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Submitting Assignment: PH1 Hypothesis Testing for Pearl Pendulum and Ball Down Ramp
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In this exercise you will test hypotheses related to the data you have taken for the pearl pendulum and for the ball down the ramp. In the process you should gain insight into the nature of the motion of the pendulum and whether your data supports the resulting predictions. You will also test an intuitive hypothesis related to symmetry, the hypothesis being that for a given ramp slope it doesn't matter whether the ramp is oriented left-right or right-left.
Typical reported completion times range from 45 minutes to 90 minutes, with times as short as 30 minutes and as long as 3 hours being reported in isolated cases.
The next several lines repeat information you have been given previously. It is repeated here in case you need it for quick reference. This section will be indented so you can easily see what you might wish to skip.
Hypothesis Testing Suppose we have observed the following time intervals: .925, .887, .938, .911, .925, .879, .941 where the time intervals are in seconds. The mean of these numbers is .915. The (mean) average deviation of the numbers from this mean is .020. The standard deviation of this distribution is .024. If these time intervals were recorded by an accurate instrument, an instrument that is accurately calibrated and without any distortion in its scale of measurement, set up and utilized in such a way that there is no systematic bias in the readings, then we expect that the time interval between the events we are measuring lies within one standard deviation of the mean. That is, we expect that the actual time interval `dt lies between (.915 sec - .024 sec) and (.915 sec + .024 sec). We could write this as an inequality .915 sec - .024 sec < `dt < .915 sec + .024 sec, meaning the same thing as .891 sec < `dt < .939 sec. We would then be able to report our result as .915 seconds +-.024 seconds.
Hypothesis Testing
Suppose we have observed the following time intervals:
.925, .887, .938, .911, .925, .879, .941
where the time intervals are in seconds.
If these time intervals were recorded by an accurate instrument, an instrument that is accurately calibrated and without any distortion in its scale of measurement, set up and utilized in such a way that there is no systematic bias in the readings, then we expect that the time interval between the events we are measuring lies within one standard deviation of the mean.
That is, we expect that the actual time interval `dt lies between (.915 sec - .024 sec) and (.915 sec + .024 sec).
We could write this as an inequality
.915 sec - .024 sec < `dt < .915 sec + .024 sec,
meaning the same thing as
.891 sec < `dt < .939 sec.
We would then be able to report our result as .915 seconds +-.024 seconds.
Pearl Pendulum
You were asked in the 'Pearl Pendulum' experiment to report the time interval between the release of your pendulum and the second 'hit' with the bracket. Specific instructions in that experiment were:
The order of events will be: click and release the pendulum simultaneously the pendulum will strike the bracket but you won't click the pendulum will strike the bracket a second time and you will click at the same instant We don't attempt to time the first 'hit', which occurs too quickly for most people to time it accurately. Practice until you can release the pendulum with one mouse click, then click again at the same instant as the second strike of the pendulum. When you think you can conduct an accurate timing, initialize the timer and do it for real. Do a series of 8 trials, and record the 8 time intervals below, one interval to each line. You may round the time intervals to the nearest .001 second. In the posted 'readable' version of your data, the information you submitted in response to this question is under the boldface heading Your report of 8 time intervals between release and the second 'hit':
click and release the pendulum simultaneously
the pendulum will strike the bracket but you won't click
the pendulum will strike the bracket a second time and you will click at the same instant
We don't attempt to time the first 'hit', which occurs too quickly for most people to time it accurately.
Practice until you can release the pendulum with one mouse click, then click again at the same instant as the second strike of the pendulum.
When you think you can conduct an accurate timing, initialize the timer and do it for real. Do a series of 8 trials, and record the 8 time intervals below, one interval to each line. You may round the time intervals to the nearest .001 second.
In the posted 'readable' version of your data, the information you submitted in response to this question is under the boldface heading
Again report those 8 time intervals, one number on each of the first 8 lines.
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#$&*
Using the data analysis program at
you can simply copy those 8 lines and click a button to find the mean and standard deviation of your results. Do this and report the resulting mean and standard deviation in comma-delimited form in the first line below.
Report in the second line the length of the pendulum in centimeters.
You were also asked to report times between alternate 'hits' starting with the second. The specific instruction was
Finally, you will repeat once more, but you will time every second 'hit' until the pendulum stops swinging. That is, you will release, click on the second 'hit', then click on the fourth, the sixth, etc.. Practice until you think you are timing the events accurately, then do four trials. Report your time intervals for each trial on a separate line, with commas between the intervals. For example look at the format shown below: .925, .887, .938, .911 .925, .879, .941 etc. In the example just given, the second trial only observed 3 intervals, while the first observed 4. This is possible. Just report what happens in the box below:
Practice until you think you are timing the events accurately, then do four trials.
Report your time intervals for each trial on a separate line, with commas between the intervals. For example look at the format shown below:
.925, .887, .938, .911
.925, .879, .941
etc.
In the example just given, the second trial only observed 3 intervals, while the first observed 4. This is possible. Just report what happens in the box below:
You were to have observed four trials and should have reported multiple time intervals in each.
In the interval below, report the time intervals you observed, one to each line. For example if your first two lines were
.925, .887, .938, .911 .925, .879, .941
your first 7 lines would be
.925 .887 .938 .911 .925 .879 .941
.925
.887
.938
.911
.879
.941
You should report only time intervals between alternate 'hits', i.e., each time interval should be between the second and fourth 'hit', the fourth and the sixth 'hit', the sixth and the eighth, etc.. If any of the time intervals in your original data did not represent such intervals, leave those intervals out:
Give the mean and standard deviation of your results in comma-delimited form below.
According to the descriptions you gave of the motion of the pendulum between release and the first 'hit', between the first and second 'hit', and between the second and fourth 'hit', would you predict that the time interval between release and the second 'hit' should be greater, less than or equal to the time interval between the second 'hit' and the fourth 'hit'?
You calculated the corresponding mean times, based on your observations. Are the results for your mean times consistent with your prediction? State you answer to this question, and how your results justify your answer, in the box below.
Using the form mean +- standard deviation, report in the first line below the result of your observations of the time from release to second 'hit'. For example, if the mean was .27 seconds and the standard deviation was .05 seconds, then you would report .27 +- .05 in your first line.
In the second line report, using the same format, the result of your observations of the time between subsequent alternate 'hits'.
If for example the result of an observation of a time interval was (.27 +- .05) sec, the lower and upper bounds on the time interval would be (.27 - .05) sec = .22 sec and (.27 + .05) sec = .32 sec.
In the first line below report the lower and upper bounds of the time interval between release and the second 'hit'. For example if your results were as in the example given here, you would report .22, .32.
In the second line do the same for the time interval between subsequent alternate 'hits'.
Sketch in your lab notebook a number line representing time intervals. Your sketch might look something like the figure below.
Sketch on your number line the interval between release and second 'hit'. For example the number-line representation of the result .27 +- .05 would be as indicated below.
Sketch also the interval between the lower and upper bounds for time interval between subsequent alternate 'hits' (i.e., second to fourth, or fourth to sixth, or sixth to eighth 'hits' etc.). Possible examples are depicted below:
Possible examples of the way a sketch might come out are depicted below:
The two intervals might be completely separate: The two intervals might overlap: One interval might even contain the other:
The two intervals might be completely separate:
The two intervals might overlap:
One interval might even contain the other:
Your two number-line intervals might overlap, or they might be completely separate.
If your number line intervals are completely separate, enter in the first line of the box below the bounds on the interval of separation.
If your number line intervals overlap, enter in the first line of the box below the bounds on the interval of overlap.
In the second line specify by the word 'separation' or 'overlap' whether the intervals are separate or overlapping.
Assuming that the actual time interval between release and second 'hit' lies somewhere in your first number-line interval, and that the actual time interval between alternate subsequent 'hits' lies somewhere in your second number-line interval, then based on your sketch:
Can you or can you not conclude that the time between release and second 'hit' is less than the time between alternate subsequent 'hits'?
Is this or is this not consistent with your original hypothesis regarding these time intervals? Explain thoroughly how your results lead you to accept or reject your hypothesis.
Based on your number-line sketch you will now develop and test another hypothesis regarding how the time intervals you observed might be related to one another.
Based on your description of the motion from release to second 'hit', and of the motion between alternate subsequent 'hits', how would you expect the two time intervals to compare? Specifically, should one time interval be double the other, should they be equal, should one by 3/2 as great as the other, or should some other ratio hold?
Give your best reasoning in answer to this question:
Answer the following three questions with three comma-delimited numbers in the first line below:
Answer the following three questions with three comma-delimited numbers in the second line below:
In terms of the pendulum experiment,
On the first observations you made of the ball rolling down the ramp, the ramp was supported on one end by a single domino.
You first made 5 observations of the ball rolling from right to left, then 5 observations from left to right.
The way the system was set up, before you performed the timing, would you have expected that there would have been a difference between the right-to-left results and the left-to-right results? That is, would you have expected that the time down the ramp should have been the same for the right-to-left setup as for the left-to-right?
That is, would you have expected that the motion of the ball from right to left have been identical but in mirror image form to the motion from left to right?
What would have been your expectation, and why?
What factors could have caused a difference in the two motions? That is, what in the setup might cause a lack of symmetry between right-to-left and left-to-right?
Using your data for the right-to-left and the left-to-right timings of the 1-domino system, you will calculate a couple of means and standard deviations. You may use any method you wish, but just copying your information into the data analysis program and clicking the Mean and Standard Deviation button might be the quickest:
Calculate the mean and standard deviation of the 5 time intervals obtained for the right-to-left motion and report in comma-delimited form in the first line of the box below.
Do the same for left-to-right motion and report in comma-delimited format in the second line of the box below.
Report your results once more, in the same order and using two lines, in the format 'mean +- std dev'.
Use a number line to sketch the corresponding intervals on a number line. Each interval will be from mean - std dev to mean + std dev, the first representing the right-to-left and the second the left-to-right timings.
Report in two lines, with the first line giving in comma-delimited format the left and right endpoints of the first number-line interval, the second line giving the same information for the second number-line interval.
If your intervals do not overlap, give in your first line the endpoints of your interval of separation. If they do give in the first line your interval of overlap.
In the second line state whether the intervals are 'separate' or whether they 'overlap'.
Do you believe your observations support the hypothesis that the left-to-right motion is not a mirror image of the right-to-left motion, i.e., a hypothesis that right-left mirror symmetry is not observed in this situation? State in terms of your data why the hypothesis is or is not supported.
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:
You may also include optional comments and/or questions.
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