similar to: Setting up hypothesis tests with the infer library?

Hello, all, We're now starting to cover hypothesis tests in my Stats 101 course. As usual in courses using the Lock5 textbook, 3rd ed., the homework answers are calculated using their StatKey application. In addition (and for no extra credit), I'm trying to solve the problems using R. In the case of hypothesis test, in addition to manually setting up randomized null hypothesis

How about calculating a 95% confidence interval about the estimated proportion in favor. The PooledInfRate package will do this for you. If confidence intervals overlap then there is no significant difference. -----Original Message----- From: R-help <r-help-bounces at r-project.org> On Behalf Of Kevin Zembower via R-help Sent: Saturday, March 29, 2025 12:10 PM To: R-help email list

Neal, thanks so much for precisely focusing on the exact question I asked. While I can follow along with the workings of your solution, I never would have thought of it myself. I'm glad to learn of all the other solutions offered, too. They broadened my understanding of what I was doing in my statistics class and how I was doing it in R. Thanks, all who contributed to this thread. Thanks,

Hello everyone,           I have a basic question regarding logical operators. > x<-seq(-1,1,by=0.02) > x   [1] -1.00 -0.98 -0.96 -0.94 -0.92 -0.90 -0.88 -0.86 -0.84 -0.82 -0.80 -0.78  [13] -0.76 -0.74 -0.72 -0.70 -0.68 -0.66 -0.64 -0.62 -0.60 -0.58 -0.56 -0.54  [25] -0.52 -0.50 -0.48 -0.46 -0.44 -0.42 -0.40 -0.38 -0.36 -0.34 -0.32 -0.30  [37] -0.28 -0.26 -0.24 -0.22 -0.20 -0.18 -0.16

I have a problem with R. I try to compute the confidence interval for my df. When I want to create the plot I have this problem: Error in xy.coords(x, y, xlabel, ylabel, log) : 'x' and 'y' lengths differ. I try this code: library(dplR) df.rwi <- detrend(rwl = df, method = "Spline",nyrs=NULL) write.table(df.rwi,file="rwi.txt",quote=FALSE,row.names=TRUE)

Hi, list- ?? I've got a large list of multi-column data and I'd like to filter it according to a threshold, such as, "show me the values that are above 0.4 for each line." For instance, if the line of data were: > line v1 v2 v3 v4 v5 v6 v7 v8 v9 1 -0.32 0.66 -0.35 -0.82 0.38 0.66 -0.02 -0.11 -0.64 I can do this > line[,line >= 0.4]

Hi all, I have this problem with lattice that xyplot() won't draw some of my axis labels if the type (i.e. the relation argument) of scales is set as free. For example, in the plot below, I would want it to also show: 1. the labels E1,...E6 below the 10th panel (i.e. 3rd row, 2 col)....just as it is now done below the 12th panel.... 2. as well as the labels (2,4,6,8) on the top of panels 1

Dear R Users,i am very new to R. I want your help on an issue regarding distance matrix and cluster analysis i had discharge data of 4 rivers(a,b,c,d) in 4 vectors each having 364 values > dput(qmu)structure(list(a = c(0.26, 0.25, 0.25, 0.25, 0.24, 0.23, 0.22, 0.21, 0.21, 0.21, 0.2, 0.19, 0.19, 0.19, 0.19, 0.18, 0.18, 0.18, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17,

Is this how a t hypothesis test is done when I don't have the actual data, but just the summarized statistics: > #Homework 9.2.6 [1] > n<-31 > xbar<-3.10 > s_x<-1.469 > m<-57 > ybar<-2.43 > s_y<-1.35 > s_pooled<- (((n-1)*s_x^2) + ((m-1)*s_y^2)) / (n + m - 2) > s_pooled [1] 1.939521 > t_obs <- (xbar - ybar) / (s_pooled * (sqrt(1/n + 1/m)))

Hi there, Here is a minimum working example: ---------------------------------------------------------------- lower = 0 upper = 1 n_bins = 50 interval = (upper - lower) / n_bins bins = vector(mode="numeric", length=n_bins) breaks = seq(from=lower + interval, to=upper, by=interval) for(idx in breaks) { bins[idx / interval] = idx } print(bins)

hi all, I want to move some routing tasks out of the router and into the controller. The goal is to make Merb feel less like mod_rewrite and give the user more control at the controller. The new Router is simple: it takes the path_info (not the whole request) then outputs a controller class and some parameters from the path matching. The rest of the routing would be done at the controller level.

Hello, I have three datasets StartSignals, MainData, StopSignals and need to compound the data for each variable in MainData over dates that fall between the Start and Stop signals. (Stop signals are common and the same to all X1:X5 variables). Please see sample below: The one way I was thinking of doing this project was to setup a nested "FOR" loop and go through the three data

I was trying to plot some data in R. I used the following code to draw a loess fit and got the output as >?lines(lowess(log(abs(t(res))), log(abs(t(synthesised)))), col="red") Error in lowess(log(abs(t(res))), log(abs(t(synthesised)))) :?? NA/NaN/Inf in foreign function call (arg 1) Then I thought to use your Limma package for background correction. Do you think it's a right

hello, i'm pete ,how can i order rows of matrix by max to min value? I have a matrix of membership degrees, with 82 (i) rows and K coloumns, K are clusters. I need first and second largest elements of the i-th row. for example 1 0.66 0.04 0.01 0.30 2 0.02 0.89 0.09 0.00 3 0.06 0.92 0.01 0.01 4 0.07 0.71 0.21 0.01 5 0.10 0.85 0.04 0.01 6 0.91 0.04 0.02 0.02 7 0.00 0.01 0.98 0.00 8 0.02

Dear list, First my information: platform i386-pc-linux-gnu arch i386 os linux-gnu system i386, linux-gnu status major 2 minor 3.1 year 2006 month 06 day 01 svn rev 38247 language R version.string Version 2.3.1 (2006-06-01) Now my question: How is it possible that a command in an R script is not

Hi guys, Suppose I have 2 data frames ie: values one 0.32 two 0.25 three 0.11 and values two 0.66 one 0.74 three 0.19 nb the first column is the row names in both cases How can I combine them on the row names column? Ie to make something like values.1 values.2 one 0.32 0.74 two 0.25 0.66 three 0.11 0.19 I guess its

Dear mailing list, I would like to use the optim() command in order to maximize the logged likelihood of the following function, where p is the parameter of interest and should be constrained between 0 and positive infinity. y = 1/2 * ((te - x)/(te - tc))^p x and y are given by x <- c(5.18, 6.28, 7.00, 7.08, 7.54, 7.90, 8.24, 8.64, 12.17, 12.89, 14.27, 15.38, 15.80, 16.46, 20.41, 21.27,

Greetings.? My wife is teaching an introductory stat class at UC Davis.? The class emphasizes the use of simulations, rather than mathematics, to get insight into statistics, and R is the mandated tool.?? A student in the class recently inquired about different approaches to sampling from a binomial distribution.? I've appended some code that exhibits the idea, the gist of which is that using

Hello R users, I've used the following help two compare two regression line slopes. Wanted to test if they differ significantly: Hi, I've made a research about how to compare two regression line slopes (of y versus x for 2 groups, "group" being a factor ) using R. I knew the method based on the following statement : t = (b1 - b2) / sb1,b2 where b1 and b2 are the two slope

I've completed my Basic Statistics course, and am pleased overall with the experience. In addition to earning a grade of 105.82% (smart-ass comment from wife: "How is that even statistically possible?"), I feel as if I have a comprehensive understanding of not only the procedures and formulas of basic statistics, but also know where the theory and numbers come from. One contribution