# Calculate the margin of error in this physics lab

• VitaminK

#### VitaminK

Homework Statement
Hey everyone,
I recently performed a lab at school where I had to drop a bouncy ball, let it bounce 4 times, and after each bounce write down the height to which the ball bounces. The procedure was repeated 4 times in total. In each of the four trials, I had variations in the observed heights. Now I have to calculate margin of error.
Relevant Equations
margin of error: sx̄ = s/√N
Bounce 1, Trial 1-4:
ha (height after bounce)

(0.74+0.67+0.69+0.73)/4=0,7075 this is my mean value.

Standard Deviation, s: 0.033040379335998 (calculated the value using an online standard deviation calculator)
Sample size N: 4
sx̄ = s/√N = 0.033040379335998/√4≈ 0.0165

At a 95% confidence interval the margin of error is 0,0165*1,96=0,03234

So I'm guessing the answer for the first bounce is ±0,032m.

Is this correct and is there an easier way to calculate? It's been awhile since I've done this type of calculation, but I don't think that's the way to do it. You have found the margin of error for the average of 4 bounces. This cannot be applied to the 1st bounce. Just look at the data from the first bounce. Would you say with 95% confidence that those fall within +- 0.032 m ?

• VitaminK
It's been awhile since I've done this type of calculation, but I don't think that's the way to do it. You have found the margin of error for the average of 4 bounces. This cannot be applied to the 1st bounce. Just look at the data from the first bounce. Would you say with 95% confidence that those fall within +- 0.032 m ?

Well, I guess not. Should I instead take the sum of all ha then divide by 16 in order to get the mean value and then calculate the margin of error?

Here is how I understand it. Starting from the same height, you made 4 independent measurements of the height after one bounce. You got a mean value of 0.71 m (you must round to 2 sig figs). You found a standard deviation of 0.03 m. Your measured value is ##\text{0.71} \pm \text{0.03 m}##. Assuming a normal (Gaussian) parent distribution, this means that if you repeat the measurement many times, about 68% of the values will be between 0.74 and 0.67 m. In your specific case it turns out that all your measured values are within that range. If you were to plot this point, you would put a data marker at 0.71 m and add an error bar 0.03 m long on either side of it.

• VitaminK and scottdave
Thank you guys for the help. I Think I got it now :)

Here is how I understand it. Starting from the same height, you made 4 independent measurements of the height after one bounce. You got a mean value of 0.71 m (you must round to 2 sig figs). You found a standard deviation of 0.03 m. Your measured value is ##\text{0.71} \pm \text{0.03 m}##. Assuming a normal (Gaussian) parent distribution, this means that if you repeat the measurement many times, about 68% of the values will be between 0.74 and 0.67 m. In your specific case it turns out that all your measured values are within that range. If you were to plot this point, you would put a data marker at 0.71 m and add an error bar 0.03 m long on either side of it.
This is not quite right. Consider, for example, if 100 measurements had been taken, still with the same variance in the set of measurements. That calculation would not yield any improvement in the reliability of the mean value, yet the uncertainty in the mean value should be reducing according to 1/√n.

What is wanted is the standard error of the mean.
https://en.m.wikipedia.org/wiki/Standard_error#Standard_error_of_the_mean

If the variance, Average(xi2) - (Average xi)2, of a set of N measurements is σ2 then the standard error of the mean is σ/√(N-1).

In the present case, the 0.033 does not represent σ in this sense because the STDEV algorithm divided by √3 instead of √4. So we should use σ=0.033*√3/√4. The standard error of the mean is then σ/√3 = 0.033/√4.

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n the present case, the 0.033 does not represent σ in this sense because the STDEV algorithm divided by √3 instead of √4. So we should use σ=0.033*√3/√4. The standard error of the mean is then σ/√3 = 0.033/√4.
Yes, of course. Thanks.

Yes, of course. Thanks.
There is a caveat even with the standard error of mean formula. The theoretical basis is that you infer the accuracy of the measurements from their scatter. E.g. with only one the scatter is unlimited and the error bar infinite.
In reality, you would always be able to set a limit on the error even for a single measurement. The observed scatter in multiple measurements then serves to refine this. So the formula is fine with a lot of measurements, but would tend to overestimate with only a few (as here).
A full treatment would require a Bayesian approach, I guess

E.g. with only one the scatter is unlimited and the error bar infinite.
That I agree with. It would be the a priori scatter in the Bayesian approach that you mentioned.
In reality, you would always be able to set a limit on the error even for a single measurement.
Are you saying that the a priori scatter can be limited without bringing in additional information?

It would be the a priori scatter in the Bayesian approach that you mentioned.
It's more complicated than that. You would need an a priori distribution of possible variances. As you collect measurements, the variance of these adjusts that distribution. In the limit, the distribution of variances converges to a spike at the observed variance.
Are you saying that the a priori scatter can be limited without bringing in additional information?
No, I mean using circumstantial information. E.g. if measuring a distance against a graduation by eye you may be able to put limits on the error based on the fineness of the graduation, parallax, etc.

I guess my understanding of "a priori" in a Bayesian-like approach is different from yours. To me it means the first guess before considering input from any additional source of information whether circumstantial or not. It's like the zeroth order term in a series expansion; all one has is a single point with no idea whether the function increases, decreases, has curvature, etc. Determining the sign of the slope and placing an upper limit to its magnitude because, say it represents the time dependence of the speed of a ball released from rest in a viscous medium is introducing new information.

I guess my understanding of "a priori" in a Bayesian-like approach is different from yours. To me it means the first guess before considering input from any additional source of information whether circumstantial or not. It's like the zeroth order term in a series expansion; all one has is a single point with no idea whether the function increases, decreases, has curvature, etc. Determining the sign of the slope and placing an upper limit to its magnitude because, say it represents the time dependence of the speed of a ball released from rest in a viscous medium is introducing new information.
If you mean a first guess for the value of a statistic, like a mean, no. With that view it is unclear how you wouid adjust your estimate based on additional info.
In the Bayesian approach you start with a guess of a probability distribution for some quantity and adjust that based on data according to Bayes' principle. If the thing you are trying to arrive at is a probability distribution then you start with a distribution of distributions.

The principle works no matter where you start. Bayes' principle is a rule for adjusting probabilities in the face of new information. That you have to start with some sort of guess, apparently without info, is just a corollary. Having adjusted it based on some observations you could then take that as your prior and go on from there. In the same way, in practice, your first guess is always based on something, maybe intuition based on life experience. Note that if you guess zero for a probability it will always be zero in this scheme. By guessing nonzero you are saying you believe it possible.

If you mean a first guess for the value of a statistic, like a mean, no.
Not what I meant, but if I tried to explain it, I suspect I will head straight into an epistemological morass so I will stop here. However I agree with what you say in #12.