--- title: "Lab 7: Normal and t-distribution" author: "ADD YOUR NAME HERE" date: now date-format: "DD/MM/YYYY HH:MM" format: pdf # html: # theme: a11y # highlight-style: a11y # self-contained: true --- #### Load R libraries ```{r} #| warning: false #| message: false library(mosaic) library(knitr) library(abd) knitr::opts_chunk$set( error = TRUE # do not interrupt in case of errors ) ``` ## Setting the seed of the random number generator Use the **set.seed()** function in R to initialize the random number generator. ```{r} set.seed(2041971) ``` ## Inference for proportions ### Inference Using Formulas #### Exercise 1 QUESTION: State the null and alternative hypotheses. Is it appropriate to conduct a one-tailed or two-tailed test? Why? ANSWER: #### Exercise 2 Under the null hypothesis that the proportion of nests that are predated does not depend on the treatment group, it is appropriate to use a pooled estimate of ˆ p for calculating the standard error. Find the pooled proportion $\hat{p}$ = overall proportion of nests that are found by predators. MAKE SURE YOU UNDERSTAND THE FORMULA USED HERE! ```{r} #| comment: NA (p.pooled<-52/112); # or (60*(13/60)+52*(39/52))/(60+52) ``` #### Exercise 3 Use the pooled $\hat{p}$ to calculate the standard error using the relevant formula. ```{r} #| comment: NA (se.pdiff.Test<-sqrt(p.pooled*(1-p.pooled)*(1/52 + 1/60))) ``` #### Exercise 4 Calculate the z-statistic, and use the normal distribution to find the p-value. ```{r} #| comment: NA (p1<-39/52) (p2<-13/60) (p.diff<-p1-p2) (z.diffprop<-(p.diff)/se.pdiff.Test) ``` ENTER CODE, below, to determine the p-value ```{r} ``` QUESTION: What does this tell you about the importance of eggshells in the ability of predators to find nests? ANSWER: #### Exercise 5 Use formulas and the normal distribution to find a 99% confidence interval for the difference in the proportion of nests that are found by predators in the two treatment groups. I have include code, below, to do this for you, except you must fill in the correct blanks when using the xqnorm function. ```{r} se.diff.CI<-sqrt(p1*(1-p1)/52 + p2*(1-p2)/60) zcrit<-xqnorm( ) # FILL IN THE CORRECT INFORMATION HERE TO GET z*! p1-p2+c(zcrit, -zcrit)*se.diff.CI ``` QUESTION: Why did we use a different SE when conducting a hypothesis test compared to the SE used to calculate the confidence interval? ANSWER: ## Inference using prop.test ### Create the data set ```{r} #| comment: NA eggdat<-data.frame(group=c(rep("Nest with shell", 52), rep("Nest without Shell",60)), result=rep(c("Predated","Not Predated", "Predated","Not Predated"), c(39,13,13,47))) tally(result~group, data=eggdat, format="count", margins=TRUE) ``` #### Exercise 1 Create a 99% confidence interval using prop.test(). ````{r} ```` ### Inference Using Simulation #### Exercise 1 Create a randomization distribution that represents the null hypothesis of no difference in the proportions associated with the 2 treatment groups (using the do function, along with 1000 simulations). ```{r} rand.dist<-do(5000)*{ phats<-diffprop(result~shuffle(group), data=eggdat) } gf_histogram(~diffprop, data=rand.dist) (pval.diffprop<-prop(~diffprop >= p1-p2, data=rand.dist)) ``` #### Exercise 2 Create a bootstrap distribution for the difference in means (using 1000 simulations), and calculate a 99% confidence interval using this distribution. ```{r} boot.diffprop<-do(5000)*{ phats<-diffprop(result~group, data=resample(eggdat, groups=group)) } gf_histogram(~diffprop, data=boot.diffprop) confint(boot.diffprop, level=0.99, method="percentile") ``` ### Exercise 3 QUESTION: How do your results from [parts 1 and 2] compare to those from Exercise 1 [parts 4 and 5]? ANSWER: ## Inference for means ### Eggshell data ```{r} hgdat<-read.csv("hgdat.csv") ``` ### Using Formulas #### Exercise 1 Hint: create dotplots for both the original mercury data and also the log-transformed data. Or, use gf_dhistogram along with gf_fitdistr(dist="dnorm") ```{r} ``` QUESTION: In Rave et al. (2014), we log transformed the mercury concentrations before testing for a difference in contaminant levels between years. Why? What assumptions do we need to meet to use the t-distriution? ANSWER: #### Exercise 2 Calculate sample means, the difference in means, and the standard deviations for the log-tranformed values in 1981 and in 2004. Also, calculate the number of observations in each group. ```{r} (mean.hg<-mean(log.Hg~year, data=hgdat)) (diff.mean<-diffmean(log.Hg~year, data=hgdat)) (favs<-favstats(log.Hg~year, data=hgdat)) ``` #### Exercise 3 Test whether the mercury levels differ in the two years. ```{r} SE.diffmean<-(sqrt(favs$sd[1]^2/favs$n[1]+favs$sd[2]^2/favs$n[2])) (tstat.diffmean<-diff.mean/SE.diffmean) ``` Use tstat and pt to calculate the p-value, below: ```{r} xpt() # FILL IN THE CORRECT INFORMATION HERE TO GET THE P-value ``` QUESTION: What is the p-value? ANSWER: #### Exercise 4 Create a 90% confidence interval for the difference in mean log-concentrations. ```{r} ``` ## Inference using t.test #### Exercise 1 Use t.test(log.Hg ∼ year, data=datasetname, conf.level=0.90) to calculate a p-value and create a 90% confidence interval. ```{r } ``` QUESTION: Interpret the confidence interval in the context of the problem. ANSWER: ## Inference Using Simulation #### Exercise 1 Create a 90% bootstrap confidence interval for the difference in means. ```{r} boot.diffmean<-do(1000)*{ diffmean(log.Hg~year, data=resample(hgdat)) } confint(boot.diffmean, level=0.90, method="percentile") ``` #### Exercise 2 The results are very similar (off by 1-2 units in the second decimal place). # Extra problems ## Confidence interval overlap versus a test for a difference in means ### Exercise 1 Calculate a 95% confidence interval for the mean of the log Hg measurements in 1981 and in 2004. Then conduct a t-test for the difference in means using conf.int = 0.95. ```{r} ``` ### Paired data ```{r} #| comment: NA data(Blackbirds) ``` #### Exercise 1 Use t.test to perform a test and compute a 95% confidence interval for the mean difference in logged before-after measurements. Interpret the results in context. ```{r} ```