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Description of Statistical Computation

Author's title

Author*The author of this computation has been verified*
R Software Modulerwasp_meanplot.wasp
Title produced by softwareMean Plot
Date of computationSun, 02 Nov 2008 03:46:31 -0700
Cite this page as followsStatistical Computations at FreeStatistics.org, Office for Research Development and Education, URL https://freestatistics.org/blog/index.php?v=date/2008/Nov/02/t1225622858g7zv4ezuds1xkwn.htm/, Retrieved Sun, 19 May 2024 12:03:26 +0000
Statistical Computations at FreeStatistics.org, Office for Research Development and Education, URL https://freestatistics.org/blog/index.php?pk=20485, Retrieved Sun, 19 May 2024 12:03:26 +0000
QR Codes:

Original text written by user:
IsPrivate?No (this computation is public)
User-defined keywords
Estimated Impact199

Tree of Dependent Computations

Family? (F = Feedback message, R = changed R code, M = changed R Module, P = changed Parameters, D = changed Data)
F     [Mean Plot] [workshop 3] [2007-10-26 12:14:28] [e9ffc5de6f8a7be62f22b142b5b6b1a8]
F    D  [Mean Plot] [Q2: seasonality] [2008-11-02 08:55:25] [4396f984ebeab43316cd6baa88a4fd40]
F R P       [Mean Plot] [Task 4: Q2] [2008-11-02 10:46:31] [54ae75b68e6a45c6d55fa4235827d5b3] [Current]
Feedback Forum
2008-11-05 16:11:46 [Ken Van den Heuvel] [reply] 
Je concludeerde hier niets.

We nemen 5% weg van de 2 staarten. Ons interval verkleint dus en we krijgen er één van 90% (100% – 2 x 5% = 90%). Door van iedere staart te nemen, vallen er dus outliers weg. Maw, de spreiding verkleint.

Uit de verscheidene plot kan je dit eveneens concluderen.

Door outliers/randwaarden weg te laten vallen zwak je de extreme waarden af, zo kan je op de mean plot zien dat de pieken bij 6 en 10 minder uitgesproken zijn. Op de periodic subseries ze je tevens dat de betrouwbaarheidsintervallen van deze data mooier aansluiten (minder afwijken dan in het oorspronkelijke geval) bij de rest van de gegevens.
2008-11-08 11:10:22 [Astrid Sniekers] [reply] 
Het antwoord is correct, maar veel te kort. We kunnen besluiten dat er een verschil is met de resultaten van vraag Q2 en Q3. Dit komt omdat we door het trimmen van de tijdreeks, de outliers uit onze tijdreeks hebben gehaald. Bijgevolg wordt de spreiding kleiner.
2008-11-11 15:46:40 [Ellen Smolders] [reply] 
Uit het antwoord kunnen we niet veel afleiden. Wanneer we de quantiles veranderen en zo de staart langs beide kanten verkleinen, vallen de outliers weg, hierdoor verkleinen we de spreiding.

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Dataset

Dataseries X:
Download CSV
Histogram
Boxplots 
109.20
88.60
94.30
98.30
86.40
80.60
104.10
108.20
93.40
71.90
94.10
94.90
96.40
91.10
84.40
86.40
88.00
75.10
109.70
103.00
82.10
68.00
96.40
94.30
90.00
88.00
76.10
82.50
81.40
66.50
97.20
94.10
80.70
70.50
87.80
89.50
99.60
84.20
75.10
92.00
80.80
73.10
99.80
90.00
83.10
72.40
78.80
87.30
91.00
80.10
73.60
86.40
74.50
71.20
92.40
81.50
85.30
69.90
84.20
90.70
100.30

Tables (Output of Computation)





Summary of computational transaction
Raw Inputview raw input (R code)
Raw Outputview raw output of R engine
Computing time2 seconds
R Server'Gwilym Jenkins' @ 72.249.127.135

\begin{tabular}{lllllllll}
\hline
Summary of computational transaction \tabularnewline
Raw Input & view raw input (R code)  \tabularnewline
Raw Output & view raw output of R engine  \tabularnewline
Computing time & 2 seconds \tabularnewline
R Server & 'Gwilym Jenkins' @ 72.249.127.135 \tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=20485&T=0

[TABLE]
[ROW][C]Summary of computational transaction[/C][/ROW]
[ROW][C]Raw Input[/C][C]view raw input (R code) [/C][/ROW]
[ROW][C]Raw Output[/C][C]view raw output of R engine [/C][/ROW]
[ROW][C]Computing time[/C][C]2 seconds[/C][/ROW]
[ROW][C]R Server[/C][C]'Gwilym Jenkins' @ 72.249.127.135[/C][/ROW]
[/TABLE]
Source: https://freestatistics.org/blog/index.php?pk=20485&T=0

Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=20485&T=0

As an alternative you can also use a QR Code:  
QR Code


The GUIDs for individual cells are displayed in the table below:

Summary of computational transaction
Raw Inputview raw input (R code)
Raw Outputview raw output of R engine
Computing time2 seconds
R Server'Gwilym Jenkins' @ 72.249.127.135


Figures (Output of Computation)

Input Parameters & R Code

Parameters (Session):
par1 = red ;
Parameters (R input):
par1 = 12 ;
R code (references can be found in the software module):
par1 <- as.numeric(par1)
(n <- length(x))
(np <- floor(n / par1))
arr <- array(NA,dim=c(par1,np+1))
ari <- array(0,dim=par1)
j <- 0
for (i in 1:n)
{
j = j + 1
ari[j] = ari[j] + 1
arr[j,ari[j]] <- x[i]
if (j == par1) j = 0
}
ari
arr
arr.mean <- array(NA,dim=par1)
arr.median <- array(NA,dim=par1)
arr.midrange <- array(NA,dim=par1)
for (j in 1:par1)
{
arr.mean[j] <- mean(arr[j,],na.rm=TRUE)
arr.median[j] <- median(arr[j,],na.rm=TRUE)
arr.midrange[j] <- (quantile(arr[j,],0.75,na.rm=TRUE) + quantile(arr[j,],0.25,na.rm=TRUE)) / 2
}
overall.mean <- mean(x)
overall.median <- median(x)
overall.midrange <- (quantile(x,0.75) + quantile(x,0.25)) / 2
bitmap(file='plot1.png')
plot(arr.mean,type='b',ylab='mean',main='Mean Plot',xlab='Periodic Index')
mtext(paste('#blocks = ',np))
abline(overall.mean,0)
dev.off()
bitmap(file='plot2.png')
plot(arr.median,type='b',ylab='median',main='Median Plot',xlab='Periodic Index')
mtext(paste('#blocks = ',np))
abline(overall.median,0)
dev.off()
bitmap(file='plot3.png')
plot(arr.midrange,type='b',ylab='midrange',main='Midrange Plot',xlab='Periodic Index')
mtext(paste('#blocks = ',np))
abline(overall.midrange,0)
dev.off()
bitmap(file='plot4.png')
z <- data.frame(t(arr))
names(z) <- c(1:par1)
(boxplot(z,notch=TRUE,col='grey',xlab='Periodic Index',ylab='Value',main='Notched Box Plots - Periodic Subseries'))
dev.off()
bitmap(file='plot5.png')
z <- data.frame(arr)
names(z) <- c(1:np)
(boxplot(z,notch=TRUE,col='grey',xlab='Block Index',ylab='Value',main='Notched Box Plots - Sequential Blocks'))
dev.off()
bitmap(file='plot6.png')
z <- data.frame(cbind(arr.mean,arr.median,arr.midrange))
names(z) <- list('mean','median','midrange')
(boxplot(z,notch=TRUE,col='grey',ylab='Overall Central Tendency',main='Notched Box Plots'))
dev.off()