Is JMP miscalculating my RMSE

Question

I plotted my observed vs predicted values, and added a line of fit with the RMSE, however, the values it's calculating/displaying is significantly lower than what I'm calculating and expecting (0.46 vs 2.07).

Here's the data:

6.3431.6071196.2596.1770046.1390.4295395.2666.053685.4518.408346.4925.2263756.525.4402725.7845.9906386.1546.075276.1035.9654316.1220.4458465.5417.0007255.1815.9314565.3555.6267065.6745.8490835.6425.8596075.1365.147445.5125.6930885.95.6900885.655.7925525.555.5137735.65.6595174.65.195817

Am I missing something?

Mark_Bailey · Answer

The error degrees of freedom equal the number of observations (23) minus the number of parameters estimated (2). This script shows how you can calculate the RMSE and compares it to the result from the Bivariate platform. I'm curious why you calculate RMSE with Excel when JMP gives you the answer.

Names Default to Here( 1 );

// duplicate example in discussion
dt = New Table( "RMSE Example",
	Add Rows( 23 ),
	New Script(
		"Source",
		Open(
			"https://community.jmp.com/t5/Discussions/Is-JMP-miscalculating-my-RMSE/m-p/583546",
			HTML Table( 1, Column Names( 0 ), Data Starts( 1 ) )
		)
	),
	New Column( "X",
		Numeric,
		"Continuous",
		Format( "Best", 12 ),
		Set Values(
			[6.343, 6.259, 6.139, 5.266, 5.451, 6.492, 6.52, 5.784, 6.154, 6.103,
			6.122, 5.541, 5.181, 5.355, 5.674, 5.642, 5.136, 5.512, 5.9, 5.65, 5.55,
			5.6, 4.6]
		)
	),
	New Column( "Y",
		Numeric,
		"Continuous",
		Format( "Best", 12 ),
		Set Values(
			[1.607119, 6.177004, 0.429539, 6.05368, 8.40834, 5.226375, 5.440272,
			5.990638, 6.07527, 5.965431, 0.445846, 7.000725, 5.931456, 5.626706,
			5.849083, 5.859607, 5.14744, 5.693088, 5.690088, 5.792552, 5.513773,
			5.659517, 5.195817]
		)
	)
);

// calculate RMSE with Bivariate platform
obj = Bivariate( Y( :Y ), X( :X ), Fit Line( 1 ) );
bivRMSE = (obj << Report)["Summary of Fit"][NumberColBox(1)] << Get( 3 );

// calculate RMSE uing direct linear regression
yData = dt:Y << Get As Matrix;
xData = dt:X << Get As Matrix;
{ estimate, se, diagnostic } = Linear Regression( yData, xData );
predY = estimate[1] + estimate[2] * xData;
df = N Row( yData ) - 2;	// error df = n - 2 for parameter estimates
manRMSE = Sqrt( Sum( (yData - predY)^2 ) / df );

// compare results
Show( bivRMSE, manRMSE );

lazzybug · Answer

I just realized I made a mistake by ignoring df. The formula above is not right, should divide by degree of residuals instead of n, and then square root of it.

Mark_Bailey · Answer

Simple question: do you take the square root of the MSE then to obtain the RMSE?

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