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Std Dev Lower and Upper 95% from oneway analysis details of calculation
Hello,
I am in JMP17.2.0 Pro
could you please explain me how is computed Std Dev Lower and Upper 95% in the example below ?
Especially why the interval is not symmetric around the StdDev value ?
Thank you very much.
dt = Open( "$SAMPLE_DATA/Big Class.jmp" );
platoneway = Oneway(
Y( :weight ),
X( :age ),
Each Pair( 1 ),
All Pairs( 1 ),
Comparison Circles( 0 ),
Means and Std Dev( 1 ),
Box Plots( 1 ),
Mean Diamonds( 1 ),
Mean Error Bars( 1 ),
Std Dev Lines( 1 ),
Comparison Circles( 0 ),
Grand Mean( 0 ),
SendToReport(
Dispatch(
{"Means and Std Deviations"},
"Std Dev Lower 95%",
NumberColBox,
{Visibility( "Visible" )}
),
Dispatch(
{"Means and Std Deviations"},
"Std Dev Upper 95%",
NumberColBox,
{Visibility( "Visible" )}
),
Dispatch(
{"Means and Std Deviations"},
"~Std Dev",
PlotColBox,
{Visibility( "Visible" )}
),
Dispatch( {}, "Means Comparisons", OutlineBox, {Close( 1 )} )
)
);
Accepted Solutions
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Re: Std Dev Lower and Upper 95% from oneway analysis details of calculation
Hi @carole : Yes, it is not symmetric. It is calculated as shown in the link below.
https://www.statology.org/confidence-interval-standard-deviation/
Why not symmetric? The answer has to with the distribution of the sample SD; it is based on a skewed distribution (the Chi-Squared distribution).
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Re: Std Dev Lower and Upper 95% from oneway analysis details of calculation
Hi @carole : Yes, it is not symmetric. It is calculated as shown in the link below.
https://www.statology.org/confidence-interval-standard-deviation/
Why not symmetric? The answer has to with the distribution of the sample SD; it is based on a skewed distribution (the Chi-Squared distribution).
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Re: Std Dev Lower and Upper 95% from oneway analysis details of calculation
Thanks a lot for the link very interesting, and your answer.
Best regards,
Carole