I want to visually inspect whether I have a similarly shaped distribution in a question having a scale dependent variable and a 3-group categorical (nominal) independent variable.
For that reason, I am trying to create a boxplot on spss v.23 but I am presented with the graph...
I am working on a pre-analysis plan and have to specify what I am going to do with outliers. I have two categorical variables (5 levels and 2 levels) and I will be performing a chi-square test for independence.
I thought of using a boxplot to detect outliers, but now I am not sure...
I am still new to SPSS and slowly getting to grips with it. At the moment, I am trying to create box-plots to visually identify extreme values in my distribution.
I have managed to get the box-plots how I want them, except for one detail: It is currently identifying the extreme...
I've got all the box plots:
boxplot( Daten$Gewicht~interaction(Daten$Dosis,Daten$Geschlecht, drop=TRUE), ylab="relative Nierengewichte", xlab="Dosisgruppen der beiden Geschlechter")
and the points overlapping the box plots:
I am using boxplots to show the distribution of some data.
However, some comments says the boxplot can not show the concentration quantity
of the data.
(For example, suppose I have data of body length ranging from 150- 169 cm of a group of girl.
The boxplot shows only the median...
If this is so obvious everyone should know I apologize for the inconvenience. I would like to add a color coded trend line to a series of box plots (graph attached)..
Any help is much appreciated..
The code I have used for the graph is as follows:
Originally, I thought this might be appropriate for the R thread, but I think this may have more general relevance.
I'm trying to decide on a data visualization option, and my right and left sides of my brain are at war. In brief, I've modeled predator densities at two sites (in a...
I have a simple question that I need your help with. I have a box plot which shows the medians, 1st and 3rd quartiles, and min/max values. I was wondering if I can get the mean and standard deviation using this information
Unfortunately I am not very statistically literate..please...