How to compare a standard deviation to a root mean square error?

Question

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1. What the problem is asking (in plain language)

You have two magnetic‑field sensors and you want to decide which one is “more accurate”.

Sensor A (Bosch BMM150) – the datasheet says the error is ± 2.5° at ± 3σ.
This tells us the standard deviation of the error distribution.
Sensor B – the specification only gives an RMSE of 2.5°.

The question: Can we tell, from these numbers alone, which sensor gives more accurate heading measurements?

To answer we must translate the two numbers into the same statistical quantity (usually a standard deviation) and see which is smaller.

2. Step‑by‑step solution

2.1 Definitions

Symbol	Meaning	Formula
(e_i)	error of the i‑th measurement (measured – true)	–
(\mu)	bias (mean error)	(\displaystyle \mu = \frac1n\sum_{i=1}^n e_i)
(\sigma)	standard deviation (spread of the errors)	(\displaystyle \sigma = \sqrt{\frac1n\sum_{i=1}^n (e_i-\mu)^2})
RMSE	Root‑Mean‑Square Error	(\displaystyle \text{RMSE}= \sqrt{\frac1n\sum_{i=1}^n e_i^{\,2}})

From the definitions we obtain the fundamental identity

[ \text{RMSE}^2 = \underbrace{\mu^{2}}{\text{bias}^2}+\underbrace{\sigma^{2}}{\text{variance}}. \tag{1} ]

Thus RMSE mixes two sources of error:

a systematic offset (bias), and
a random spread (standard deviation).

If the sensor is unbiased ((\mu=0)), then

[ \text{RMSE}= \sigma . \tag{2} ]

Otherwise the RMSE is larger than the standard deviation.

2.2 Convert the Bosch specification to a standard deviation

The datasheet states:

“Accuracy ± 2.5° at ± 3σ”

Interpretation: the error distribution is (approximately) normal, and 99.7 % of the errors lie within ± 2.5°. For a normal distribution

[ \text{range for } \pm 3\sigma = \pm 3\sigma . ]

Therefore

[ 3\sigma = 2.5^\circ \quad\Longrightarrow\quad \sigma_{\text{Bosch}} = \frac{2.5^\circ}{3}=0.8333^\circ . \tag{3} ]

So the Bosch sensor’s standard deviation is about 0.83° (assuming the error is unbiased, which is the usual assumption when a “± 3σ” accuracy is quoted).

2.3 What does an RMSE of 2.5° tell us about Sensor B?

The specification gives

[ \text{RMSE}_{\text{B}} = 2.5^\circ . \tag{4} ]

Using (1),

[ \sigma_{\text{B}} = \sqrt{\text{RMSE}{\text{B}}^{2} - \mu{\text{B}}^{2}} . \tag{5} ]

Two possibilities:

Situation	Consequence
(a) Sensor B is unbiased ((\mu_{\text{B}}=0))	(\sigma_{\text{B}} = \text{RMSE}_{\text{B}} = 2.5^\circ).
(b) Sensor B has a non‑zero bias ((	\mu_{\text{B}}	>0))	(\sigma_{\text{B}} = \sqrt{(2.5^\circ)^2 - \mu_{\text{B}}^{2}} \;<\; 2.5^\circ). The larger the bias, the smaller the random spread.

Because the bias is unknown from the specification, the best‑case random spread for Sensor B is obtained when all of the RMSE comes from bias (i.e. (\sigma_{\text{B}}=0)). The worst‑case random spread occurs when the sensor is unbiased, giving (\sigma_{\text{B}} = 2.5^\circ).

Thus the maximum possible standard deviation for Sensor B is 2.5°, which is far larger than the Bosch sensor’s 0.83°. Even if Sensor B had a large bias and a smaller σ, its overall error (RMSE) would still be 2.5°, meaning that on average its absolute error is larger than Bosch’s.

2.4 Direct comparison

Sensor	Known quantity	Implied (\sigma) (if unbiased)
Bosch (A)	± 2.5° at 3σ	(\sigma_{\text{A}} = 0.83^\circ)
Other (B)	RMSE = 2.5°	(\sigma_{\text{B}} = 2.5^\circ) (maximum)

Because 0.83° < 2.5°, the Bosch sensor’s random error is much smaller. Even allowing for an unknown bias in Sensor B, the overall error (RMSE) of 2.5° is larger than the typical error (≈0.83°) of the Bosch sensor.

Conclusion: The Bosch BMM150 sensor is more accurate (i.e., it provides tighter, less noisy heading estimates) than a sensor whose only specification is RMSE = 2.5°, unless the latter sensor has a systematic bias that is somehow acceptable for the application.

If the application cares only about the average magnitude of error, the Bosch sensor’s expected absolute error (≈0.83° × (\sqrt{2/\pi}) ≈ 0.66° for a normal distribution) is still smaller than 2.5°.

3. Final answer

The Bosch BMM150 sensor, with a standard deviation of about 0.83°, is more accurate than a sensor whose RMSE is 2.5°. The RMSE of 2.5° implies a standard deviation of at most 2.5° (if the sensor is unbiased) and, even in the most favorable bias scenario, its average error is still larger than the Bosch sensor’s typical error.

4. Common mistakes

Mistake	Why it’s wrong	How to avoid it
Treating “± 2.5° at ± 3σ” as “± 2.5° = 1σ”.	The 2.5° corresponds to three standard deviations, not one.	Divide by 3 to obtain σ (as done in Eq. 3).
Equating RMSE directly to σ without checking bias.	RMSE = √(bias² + σ²); a non‑zero bias makes RMSE larger than σ.	Use Eq. (1); if bias is unknown, state the assumption (usually “unbiased”) and discuss the effect.
Comparing a confidence interval (± 3σ) to an average error (RMSE).	They describe different aspects (tail probability vs. mean squared error).	Convert both numbers to the same metric (σ or RMSE) before comparing.
Assuming the error distribution is normal when it isn’t.	The σ‑to‑confidence‑interval relation (± 3σ ≈ 99.7 %) only holds for a normal distribution.	Verify the assumption or note that the conversion is an approximation.
Ignoring the units or mixing degrees with radians.	A numeric comparison must be in the same units.	Keep all angles in the same unit (degrees here).

By keeping these points in mind, you can correctly compare specifications that use different error metrics.

Original question: How to compare a standard deviation to a root mean square error? on Cross Validated (Stats Stack Exchange), licensed CC BY-SA.