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July 3 almost midnight again!!
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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 brief discussion/description/explanation:
I used thread and a washer. It measured exactly 36cm.
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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 brief discussion/description/explanation:
1 17.12 17.12
2 18.233 1.113
3 19.178 0.945
4 20.189 1.011
5 21.153 0.964
6 22.207 1.054
7 23.211 1.004
8 24.196 0.985
9 25.244 1.048
10 26.298 1.054
11 27.173 0.875
12 28.084 0.911
13 29.255 1.171
14 30.26 1.005
15 31.164 0.904
16 32.135 0.971
17 33.16 1.025
18 34.212 1.052
19 35.194 0.982
20 36.277 1.083
21 37.257 0.98
22 38.374 1.117
23 39.245 0.871
24 40.279 1.034
25 41.331 1.052
26 42.295 0.964
27 43.329 1.034
28 44.392 1.063
29 45.422 1.03
30 46.406 0.984
The above numbers represents the pendulum experiment I just completed. I measured 30 cycles using the TIMER program. Every time the pendulum was at the start position, I would click the button to time one oscillation. I did this 30 times.
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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 brief discussion/description/explanation:
0.976
The above number is the mean of my intervals from above. I found the mean by subtracting the first clock time from the last and dividing by 30.
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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 brief discussion/description/explanation:
3 .945
8 .985
13 1.17
18 1.05
23 .871
28 1.06
The above is a sample of my intervals by reporting the most accurate intervals.
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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 brief discussion/description/explanation:
1.014
This is the mean of the above six intervals. I found the mean by adding the six intervals and dividing by the number of intervals.
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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 brief discussion/description/explanation:
0.038
The mean from my original 30 trials was this much more than my mean from my 6 sample 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 brief discussion/description/explanation:
No, I would not expect the means to be exactly the same. I think they would only be the same if I had no human error in the click of the timer and the swing of the pendulum. These are the same things that would make the results different. This experiment counted greatly on my ability to click the timer and keep the pendulum swinging at a constant pace. I would expect the two means to be within the range of the 30 intervals. If none of them were greatly skewed then the means should be close.
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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 how the given information tells us that the first sampled time deviates by .03 seconds from the mean?
You would take the mean time interval 1.23 s and find the difference of the first time interval which was 1.20s. You are seeing how far apart these two numbers are. This deviates from the mean by .03s.
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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 brief discussion/description/explanation:
.03
.02
.04
.01
.04
.05
My numbers above are the standard deviation from the mean for 6 different times listed in seconds. I found this by finding the absolute value of the difference.
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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 brief discussion/description/explanation:
1.014
.069
.029
.157
.038
.144
.049
The first number is the mean from the six samples of my pendulum experiment. The next six numbers are the deviation from the mean for those same numbers. I obtained the deviation from the mean by finding the absolute value of the difference of the mean and sample.
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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 brief discussion/description/explanation:
.081
The above is the mean of the sample deviation from the mean. I first took 6 samples from my population of 30 times in seconds. I found the mean of the sample then the deviation from the six samples. I used the deviation of the mean to find its mean, which is the number above! Wow! That’s wordy!
@&
Good words, though.
*@
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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 brief discussion/description/explanation:
The mean of the deviation is very important to tell us how spread our experiment results are. If we have a high mean deviation our data is more spread out. If our low mean deviation then our data is more accurate.
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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 brief discussion/description/explanation:
0.00483
0.00087
0.02449
0.00141
0.02059
0.00235
The above numbers are the squared deviation of the mean from my six samples.
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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 brief discussion/description/explanation:
0.05454
0.01091
The first line contains the total of my squared deviations from the mean. The second line contains the average of this number. I added the six squared deviations from the mean to obtain the first number then I divided by five to get the second number.
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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 brief discussion/description/explanation:
0.01091
0.104
The first line is the average squared deviation. The second line is the square root of the average squared deviation which gives us the standard deviation.
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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 brief discussion/description/explanation:
Standard deviation shows how much variation from the average. To find standard deviation we find the deviation from the mean then we square those numbers. Then find the AVERAGE of the squared deviation from the mean. Take the square root of this number to get 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 brief discussion/description/explanation:
I do not understand the second part of this problem: standard deviation of the sample of 6) / sqrt(6). I do not know what the sqrt(6) means.
@&
You calculated the standard deviation of your sample of 6.
So to get the standard deviation of the sample of 6 / sqrt(6), you divide that standard deviation by sqrt(6)
*@
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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 brief discussion/description/explanation:
One way to get a random sample is to just pull the 6 numbers out of a hat.
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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?
2 hours
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&#Good responses. See my notes and let me know if you have questions. &#