Error Analysis involving random errors - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-09-17T19:22:45Z https://physics.stackexchange.com/feeds/question/260809 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/260809 1 Error Analysis involving random errors user104014 https://physics.stackexchange.com/users/104014 2016-06-05T06:25:00Z 2016-06-05T16:12:10Z <p>The question goes like this:</p> <blockquote> <p>In an experiment, the time period of an oscillating object in five successive measurements is found to be $0.52$s, $0.56$s, $0.57$s, $0.54$s, $0.59$s. The least count of the watch used for the measurement of time period is $0.01$s. What is the percentage error in measurement of time period $T$.</p> </blockquote> <p><strong>My attempt:</strong> The maximum error in measurement of T due to limited precision of the measuring instrument is the least count <em>i.e.</em> $0.01$s. Also the mean of measured values is $$\frac { 0.52+0.56+0.57+0.54+0.59 }{ 5 } =0.556$$which when rounded off to 2 significant figures is $0.56$. Also the standard deviation can be calculated after rounding off as $0.02$, which can be a good estimate to random error. Hence, the value of $T$ can be written as $0.56\pm (0.01+0.02)=0.56\pm 0.03$s. Hence the percentage error should be $$\frac { 0.03 }{ 0.56 } \times 100\approx 5.357$$ Hence the percentage error should be $5.357$%.<br><br> But the answer given in the book is $3.57$%. How is this possible? Where did I commit a mistake?</p> https://physics.stackexchange.com/questions/260809/-/260815#260815 2 Answer by Farcher for Error Analysis involving random errors Farcher https://physics.stackexchange.com/users/104696 2016-06-05T07:31:20Z 2016-06-05T07:31:20Z <p>I think you are confusing systematic and random errors.<br> Your experimental results can give you no idea about the systematic error.<br> For example it might be that your timing device is calibrated incorrectly and when the correct time is 1.00 seconds then your timing device gives a reading of 1.10 seconds; when the correct time is 2.00 seconds the timing device gives a reading of 2.20 seconds.<br> Repeating readings or the smallest subdivision of your scale will not give you an indication of what the systematic error is.<br> You could only find that error by checking the calibration of your timing device against a reliable standard.</p> <p>So in this example you have found an estimate of the random error by evaluation the standard deviation and that is the best you can do.</p> https://physics.stackexchange.com/questions/260809/-/260825#260825 1 Answer by valerio for Error Analysis involving random errors valerio https://physics.stackexchange.com/users/115736 2016-06-05T08:35:51Z 2016-06-05T16:12:10Z <blockquote> <p>The least count of the watch used for the measurement of time period is $0.01$ s</p> </blockquote> <p>This information is just telling you to round off to the second decimal place, as you correctly did.</p> <p>The sample mean is $\mu = 0.56$ and the sample standard deviation is $\sigma = 0.02$. The answer the text is referring to is</p> <p>$$\frac \sigma \mu = 0.0357 = 3.57 \%$$</p> <p>But I would say that this is not entirely correct. The standard error is not $\sigma$, but </p> <p>$$\frac \sigma {\sqrt N}$$</p> <p>Where $N$ is the number of measurements. In our case,</p> <p>$$\frac \sigma {\sqrt N}=0.009$$</p> <p>So the real percentage error should be</p> <p>$$\frac{0.009}{0.56} = 0.0161 = 1.61 \%$$</p> <p><strong>Update: a more careful discussion</strong></p> <p>As requested, I will try to explain more why we don't need to explicitly include the resolution of the instrument ($0.01/2$) in our calculation.</p> <p>In my previous discussion I explained why the solution reported in your text was $35.7 \%$, but actually that reasoning is not really correct.</p> <p>The sample mean of your data set is not really $\mu=0.56$, but $\mu=0.0556$, as you correctly wrote. But since they (incorrectly) used the standard deviation, $0.02$, as standard error, we have to round off the mean and write our result as</p> <p>$$0.56 \pm 0.02$$</p> <p>Because it would clearly be silly to write</p> <p>$$0.556 \pm 0.02$$</p> <p>because if we are not sure of the second decimal place we bother writing the third?</p> <p>But if the correct standard error is used, we get</p> <p>$$0.556 \pm 0.009$$</p> <p>You may notice a strange thing: <em>the number of significative digits has increased</em>, even if our instrument had a resolution of only $0.01/2=0.005$. This is a property of the mean and it is why we use the mean in the first place: <em>via the mean operation, we can increase the number of significative digits and circumvent the limitations of our instrument</em>.</p> <p>Take for example the case in which we have two measurements: $2$ and $7$, with resolution of $0.5$ clearly. The mean is $9/2=4.5$, so we have gained one significative place. </p> <p>You can then see that with an infinite number of measurement our result becomes exact, <em>regardless of the resolution of the instrument</em>, because of the $\sqrt N$ term in the denominator of the standard error.</p>