Error Anaylsis

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course Phys201

4/19 @8:26pm

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Most students report that it takes them between 1 hour and 2 hours to complete this experiment, with some as short as 45 minutes and some reporting 3 hours or even more. If you already know how to calculate the mean and standard deviation of a set of numbers, you will probably complete the experiment more quickly than that. Most students are not familiar with this calculation and do require an hour or two. However it is not difficult to calculate the standard deviation, and the instructions given here will lead you through the process.

Using the washer that came with your lab kit and a piece of thread to make a pendulum. If you don't have a washer, you may use a stack of about 4 CDs or DVDs. Hold the string so that the length of the pendulum, from the point at which you are holding it to the middle of the washer (or CD stack), is equal to the distance between your wrist and your elbow. Measure this length in cm, accurate to the nearest millimeter, and report it in the box below:

Your response (start in the next line):

24.4cm

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You are going to hold this pendulum with one hand as it oscillates, and you are going to click the TIMER program with every full cycle of its oscillation.

As the pendulum oscillates it moves back and forth between one extreme position and the other, stopping for an instant at each extreme position before beginning to move back to its middle position. The middle position is called its equilibrium position, because if you place the pendulum at this position and release it, it will stay there; and if you move it to any other position, it will tend to move back toward the equilibrium position.

A full cycle is from one extreme position, which we'll call the 'beginning position', through equilibrium, to the other extreme position, back through equilibrium and finally back to the beginning position.

You will release the pendulum from some point away from equilibrium, then click the TIMER with every return to this initial position. Your release point should not be at a distance from equilibrium which is more than 10% of the length of the pendulum. You will want the pendulum to complete at least 30 full cycles.

The pendulum will tend to lose a little of its energy, and hence will tend not to return all the way to its original point with every oscillation. So you should 'nudge' the pendulum a little bit, when necessary, to keep it swinging back to about the original release position.

You should keep in mind that the way you handle this pendulum will have a small but significant effect on its motion, and that while the results you get here will be close to the ideal results for a pendulum of this length, they will deviate slightly. We are interested here in measuring the various deviations that occur in this experiment.

Release the pendulum and allow it to go through at least 30 oscillations, clicking the timer with each return to the release position. Run through this at least a few times until you are sure you are consistently clicking very close to the instant the pendulum briefly stops at its extreme position.

When you have a good 'run' of this procedure, copy the output of the TIMER program into the box below: Below the TIMER output give a brief description of what it represents.

Your response (start in the next line):

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output of TIMER for 30 oscillations:

Your response (start in the next line):

0 8.5 8.5

1 9.457031 .9570313

2 10.55078 1.09375

3 11.44922 .8984375

4 12.57031 1.121094

5 13.61328 1.042969

6 14.5625 .9492188

7 15.55078 .9882813

8 16.66016 1.109375

9 17.64063 .9804688

10 18.66016 1.019531

11 19.72266 1.0625

12 20.80859 1.085938

13 21.8125 1.003906

14 22.92188 1.109375

15 23.98438 1.0625

16 25.04688 1.0625

17 26.29297 1.246094

18 27.23438 .9414063

19 28.19922 .9648438

20 29.11328 .9140625

21 30.09375 .9804688

22 31.20313 1.109375

23 32.29688 1.09375

24 33.27734 .9804688

25 34.34766 1.070313

26 35.42578 1.078125

27 36.48828 1.0625

28 37.44531 .9570313

29 38.47266 1.027344

30 39.55078 1.078125

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Your data will indicate at least 30 time intervals.

What is the mean of those time intervals? The mean is the 'average', which can be obtained by either adding up all the time intervals and dividing by the number of intervals, or by subtracting the first clock time from the last and dividing by the number of intervals.

Report the mean of your 30 time intervals on the first line of the box below, and starting on the second line state that this is the mean of the intervals you previously reported and explain how you obtained that mean.

Your response (start in the next line):

This is the mean of my 30 time intervals which I obtained by adding up all of the time intervals and divided by the number of intervals.

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mean of intervals: 1.04

how mean was obtained: This is the mean of my 30 time intervals which I obtained by adding up all of the time intervals and divided by the number of intervals.

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Now you will sample your time intervals by writing down the 3rd, 8th, 13th, 18th, 23d and 28th intervals. You know from before that those intervals are at best accurate to .01 second, so you can round off your intervals to 3 significant figures.

Report your intervals in the first 6 lines of the box below. Briefly identify the meaning of your numbers starting the 7th line, and if you wish add more comments or observations:

Your response (start in the next line):

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your six intervals:

3 11.44922 .898

8 16.66016 1.11

13 21.8125 1.01

18 27.23438 .941

23 32.29688 1.09

28 37.44531 .957

The numbers in the first column are the event numbers. The second column represents the clock time. The third column represents the time interval which is the difference between one time interval to the next.

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Find the mean of those intervals and report it below on the first line, with an identifying phrase and an explanation of your method for finding the mean starting on the second line.

Your response (start in the next line):

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mean of your six intervals:

1.001

To find the mean, I added up the 6 time intervals, then I divided by six giving me 1.001.

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By how much did the mean you found deviate from the mean of your 30 trials?

Report this number in the first line. Identify the meaning of the number and explain how you obtained it starting in the second line.

Your response (start in the next line):

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deviation of mean from original mean:

0.39

The mean is going to be your average. The mean of the 6 intervals would be the average time difference between the oscillations on each trial. I took the mean of the 30 intervals and got an answer of 1.04. I then took the mean of 6 intervals and got an answer of 1.001. Then I subtracted 1.04-1.001 to get an answer of .039 which would be how much the mean of the 30 trials deviated from the mean of the 6 trials.

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Would you expect the mean of your 6 trials to be exactly equal to the mean of the 30 trials?

What factors would tend to make the two results the same?

What factors would tend to make the two results different?

How nearly the same and how different would you expect the two means to be?

Your response (start in the next line):

The mean of your six trials should be extremely close to the mean of the 30 trials. There could be a slight variation as the pendulum could be swinging faster or slower at those points but overall it should be almost exact. If you rounded the numbers then the two results would be exactly the same. The two results could be different depending on how fast or slow the pendulum was swinging at that point or what events you chose to include. I would say it would be about +/-.5 the same or different between the two means.

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Within your sample there are deviations. Not all your time intervals will be the same.

We will proceed to analyze those deviations:

If for example you obtained a mean time interval of 1.23 seconds from sampled time intervals of 1.20, 1.21, 1.27, 1.22, 1.19 and 1.28 seconds, then your first sampled time would deviate by .03 seconds from the mean.

Can you explain why we say that the first sampled time deviates by .03 seconds from the mean?

If you add1.20, 1.21, 1.27, 1.22, 1.19 and 1.28 seconds and divide by 6 you get a mean of 1.23sec. The first sampled time is 1.20sec. If we take the difference by subtracting the mean 1.23sec minus 1.20sec we get an answer of .03 seconds. This .03 seconds is how far we are from the mean. It is how much 1.20sec differentiates from 1.23sec.

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The reason we say this, in case you didn't correctly state it, is that the first time interval is 1.20 s, the mean is 1.23 s, and the deviation between two numbers is how far they are apart. These numbers are .03 s apart.

More formally, the deviation of one number from another is the absolute value, or magnitude, of their difference. So the deviation of 1.20 from 1.23 is | 1.20 - 1.23 | = | -.03 | = .03.

List the deviations of 1.20, 1.21, 1.27, 1.22, 1.19 and 1.28 seconds from the 1.23 second mean. List one deviation in each of the first six lines, with a brief explanation of what your numbers mean and how you obtained them starting in the seventh line.

Your response (start in the next line):

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six deviations of sample data:

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Now find the deviations of your six sampled time intervals from the mean of those six intervals.

Give the mean of your six sampled intervals in the first line, then in each of the next six lines give the deviations from the mean. Give these results in the order in which the intervals were reported above. Starting on the 8th line, give a brief explanation of what your numbers mean and how you obtained them; you may also include any clarifications or comments you wish.

Your response (start in the next line):

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mean of sampled intervals:

1.23s

deviation of first sampled interval from sample mean:

.03s

deviation of second sampled interval from sample mean:

.02s

deviation of third sampled interval from sample mean:

-.04s

deviation of fourth sampled interval from sample mean:

.01s

deviation of fifth sampled interval from sample mean:

.04s

deviation of sixth sampled interval from sample mean:

-.05s

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You now have a list of six deviations. Some deviations are bigger than others.

On the average, how big are your deviations?

This question is easily enough answered by adding up the six deviations, and dividing by six.

That is, we can find the mean of the deviations.

Give your result for the mean of the deviations in the first line. Starting in the second line with a brief explanation of what your numbers mean and how you obtained them, explain how you found the mean of the deviations, and explain as best you can why it makes sense to call this quantity the 'mean of the deviations of the sample data from their mean'.

Your response (start in the next line):

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mean of deviations:

.002

explanation of meaning:

My number is the average of the 6 deviations. I obtained this by adding the 6 deviations and dividing by 6 the answer being .002. It makes sense to call this quantity the 'mean of the deviations of the sample data from their mean'. Because it involves the mean of the deviations we calculated from the sample data’s mean which was given to us.

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The mean of the deviations is a significant quantity. It gives us a good idea of how 'spread out' our observations are.

A sample with a high mean deviation is more 'spread out' than a sample with a low mean deviation.

The less 'spread out' a sample is, the more precisely we tend to think its mean represents our observations.

Explain this in your own words.

Your response (start in the next line):

When we find the mean of deviations we get a good idea of how much our observations deviate. When we get a high mean deviation, the sample is considered being more spread out than when we have a sample with a lower mean deviation. Furthermore, the less spread out a sample is the closer we think the sample’s mean is accurate to our observations.

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As significant as the mean deviation is, there is another measure that is more significant for our purpose of analyzing and drawing conclusions from data. The standard deviation is closely related to the Normal Distribution (the 'bell-shaped curve' of wide repute) in a way the mean deviation is not.

It isn't difficult to calculate the standard deviation. We've already done most of the work, and we can use everything we've calculated so far (except the mean deviation).

To get the standard deviation we just square our deviations, average them and then take the square root. The averaging is sometimes a little different than the calculation of the mean, but there's a very easy rule to follow when you do the averaging process.

Let's do this one step at a time:

First square all your deviations and list them in the box below, one squared deviation to a line. For example, of your first deviation was .03, you would square it to get .03^2 = .03 * .03 = .0009, and the number on the first line of the box would be .0009.

List your six squared deviations, one to a line, followed by a brief explanation of what your numbers mean and how you obtained them

Your response (start in the next line):

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squared deviation of first sample point:.0009

squared deviation of next sample point:.0004

squared deviation of next sample point:.0016

squared deviation of next sample point:.0001

squared deviation of next sample point:.0016

squared deviation of next sample point:.0025

explanation:

The numbers above represent the squared deviation. I obtained these numbers by squaring each sample point ex (-.05)^2=.0025 or multiplying it by itself ex (-.05)(-.05)=.0025

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Now we're going to find an average of the squared deviations.

This average is very like a mean, but with one small change.

If we calculate the mean of six numbers, we add them and divide by 6.

When we're finding the 'average' of the squared deviations, any time we have fewer than 30 squared deviations to 'average', we divide by 1 less than that number.

So in this case we'll be dividing by 5 rather than 6.

Add up your six squared deviations, and instead of dividing by 6, divide by 5 to get your 'average' squared deviation.

Report the sum of your squared deviations in the first line below, and your 'average' squared deviation in the second line. Explain the meanings of these numbers how you arrived at these results, starting in the third line.

Your response (start in the next line):

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sum of squared deviations:.0071

'average' of squared deviations:.00142

explanation:

.0071 is the sum of the squared deviations. This means that we took all 6 of the squared deviations of the sample points, added them up and got that number. .00142 represents the average of the squared deviations. To get this number I took the sum of the squared deviations and divided it by 5 because we have less than 30 squared deviations to average. I then got an answer of .00142.

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OK, the rest is easy.

You now have your 'average' squared deviation.

We want to get the standard deviation.

All you have to do is take the square root of the 'average' squared deviation.

That's the square root of the number you just got done calculating.

What you get when you take the square root is the standard deviation.

Report the 'average' squared deviation (again) on the first line below.

Report the standard deviation on the second line.

Explain starting in the third line how you got the standard deviation, including a brief explanation of what your numbers mean and how you obtained them

Your response (start in the next line):

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'average' of squared deviation:.00142

standard deviation:.0377

explanation:.00142 is the average of squared deviation. To get this number I took the sum of the squared deviations and divided it by 5 because we have less than 30 squared deviations to average. I then got an answer of .00142. .0377 represents the standard deviation. To get this number You take the square root of the average squared deviation. So I took the square root of .00142 and got a standard deviation of .0377.

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Explain in the box below your understanding of the following statement:

The standard deviation of a set of numbers is the square root of the 'average' of the squared deviations of those numbers from their mean.

Include in your explanation any qualifications associated with the word 'average'.

Your response (start in the next line):

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explanation: My understanding of the process is probably best broken down into steps.

1. You have a set of numbers and you are going to determine their mean by averaging them (adding them all up then dividing by amount of numbers there are)

2. You will find squared deviations of these numbers next by taking those averaged numbers and squaring them ex .03^2=.0009.

3. Next, you add those up averaged numbers up and divide by 5 where you will obtain an average of the squared deviations.

4. Finally, you have your average of the squared deviations so you take the square root of that which should give you the standard deviation.

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We'll see more about this later, and this statement is nearly but not completely accurate (the completely accurate statement involves also the standard deviation of the entire 30-interval data set), but here it is:

We expect that due to statistical fluctuations, the mean of a random sample of 6 observations will differ somewhat from the mean of all 30 observations from which the sample was taken.

There is a fairly high probability that the difference in these two means will be less than the quantity

(standard deviation of the sample of 6) / sqrt(6).

Calculate this quantity, using the standard deviation you have just calculated.

In the box below, report in the first line the mean of the 30 time intervals you observed.

In the second line report the mean of your 6 sampled time intervals.

In the third line report the standard deviation of your 6 sample time intervals.

In the fourth line report the result of your most recent calculation, i.e., report (standard deviation of the sample of 6) / sqrt(6).

In the fifth line report the difference of your 6-sample mean and the 30-interval mean.

In the sixth line report the word 'yes' or the word 'no', depending on whether the difference of the two means is or is not less than (standard deviation of the sample of 6) / sqrt(6).

Starting in the 7th line give a brief explanation of what your 'yes' or 'no' tells you about the situation.

Your response (start in the next line):

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mean of 30 intervals:1.04

mean of 6 sampled intervals:1.001

standard deviation of sampled intervals:.0377

std dev / sqrt(6):.015

'yes' or 'no':no

explanation: The difference of the 2 means would be calculated by doing 1.04-1.001=.039. The standard deviation of the sample/ sq rt of 6 would be .0377/2.45 which equals .015. .039 is larger than .015.

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The above statements were based on a random sample of 6 of the 30 trials.

Was the sample taken of the 6 means actually a random sample?

Is there a way to obtain a more random sample of the 6 time intervals?

What do you think would be the best way to get a truly random sample of 6 of the 30 time intervals?

Answer these questions in the box below.

Your response (start in the next line):

I don’t think this was representative of a random sample because you are taking means or averages. Typically you do not do this in a random sample because the numbers are truly completely random. If you increased your numbers it probably would be easier to get a random sample. With 6 numbers you could put the numbers into an Excel file and do a random draw for 6 time intervals.

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Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:

Approximately how long did it take you to complete this experiment?

4 hrs

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You may also include optional comments and/or questions.

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