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course PHY
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.
Your hypothesis:
In this experiment, which uses the TIMER program, you are going to click the mouse as quickly as possible with the index finger of your dominant hand, then you are going to click it as quickly as possible with the fist of your non-dominant hand.
• Do you think the index finger is be 'quicker' than the fist?
• Do you think the fist is be 'quicker' than the index finger?
• Do you think the index finger and the first are equally quick?
State which you think is the case:
Your answer (start in the next line):
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My index should be quicker than my fist.
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• The statement you just made is your hypothesis for this experiment.
It should take you only a few minutes to get your data for this experiment:
• Open the TIMER program and do 10 clicks, as fast as possible, using the index finger of your dominant hand (i.e., your right hand if you are right-handed, your left hand if you are left-handed).
• Do this until you have managed 10 good, quick clicks, with no 'misfires'.
• Copy the relevant portion of the TIMER output into the data analysis program and eliminate everything but the 10 time intervals, one to each line.
• Find their mean and standard deviation, and note these results.
• Copy your 10 time intervals into the space below:
Your answer (start in the next line):
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0.144
0.175
0.161
0.184
0.16
0.159
0.168
0.161
0.136
0.184
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Now repeat, but instead of the index finger of your dominant hand use the fist of your non-dominant hand. Use your fist gently. Don't hit the mouse hard enough to damage it or cause it to start moving around.
Copy your 10 time intervals into the space below:
Your answer (start in the next line):
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0.159
0.145
0.175
0.184
0.145
0.16
0.184
0.16
0.16
0.24
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Report the mean and standard deviation of your index-finger data in the first line below, in comma delimited format. Report your fist data in the second line, in the same format.
Your answer (start in the next line):
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0.161, 0.015540628
0.16, 0.027836028
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Using the form mean +- standard deviation, report in the first line below the result of your index-finger observations.
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 'fist' observations.
Your answer (start in the next line):
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0.161± 0.016
0.16± 0.028
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You will now report the same information by reporting upper and lower bounds.
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
• lower bound - (.27 - .05) sec = .22 sec and
• upper bound (.27 + .05) sec = .32 sec.
In the first line below report the lower and upper bounds of the 'index finger' results. 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 your 'fist' results.
Your answer (start in the next line):
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LB= 0.145, UB= 0.177
LB=0.132, UB=0.188
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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 mean - standard deviation and mean + standard deviation from your 'index finger' results. For example the number-line representation of the result .27 +- .05 would be as indicated below. The interval is shaded (here it is shaded in blue) and set off with parentheses.
Sketch also, on the same number line, the interval corresponding to your 'fist' results.
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:
So, your two number-line intervals might overlap, or they might be completely separate.
For example, if your two results were .27 +- .05 and .37 +- .03, then one of your number-line intervals would run from .22 to .32 and the other would run from .34 to .41.
• In this case we would say that the two number-line intervals are separated by the number-line interval from .32 to .34. We will call this the interval of separation, and the bounds on this interval of separation are .32 and .34.
On the other hand if your two results were .27 +- .06 and .37 +- .06, then one of your number-line intervals would run from .21 to .33 and the other would run from .31 to .43.
• In this case we would say that the two number-line intervals overlap on the number-line interval from .31 to .33. We will call this the interval of overlap, and the bounds on this interval of overlap are .31 and .33.
Report the nature of your intervals below:
• If your number line intervals are completely separate, enter in the first line of the space below the bounds on the interval of separation.
• If your number line intervals overlap, enter in the first line of the space 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.
Your answer (start in the next line):
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LB= 0.145, UB= 0.177
Overlap
Fist completely contains Index
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The results you obtained in this experiment are indications of what we will call 'finger repeat time' and 'fist repeat time'. Let's assume that you have an actual neurologically controlled repeat time for your index finger, and one for your fist. This is a very questionable assumption, but for the purposes of our analysis here let's make it.
The 10 results from each trial, on which you based your analysis here, comprise a limited sample of your actual repeat times.
Assume that you have a specific 'actual' index finger repeat time (another questionable assumption).
• Assume furthermore that it is represented somewhere in the number-line interval you obtained for your index-finger results, and
• Similarly assume that your 'actual' fist repeat time is represented somewhere in its number-line interval:
Based on these assumptions:
• Is it possible for your finger time to be less than you fist time?
• Is it possible for your fist time to be less than you finger time?
• Is it possible for both times to be identical, within the limits of accuracy of the TIMER?
• Can you or can you not conclude that your index-finger response time is different from your fist response time?
Your answer (start in the next line):
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It is not possible for finger time to be less than fist time since the LB on fist time is less than LB on finger and vice versa for the fist time, but it is possible for them both to be the same since they have overlapping intervals
According to this test, because of the overlapping interval anything is possible. We can draw conclusions, according to the test, only if the intervals are separate.
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Based on the preceding set of questions, do your results support or fail to support your original hypothesis? Explain thoroughly how your results lead you to accept or reject your hypothesis.
Your answer (start in the next line):
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My hypothesis is neither confirmed nor denied, necessarily, by these results but there does exist the possibility that it could be denied since the fist time could, in fact, be faster than the index but not vice versa.
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&#Good responses on this lab exercise. See my notes and let me know if you have questions.
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