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Description of Statistical Computation

Author's title

Author*The author of this computation has been verified*
R Software Modulerwasp_hypothesismean1.wasp
Title produced by softwareTesting Mean with known Variance - Critical Value
Date of computationTue, 11 Nov 2008 13:06:59 -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/11/t1226434062cq5d4tgn48ulp0s.htm/, Retrieved Sun, 19 May 2024 10:09:27 +0000
Statistical Computations at FreeStatistics.org, Office for Research Development and Education, URL https://freestatistics.org/blog/index.php?pk=23918, Retrieved Sun, 19 May 2024 10:09:27 +0000
QR Codes:

Original text written by user:
IsPrivate?No (this computation is public)
User-defined keywords
Estimated Impact137

Tree of Dependent Computations

Family? (F = Feedback message, R = changed R code, M = changed R Module, P = changed Parameters, D = changed Data)
F     [Testing Mean with known Variance - Critical Value] [Q1] [2008-11-04 11:50:48] [c5a66f1c8528a963efc2b82a8519f117]
F         [Testing Mean with known Variance - Critical Value] [] [2008-11-11 20:06:59] [f6a332ba2d530c028d935c5a5bbb53af] [Current]
Feedback Forum
2008-11-14 09:56:42 [Ken Van den Heuvel] [reply] 
Je stelt:'Indien we echter wel rekening zouden houden met het contractueel bepaalde vetgehalte (van 15%) dan ligt de gemiddelde waarde van de stalen er wel boven. In deze situatie zou er wel een klacht ingediend moeten worden.'

Je houd hier geen rekening met het volgende:
de contractuele variatie is 1,2%, in principe mogen de leveringen dus tussen 13,8% en 16,2% vet bevatten.

Dat is de hele reden waarom we deze berekeningen maken, om te zien of het gemiddelde van 15,46% acceptabel is binnen de grenzen van het contract. Dit is hier het geval, je stelling is maw fout.

Tevens stel je dat: 'We nemen one-tailed omdat er maar 1 maximale grens bepaald is nl. dat het vetgehalte niet hoger mag zijn dan de critical value indien de type I error 5% bedraagt.'

Omdat in deze vraag niets gespecifieerd werd mbt vermoedens van fraude in welke richting dan ook (te veel of te weinig vet). Had het gebruik van de 2zijdige methode rechtvaardigbaar geweest en zelfs passender bij deze algemene vraagstelling.
Je antwoord is wel juist, je argumentie waarom je de 1zijdige methode gebruikt is dit niet.
2008-11-19 19:58:11 [An De Koninck] [reply] 
Het antwoord van de student is correct, maar hij kon er nog bij vermelden dat dit komt doordat de sample mean in het betrouwbaarheidsinterval ligt.

Ken, je opmerking dat de test ook tweezijdig kon zijn is wel correct. In principe zou dit zo kunnen, maar in de opgave staat wel het volgende: the supplier is required to pay a penalty fee when the pork's fat content deviates TOO MUCH. Hier staat het dus eigenlijk al dat je maar een 1-sided test moet gebruiken...

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Dataset

Tables (Output of Computation)





Summary of computational transaction
Raw Inputview raw input (R code)
Raw Outputview raw output of R engine
Computing time1 seconds
R Server'Sir Ronald Aylmer Fisher' @ 193.190.124.24

\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 & 1 seconds \tabularnewline
R Server & 'Sir Ronald Aylmer Fisher' @ 193.190.124.24 \tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=23918&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]1 seconds[/C][/ROW]
[ROW][C]R Server[/C][C]'Sir Ronald Aylmer Fisher' @ 193.190.124.24[/C][/ROW]
[/TABLE]
Source: https://freestatistics.org/blog/index.php?pk=23918&T=0

Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=23918&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 time1 seconds
R Server'Sir Ronald Aylmer Fisher' @ 193.190.124.24






Testing Mean with known Variance
sample size27
population variance0.012
sample mean0.1546
null hypothesis about mean0.15
type I error0.05
critical value (one-tailed)0.184676559191704
confidence interval (two-tailed)(sample mean)[ 0.113280331179696 , 0.195919668820304 ]
conclusion for one-tailed test
Do not reject the null hypothesis.
conclusion for two-tailed test
Do not reject the null hypothesis

\begin{tabular}{lllllllll}
\hline
Testing Mean with known Variance \tabularnewline
sample size & 27 \tabularnewline
population variance & 0.012 \tabularnewline
sample mean & 0.1546 \tabularnewline
null hypothesis about mean & 0.15 \tabularnewline
type I error & 0.05 \tabularnewline
critical value (one-tailed) & 0.184676559191704 \tabularnewline
confidence interval (two-tailed)(sample mean) & [ 0.113280331179696 ,  0.195919668820304 ] \tabularnewline
conclusion for one-tailed test \tabularnewline
Do not reject the null hypothesis. \tabularnewline
conclusion for two-tailed test \tabularnewline
Do not reject the null hypothesis \tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=23918&T=1

[TABLE]
[ROW][C]Testing Mean with known Variance[/C][/ROW]
[ROW][C]sample size[/C][C]27[/C][/ROW]
[ROW][C]population variance[/C][C]0.012[/C][/ROW]
[ROW][C]sample mean[/C][C]0.1546[/C][/ROW]
[ROW][C]null hypothesis about mean[/C][C]0.15[/C][/ROW]
[ROW][C]type I error[/C][C]0.05[/C][/ROW]
[ROW][C]critical value (one-tailed)[/C][C]0.184676559191704[/C][/ROW]
[ROW][C]confidence interval (two-tailed)(sample mean)[/C][C][ 0.113280331179696 ,  0.195919668820304 ][/C][/ROW]
[ROW][C]conclusion for one-tailed test[/C][/ROW]
[ROW][C]Do not reject the null hypothesis.[/C][/ROW]
[ROW][C]conclusion for two-tailed test[/C][/ROW]
[ROW][C]Do not reject the null hypothesis[/C][/ROW]
[/TABLE]
Source: https://freestatistics.org/blog/index.php?pk=23918&T=1

Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=23918&T=1

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


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

Testing Mean with known Variance
sample size27
population variance0.012
sample mean0.1546
null hypothesis about mean0.15
type I error0.05
critical value (one-tailed)0.184676559191704
confidence interval (two-tailed)(sample mean)[ 0.113280331179696 , 0.195919668820304 ]
conclusion for one-tailed test
Do not reject the null hypothesis.
conclusion for two-tailed test
Do not reject the null hypothesis


Figures (Output of Computation)

Input Parameters & R Code

Parameters (Session):
par1 = 27 ; par2 = 0.012 ; par3 = 0.1546 ; par4 = 0.15 ; par5 = 0.05 ;
Parameters (R input):
par1 = 27 ; par2 = 0.012 ; par3 = 0.1546 ; par4 = 0.15 ; par5 = 0.05 ;
R code (references can be found in the software module):
par1<-as.numeric(par1)
par2<-as.numeric(par2)
par3<-as.numeric(par3)
par4<-as.numeric(par4)
par5<-as.numeric(par5)
c <- 'NA'
csn <- abs(qnorm(par5))
csn2 <- abs(qnorm(par5/2))
if (par3 == par4)
{
conclusion <- 'Error: the null hypothesis and sample mean must not be equal.'
conclusion2 <- conclusion
} else {
cleft <- par3 - csn2 * sqrt(par2) / sqrt(par1)
cright <- par3 + csn2 * sqrt(par2) / sqrt(par1)
c2 <- paste('[',cleft)
c2 <- paste(c2,', ')
c2 <- paste(c2,cright)
c2 <- paste(c2,']')
if ((par4 < cleft) | (par4 > cright))
{
conclusion2 <- 'Reject the null hypothesis'
} else {
conclusion2 <- 'Do not reject the null hypothesis'
}
}
if (par3 > par4)
{
c <- par4 + csn * sqrt(par2) / sqrt(par1)
if (par3 < c)
{
conclusion <- 'Do not reject the null hypothesis.'
} else {
conclusion <- 'Reject the null hypothesis.'
}
}
if (par3 < par4)
{
c <- par4 - csn * sqrt(par2) / sqrt(par1)
if (par3 > c)
{
conclusion <- 'Do not reject the null hypothesis.'
} else {
conclusion <- 'Reject the null hypothesis.'
}
}
c
conclusion
load(file='createtable')
a<-table.start()
a<-table.row.start(a)
a<-table.element(a,hyperlink('ht_mean_knownvar.htm','Testing Mean with known Variance','learn more about Statistical Hypothesis Testing about the Mean when the Variance is known'),2,TRUE)
a<-table.row.end(a)
a<-table.row.start(a)
a<-table.element(a,'sample size',header=TRUE)
a<-table.element(a,par1)
a<-table.row.end(a)
a<-table.row.start(a)
a<-table.element(a,'population variance',header=TRUE)
a<-table.element(a,par2)
a<-table.row.end(a)
a<-table.row.start(a)
a<-table.element(a,'sample mean',header=TRUE)
a<-table.element(a,par3)
a<-table.row.end(a)
a<-table.row.start(a)
a<-table.element(a,'null hypothesis about mean',header=TRUE)
a<-table.element(a,par4)
a<-table.row.end(a)
a<-table.row.start(a)
a<-table.element(a,'type I error',header=TRUE)
a<-table.element(a,par5)
a<-table.row.end(a)
a<-table.row.start(a)
a<-table.element(a,hyperlink('ht_mean_knownvar.htm#overview','critical value (one-tailed)','about the critical value'),header=TRUE)
a<-table.element(a,c)
a<-table.row.end(a)
a<-table.row.start(a)
a<-table.element(a,'confidence interval (two-tailed)
(sample mean)',header=TRUE)
a<-table.element(a,c2)
a<-table.row.end(a)
a<-table.row.start(a)
a<-table.element(a,'conclusion for one-tailed test',2,header=TRUE)
a<-table.row.end(a)
a<-table.row.start(a)
a<-table.element(a,conclusion,2)
a<-table.row.end(a)
a<-table.row.start(a)
a<-table.element(a,'conclusion for two-tailed test',2,header=TRUE)
a<-table.row.end(a)
a<-table.row.start(a)
a<-table.element(a,conclusion2,2)
a<-table.row.end(a)
a<-table.end(a)
table.save(a,file='mytable.tab')