qnW0T5@h|x0AF~gD$9NZk77T@O13=F}t02zR`1{Sb9JV1JBV;fT^ zGa&1yNfAgdW^UzV?C^QE(swczF*dX{G6wST!8$rQ80%ZZx`8xmDMaG2Bl|p7&p7U& zo3kXS%3tIaqQ77=+~wU7+It2$PzJqz%-FhEOFc{#wPtAtgfBn%ydG!HPM8&V*H#=m!xek?Y&p<~tR5 zjNAp%qak$s^_bi29*A%nzr}4~yY^T!4E^)Lu=ry!_n$)_&Z&qJBt*U-fzUnk+?kAN z=ZHsi0Y5+{(f1%VPp4sDr>p*r5>!!L2g%2{+mf)arf4H}pe?5Gl&Ndes6&JE&I@lK z{2F|1QC=N}Uem?Hr*b`9!=g$3%OVKPi8Xtoy8Y z!vkEH{Se883AzKmYGlJQqTtwpJ0l^aGNd^M!fg>uhF^n)CJ8PZ@!a)|q`+~=5|woSbGD2Q~32DO5y z`gl#E_5hu*vN0{CPT6PLyo>ttTtBJ_+Hs8|OyZn=ZZJYNu6K@xE3%tb6)qZO1}F#ztC(dLWQTPrxhp+rDrq+`e1*K1*@7qXTT0 z$&-VO6yHheIP;W?#F78rR+|-h2$V4xJB^)!{nbd8$bw)#5f!HM$;sx%zNIs!Wom zmBiqHncmZbDSm7Wkik@EF2bac>nb`k#jeJ$j}uN(QnSe{!*z$=%SD)FHbapMA25LI z{qrZ`WrRx>3l$H2OrBjA?N}Eh6ynH&m*DPFZe@XzFs_JzqB?mdioPJa4qjC=`7eY3 z=`;6LzfH7PZkFDNg&Qqsq3~I^*+~gDK?+kRY$r2~Iczd`l^|bfr$ynY8MDTP5=$9R zQ~`EF*sIfkU4h$EUK~9?F=+(FMYEYE^f)FDIUE=Te>T}W<+$bmf@~X4jcP4r_H%8c$ic07ci*F9K6!>a+gflIb|=mw14Y}$s9$SB%R0gNac3@9e@d+KFkZ4)=>;{uh~byn zGjC_yuj=`}0A6LYO(jL78SBjZDi!O{B0oDUr=@nx955C_iT+B~O4mw0sNX>gb-f;H zmIi)@<|`&jMNy0#3X}vgNGiyNb_pThQc{ZgG;yyXyU93m&P*(3!~_#KbeDw7?8o#V zA`&D+*(|b*M&-B~6i$O;*3g~Is4e%aN4T6e>k-eD$W#QdzN}(rCp6r$spFAzK|PrH zmprdiK#L35KHq0Mn=NOcJkQ9?Xk2R0AkJue!Fga9NJB)`!)ZI48B*!dw4Y#FYK8YJ z4JrzlT1R=|OogI~S3X5KOT<9%>V6Ka%=|*F7rnXJ_Jc+CE@~u{Bwq+Gi=Ozr>8CJW zmfS;l{EW)d&{a~r%UZ!-!^?IU(AC>69|R?J1|fk-A0f+LaYWu<4-KqT+aMdXv)FUX zkU*Qb&o3?1(kvf{rmyAv5Y5`a*saHp5>0dc*__R{59^cawnLx^rGyuYq8J%lSNQ(< zWWGS;GcXq<-F)k6zZbgA!#EmzNoMdnHT5?;E+63rlZz% ze+jE7i)uk^ZY^6AMGaBT<2I|=+Rw0p{0@Dw{Lm2;n}qeBHB(J0L_6@Jn^%^GkpU!U zl7^!Cw4pQzIE*_iXo$Z`094OlipG7edWg!WXbnC(%; z*Q$UDI+P4u7`(hVQ>I`|p{Nlac-J)xT|~ z|0T35yW1H9>E-oJ|2aAs+c*K4|0~QZ8avuLI~W=}0@?n{AZ%;n^m*M8_!kX6IY7qP z$Xs8@)(xn|@M&OUWCF5tGU>p6qQHOr{O$KILdZMV8Y&q(0kuAziHHH|m5tq;fI2{W zVOuL(2PHdwLu24y(hzoJ1hW4#JRjdDVSIZ0$1KhS@5Wbd3oB?6M( z;d;OT=eQ*>z!kzJ{p3MIg2-G%AjHs7C{>M=dqBiPi3~;2;Q|8;mSJv?<%fC|L_`lZ zqAX!9D_L(hGv0?awpNasmsdvBA^RzoxJJkB@#pR4;OTPM(t}f-ST9POCP@RMC`&OkN{ZGMK&HEGOPp8 zB&GU69HtNOvAbkSrVvIZ%siN%6-0Xj)m9J1ht%z(X{luUdAJ%=I2|n6QSI`Z8KgV; zQQ&#nX{6hx`Cd@Qoyw9KfP+Vh`T3_G(f|hZblr&~vOKGBCQ)>pi|`vz0Cds-XNLn~ zTz^NutC0yJHFU#NzP2dMX{ z7BVC3NLPnhWqh?`ER)BdqnEF6+7UK8(t<=vhY#?vMh>JuX>jtlf~w1^bqj_b z0?#vKCD>GfWqHE#ym!P-xNXonfiJmGGe8!|cM$I&lr+$JeS}|tCcl*$ZhUC*zVU4l zH;f-pSUqby-&!#F5S#n3cJQtMG{L%kiO>Qg&|tG->d<)CB0_QMBp@h+mhnUdprxW2 zafk(Yt?(U<0b1POdRDh5_iDX z5cn=vMUn&5B}Gc$mrtS~Lznx8FDkSm%lWNUz+J#yxKI(H#AZRgEUhg|Q?66Q`XKc)!T$DsGbtcX(>l^E3?6?@Ba*gCy8y~b8; z{{5qaGWnUxhRQ3MDA^<#BssQBTE%t|eKD)dx=e0?!Gg>}-(tn$b&;+-db)VpHG4Ws zm*r#sE%cII+VLx0df%&kS%?FH$hGwYI zu*p8+nr}~wtR_55GC?wWR7UC7!ltrb+jPS;yyc-88A~~9`|!`2bq$RQv87zjex4Dp zeD};}YB=6V^~fyD4NP5{a~dVuQ<|?Dlo}A~)(s~$Fa4_n8J0SR-oqgldxrH>MH`2W zm8G;>F5Tze+!;DG8<_2~?b8p44@9qIuZ&Q

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- - - - - - - - - - - - - - -
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PART 1 REGRESSION VS ANOVA

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-

An ANOVA-design and a regression-design experiment

-
# playing with the data 
-
-rm(list=ls())
-#setwd("/Users/emilynonnamaker/Box/personalStuff/School/PhD/biocomputingFall2018/biocomputing_StatsGroupProject")
-library(ggplot2)
-library(arm)
-
## Loading required package: MASS
-
## Loading required package: Matrix
-
## Loading required package: lme4
-
## 
-## arm (Version 1.10-1, built: 2018-4-12)
-
## Working directory is C:/Users/JMac/Documents/intro.biocomp_excercise1/biocomputing_StatsGroupProject
-
library(tidyr)
-
## 
-## Attaching package: 'tidyr'
-
## The following object is masked from 'package:Matrix':
-## 
-##     expand
-
library(reshape2)
-
## 
-## Attaching package: 'reshape2'
-
## The following object is masked from 'package:tidyr':
-## 
-##     smiths
-
antibiotics <- read.csv("antibiotics.csv")
-sugar <- read.csv("sugar.csv")
-
-#playing around - check out the data
-boxplot(growth ~ trt, data=antibiotics)
-

-
lm <- lm(growth ~ sugar, data=sugar)
-aov <- aov(growth ~ trt, data=antibiotics)
-
-# how a normal person would do this
-summary(aov)
-
##             Df Sum Sq Mean Sq F value   Pr(>F)    
-## trt          3  714.3  238.08   38.74 1.89e-06 ***
-## Residuals   12   73.7    6.15                     
-## ---
-## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
-
posthoc <- TukeyHSD(aov, "trt", conf.level=0.95) #a, b, ac, bd. 
-posthoc
-
##   Tukey multiple comparisons of means
-##     95% family-wise confidence level
-## 
-## Fit: aov(formula = growth ~ trt, data = antibiotics)
-## 
-## $trt
-##                  diff        lwr       upr     p adj
-## ab2-ab1     13.410872   8.206514 18.615230 0.0000306
-## ab3-ab1      4.226769  -0.977589  9.431128 0.1276540
-## control-ab1 16.496238  11.291879 21.700596 0.0000036
-## ab3-ab2     -9.184103 -14.388461 -3.979744 0.0010300
-## control-ab2  3.085366  -2.118993  8.289724 0.3374873
-## control-ab3 12.269468   7.065110 17.473827 0.0000736
-
-
-

Now to do the actual assingment:

-
# set up your data: 
-
-N <- length(antibiotics$growth)
-y <- antibiotics$growth
-x<- antibiotics$trt
-antibiotics$x1 <- ifelse(x=="ab1", 1, 0) #dummy variables
-antibiotics$x2 <- ifelse(x=="ab2", 1, 0) #dummy variables
-antibiotics$x3 <- ifelse(x=="ab3", 1, 0) #dummy variables
-
-#Build null model 
-nullmod<-function(p,x,y){
-  B0=p[1]
-  sigma=exp(p[2])
-  
-  expected=B0
-
-null=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE))
-return(null)
-}
-
-#Build full model
-fullmod<-function(p,x,y){
-B0=p[1]
-B1=p[2]
-B2=p[3]
-B3=p[4]
-sigma=exp(p[5])
-
-expected=B0+B1*x[,3]+B2*x[,4]+B3*x[,5]
-
-full=-sum(dnorm(x=y,mean=expected,sd=sigma,log = TRUE))
-return(full)
-}
-
-# Check fit
-nullguess <- c(1, 2)
-fullguess <- c(18, -12, -3, -4, 1)
-fitnull=optim(par=nullguess,fn=nullmod,y=antibiotics$growth) # converges
-fitfull=optim(par=fullguess,fn=fullmod,x=antibiotics,y=antibiotics$growth) # converges
-
-# Get t.statistic . 
-t.stat <- 2*(fitnull$value - fitfull$value)
-t.stat # 37.90134
-
## [1] 37.90134
-
# get degrees of freedom
-df <- length(fitfull$par) - length(fitnull$par)
-
-# and p value
-1-pchisq(t.stat, df=df) # 2.965739e-08
-
## [1] 2.965739e-08
-

Looks like there’s a significant difference between the null and full anova model (t = 25.69, p < .001), suggesting that there are differences between treatments.

-
-
-

Graphical visualization!

-

Using a posthoc Tukey test (see hidden code) we can find significant differences between groups, and add these differences as letters to our visualization (see graph below)

-
p <- ggplot(antibiotics, aes(x=trt, y=growth, fill=trt)) + 
-  geom_boxplot()
-p + geom_jitter(shape=16, position=position_jitter(0.2), alpha=0.3) + theme_classic() + labs(x = "Treatment", y= "Growth") + annotate("text", x = c(1, 2, 3, 4), y = c(23, 23, 23, 23), label = c("a", "b", "ac", "bd"))
-

-
-
-

Regression

-

Build the null model:

-
lmnullmodreg<-function(p,x,y){
-  B0=p[1]
-  sigma=exp(p[2])
-  
-  expected=B0
-
-null=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE))
-return(null)
-}
-

Now build the extension:

-
lmfullmodreg<-function(p,x,y){
-  B0=p[1]
-  B1=p[2]
-  sigma=exp(p[3])
-  
-  expected=B0+B1*x
-
-  null=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE))
-  return(null)
-}
-

Give it a guess:

-
lmnullregguess <- c(1, 1)
-lmfullregguess <- c(1, 2, 3)
-lmfitregnull=optim(par=lmnullregguess,fn=lmnullmodreg,x=sugar$sugar,y=sugar$growth) # converges
-lmfitregfull=optim(par=lmfullregguess,fn=lmfullmodreg,x=sugar$sugar,y=sugar$growth) # converges
-

Now to compare:

-
# Get t statistic
-t.statreg <- 2*(lmfitregnull$value - lmfitregfull$value)
-t.statreg # 39.92512
-
## [1] 39.92512
-
# get degrees of freedom
-dfreg <- length(lmfitregfull$par) - length(lmfitregnull$par)
-
-#and p value
-1-pchisq(t.stat, df=dfreg) # 7.441429e-10, woot
-
## [1] 7.441429e-10
-

Comparing between the null model of no effect of sugar concentration on growth to the model taking this affect into account, we can see that sugar concentration does have a significant affect on growth (t = 39.92512, p < .0001). Looking at the graph below, we can tell that sugar concentration has a positive affect, with higher concentrations of sugar increasing growth.

-
-
-

Visualization!

-
a <- ggplot(data=sugar,aes(x=sugar,y=growth))
-a + geom_point() + coord_cartesian() + labs( x = "sugar concentration", y = "growth") + theme_classic() + geom_smooth(method = "lm")
-

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- - - - - - - - From 752dead838d042dc4201849cc8a0ea0067b94124 Mon Sep 17 00:00:00 2001 From: hkodak <43554183+hkodak@users.noreply.github.com> Date: Thu, 6 Dec 2018 16:00:37 -0500 Subject: [PATCH 16/16] Delete statPowerAnalysis.html --- statPowerAnalysis.html | 335 ----------------------------------------- 1 file changed, 335 deletions(-) delete mode 100755 statPowerAnalysis.html diff --git a/statPowerAnalysis.html b/statPowerAnalysis.html deleted file mode 100755 index d6cae55..0000000 --- a/statPowerAnalysis.html +++ /dev/null @@ -1,335 +0,0 @@ - - - - - - - - - - - - - -Power Analysis - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
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Part 2 Statistical Power Analysis

-

Creating the models:

-
# Null model
-nullmod<- function(p,x,y){
-  B0=p[1]
-  sigma=exp(p[2])
-  
-  expected=B0
-  
-  null=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE))
-  return(null)
-}
-
-# Regression model
-lin<-function(p,x,y){
-  B0=p[1]
-  B1=p[2]
-  sigma=exp(p[3])
-  
-  expected=B0+B1*x
-
-  lmmod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE))
-  return(lmmod)
-}
-
-# T-test model
-ttest<-function(p,x,y){
-  B0=p[1]
-  B1=p[2]
-  sigma=exp(p[3])
-  x1=x
-  expected=B0+B1*x1
-
-  tmod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE))
-  return(tmod)
-}
-
-# ANOVA - four groups
-anovamod<-function(p,x,y){
-B0=p[1]
-B1=p[2]
-B2=p[3]
-B3=p[4]
-sigma=exp(p[5])
-
-expected=B0+B1*x[,1]+B2*x[,2]+B3*x[,3]
-
-anova=-sum(dnorm(x=y,mean=expected,sd=sigma,log = TRUE))
-return(anova)
-}
-
-# Now an ANOVA with 8 groups
-eightanovamod<-function(p,x,y){
-  B0=p[1]
-  B1=p[2]
-  B2=p[3]
-  B3=p[4]
-  B4=p[5]
-  B5=p[6]
-  B6=p[7]
-  B7=p[8]
-  sigma=exp(p[9])
-  
-  expected=B0+B1*x[,1]+B2*x[,2]+B3*x[,3]+B4*x[,4]+B5*x[,5]+B6*x[,6]+B7*x[,7]
-  
-  eightanovamod=-sum(dnorm(x=y,mean=expected,sd=sigma, log = TRUE))
-  return(eightanovamod)
-}
-
-

Now do your for loops!

-
# Set up an empty data frame in which to store your p-values for each sigma and model.
-dfp <- setNames(data.frame(matrix(ncol = 5, nrow = 8)), c("Sigmas", "Regression", "T.test", "Anova",  "Eight.Anova"))
-
-# Set up your vector of sigmas to loop through
-sigmas <- c(1,2,4,6,8,12,16,24)
-
-# Write your nested for loop:
-for (i in 1:length(sigmas)){ # sigma runs
-  # set up your vectors for your p-values from each monte-carlo run. 
-  reg.p <- numeric(10) # changes each time you use a new sigma
-  t.p <- numeric(10)
-  anova.p <- numeric(10)
-  eightanova.p <- numeric(10)
-  for (j in 1:10){ # monte carlo runs
-    x <- runif(24, 0, 50) # build set of x's
-    e <- rnorm(24, 0, sd = sigmas[i]) # put in your error
-    y <- 10 + 0.4*x + e # write your linear equation
-    df <- cbind(x, y) # bind x and y into a data frame
-    df <- as.data.frame(df)
-    df <- df[order(df$x),] # order for later sorting. 
-    
-    # dummy coding:
-    
-    # For t-test
-    tdummy <- matrix(0, 24, 1)
-    tdummy[13:24,1] <- 1
-  
-    # For anova
-    dummy <- matrix(0, 24, 3)
-    dummy[7:12,1] <- 1
-    dummy[13:18,2] <-1
-    dummy[19:24,3] <-1
-    
-    #for eight-level anova
-    edummy <- matrix(0, 24, 7)
-    edummy[4:6,1]<-1
-    edummy[7:9,2]<-1
-    edummy[10:12,3]<-1
-    edummy[13:15,4]<-1
-    edummy[16:18,5]<-1
-    edummy[19:21,6]<-1
-    edummy[22:24,7]<-1
-    
-    # guesses
-    nullmodguess <- c(1,2) # 2
-    lmmodguess <- c(1, 2, 3) # 3 
-    tmodguess<-c(1,1,1,1,1) # 5
-    aovguess<- c(9,10,-6,8,2) # 5 
-    eightaovguess<-c(1, 12.5,15,17.5,20,22.5,25,27.5,30) # 9
-    
-    #optims
-    nullfitmod=optim(par=nullmodguess,fn=nullmod,x=df$x,y=df$y) # converges
-    lmfitmod=optim(par=lmmodguess,fn=lin,x=df$x,y=df$y) # converges 
-    tfitmod=optim(par=tmodguess,fn=ttest,x=tdummy,y=df$y) # converges
-    anovafitmod=optim(par=aovguess,fn=anovamod,x=dummy,y=df$y, control=list(maxit=1e5)) # converges
-    eightanovafitmod=optim(par=eightaovguess,fn=eightanovamod,x=edummy,y=df$y, control=list(maxit=1e5)) # converges
-   
-    # degrees of freedom 
-     degfreg <- length(lmfitmod$par) - length(nullfitmod$par) 
-     degft <- length(tfitmod$par) - length(nullfitmod$par) 
-     degfanova <- length(anovafitmod$par) - length(nullfitmod$par) 
-     degfeightanova <- length(eightanovafitmod$par) - length(nullfitmod$par) 
-     
-     # test stats
-    t.statreg <- 2*(nullfitmod$value - lmfitmod$value)
-    t.statt <- 2*(nullfitmod$value - tfitmod$value)
-    t.statanova <- 2*(nullfitmod$value - anovafitmod$value)
-    t.stateightanova <- 2*(nullfitmod$value - eightanovafitmod$value)
-    
-    # pvalues 
-    reg.p[j] <- 1-pchisq(t.statreg, df=degfreg)
-    t.p[j]<- 1-pchisq(t.statt, df=degft)
-    anova.p[j]<- 1-pchisq(t.statanova, df=degfanova)
-    eightanova.p[j]<- 1-pchisq(t.stateightanova, df=degfeightanova)
-  }
-  
-  # the mean p values for each of ten monte carlo runs, and each of 8 sigmas into a data frame.
-  dfp[,1]<- c(1, 2, 4, 6, 8, 12, 16, 24)
-  dfp[i,2] <- mean(reg.p)
-  dfp[i,3] <- mean(t.p)
-  dfp[i,4] <- mean(anova.p)
-  dfp[i,5] <- mean(eightanova.p)
-}
-

Now reshape your dataframe of p-values into something you can plot!

-
library(reshape2)
-mframe <- melt(dfp)
-
## No id variables; using all as measure variables
-
mframe <- mframe[9:40,]
-allsigmas <- c(1,  2,  4,  6,  8, 12, 16, 24, 1,  2,  4,  6,  8,12, 16, 24, 1,  2,  4,  6,  8, 12, 16, 24, 1,  2,  4,  6,  8, 12, 16, 24)
-mframe$sigma <- allsigmas
-

Scatterplot with each model as a different color! Linear model should be best

-

NOTE 8 LEVEL ANOVA STILL NOT WORKING

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-

Visualize!!

-
library(ggplot2)
-mframe <- as.data.frame(mframe)
-p <- ggplot(mframe, aes(x=mframe$sigma, y=mframe$value, colour=variable)) + geom_point (size=2) + xlab("Sigma Values") + ylab("Mean p-values") 
-p
-

-

The regression model performs best across the sigmas, as evidenced by it’s low p-value scores. The eight level anova should perform close to the regression, but there’s something wrong with our code. The four level anova performs in between the t-test and the regression. Finally the t-test, as expected, performed worse than either the regression or the four level ANOVA (and should perform worse than the eight-level as well). This all makes sense, as our data is linearly correlated, so a linear regression should fit best, followed by models that break up the data into larger numbers of groups (closer to a linear model).

-

It’s clearly best to try to fit linear data to a linear model, but when resources are constrained, it seems the best strategy is to fit as many groups as possible. In this case, you need at least four experimental units (four way anova), to detect a significant effect.

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