http://www.vhcc.edu/dsmith > General Information > (scroll to bottom of page and click

on Access Your Information) then when the first menu comes up clicking on 15, then when

the new menu appears on 51, then when the third menu appears on Data Program, which will

be near the top of the page.

Analyze 30-interval data

In the space below include a copy of your data from the Error Analysis I experiment.

Then state the mean of your 30 time intervals, the mean and standard deviation of your

sample of 6 intervals, and whether the difference of the means is less tha the standard

deviation of your sample divided by sqrt(6).

your response &&&&&&&&&&&&&&&&&&

(start in the next line):

1 122.0508 122.0508

2 123.082 1.03125

3 124.043 .9609375

4 125.1133 1.070313

5 126.2109 1.097656

6 127.2969 1.085938

7 128.3242 1.027344

8 129.3789 1.054688

9 130.457 1.078125

10 131.5469 1.089844

11 132.6328 1.085938

12 133.6836 1.050781

13 134.793 1.109375

14 135.7539 .9609375

15 136.8672 1.113281

16 137.9453 1.078125

17 138.9805 1.035156

18 140.0664 1.085938

19 141.1875 1.121094

20 142.1602 .9726563

21 143.1133 .953125

22 144.0664 .953125

23 145.082 1.015625

24 146.1953 1.113281

25 147.3945 1.199219

26 148.4023 1.007813

27 149.5703 1.167969

28 150.5938 1.023438

29 151.6914 1.097656

30 152.707 1.015625

31 154.0508 1.34375

32 154.9531 .9023438

Mean of trials: 1.061364516

Mean of samples: 1.04 Standard deviation of samples: .05

The square root of 6 is 2.45; The difference of the means is .02; .05/2.45 = .02. The

difference of the means is the same as the standard deviation divided by the square root

of 6.

#$&*

You may download the data program or run it directly from the site. Note the following:

The program is in an ongoing process of development and some of the buttons might not

work. However the operations you are instructed to perform below have been tested and do

work.

When you run the program you might encounter some message boxes at the beginning; these

boxes have been inserted to prompt the inclusion of some additional features in the

program. If you do encounter these message, you may safely just click through them

messages until just the form appears.

The program is easy to use and is very efficient for its purpose. Additional features

will be added as needed.

Run the program and click through any extraneous messages. (If necessary you might need

to click on the maximize button to maximize the size of the form and make all the

buttons visible, but this should not be an issue.)

Delete all information in the textbox (you can use the Clear button near the lower right

corner of the box), and copy your TIMER data into the box. You may use either the data

you have retained from the TIMER program or the data as posted on your access page (data

should be posted if you submitted the program in a timely fashion).

Your data will be in 3 columns. Manually delete all the information except the 30 time

intervals, so there are 30 lines each with a single number in the textbox, the number

representing the time interval in seconds (if your original data is in a spreadsheet you

could just copy the single column corresponding to the time intervals).

Copy and paste these 30 lines into a separate text editor or word-processing program so

you can use them again later.

Click on the Mean and Standard Deviation button. A message box will appear asking you

to confirm that your data is entered in the necessary format. Then the program will

very quickly display the mean and standard deviation of that distribution.

What are the mean and standard deviation of your 30 time intervals, as reported by the

program? Report

below, using two tab-delimited numbers in the first line. Starting in the next line

give a brief explanation of what your numbers mean and how you obtained them. After

that explanation, include a copy of your data set for reference.

your response &&&&&&&&&&&&&&&&&&

(start in the next line):

1.067 .07944

There are 30 numbers in your distribution.

Their sum is 32.00

Their mean is 32.00/ 30 = 1.067.

The (mean) average deviation from the mean is 0.05806.

The sum of the squared deviations is .1893.

The standard deviation is therefore sqrt( .1893/ 30) = .07944.

1.03125

.9609375

1.070313

1.097656

1.085938

1.027344

1.054688

1.078125

1.089844

1.085938

1.050781

1.109375

.9609375

1.113281

1.078125

1.035156

1.085938

1.121094

.9726563

.953125

.953125

1.015625

1.113281

1.199219

1.007813

1.167969

1.023438

1.097656

1.015625

1.34375

#$&*

Investigate 'first differences' of 30-interval data

Now restore your original 30 time intervals to the box. You will have to do this

manually, clearing the contents of the box and then copying and pasting the data from

the text editor or word processor where you stored it before. Make sure your data also

stays in that location, because you'll need it at least once again.

Click on the First Difference button. You will see a report of the differences between

your successive time intervals.

Give the first three differences

below, in the first line in comma-delimited format.

Starting at the second line answer the two questions:

Are all the differences between your time intervals all different, or do some occur more

than once?

Where have you see this information before and what does it mean?

your response &&&&&&&&&&&&&&&&&&

(start in the next line):

-.07031. .1094, 0.02734

The differences are all different. I do not recall finding the differences between the

elapsed times before; however it looks like it is taking the elapsed time for each cycle

and finding the differences.

#$&*

Sum your 30 time intervals and speculate on meaning

Restore your original 30 intervals to the box. Click on the Running Sum button.

Scroll down and take a quick look at the entire report.

Give your first three running sums

below, in the first line in comma-delimited format.

Starting at the second line, explain how you think these numbers were calculated from

the time intervals, and what these numbers might mean.

your response &&&&&&&&&&&&&&&&&&

(start in the next line):

1.031, 1.992, 3.063

They added the time intervals to obtain the elapsed time for the cycles. This means

that for 2 cycles, it took a total of 1.992 seconds. Three cycles took a total of 3.063

seconds.

#$&*

Analyze the first difference of the running sums, and the first difference of this

result

Delete everything but the single-column report of the running sums, so the data box

contains just the running sums with one sum on each line, and click on the 'first

difference' button.

Report your first three new numbers

below, in the first line in comma-delimited format.

Describe what you see and what might be the meaning of the new numbers.

Suggestion: look at your original 30 time intervals.

How do you think the new numbers were calculated, where have you seen them before, and

why do they come out the way they do?

your response &&&&&&&&&&&&&&&&&&

(start in the next line):

.961, 1.071, 1.097

These are the 2nd, 3rd and 4th numbers of the original data, rounded to the nearest

thousandth.

#$&*

Again isolate only the single-column report and again click on First Differences.

Report your first three new numbers

below, in the first line in comma-delimited format.

How do you think the new numbers were calculated, where have you seen them before, and

why do they come out the way they do?

your response &&&&&&&&&&&&&&&&&&

(start in the next line):

.961, 1.071, 1.097

These seem to be the differences of the differences. You get them by subtracting the

two numbers that are above/below each other.

These are numbers 3, 4, and 5 from my original data.

#$&*

Find difference quotients for a new set of data and speculate on the meaning of the

difference quotient

Clear the box then copy the following 4 lines into the textbox:

0, 0

10,10

20,25

30,45

Click on the Difference Quotient button.

Report

below the three new numbers you see, reporting your numbers in the first line in comma-

delimited format.

In the second line speculate on how the program might have calculated these numbers.

your response &&&&&&&&&&&&&&&&&&

(start in the next line):

1, 1.5, 2

The top row subtracted from the row below it: 10-0 = 10 and 10 - 0 = 10. 10/10 = 1

The 2nd row subtracted from the row below it: 20-10=10 and 25-10 = 15. 15/10=1.5

The third row subtracted from the row below it: 30-20 = 10 and 45-25 = 20. 20/10=2

#$&*

The information in the table

0, 0

10,10

20,25

30,45

represents the position of an object rolling down an incline vs. clock time, with

position in meters and clock time in seconds. Recall that according to our 'y vs. x'

convention, in a position vs. clock time table the clock time is in the first column.

How far did the object travel in the first time interval? How much time elapsed while it

traveled through this distance? What therefore was its average speed during this time

interval? Report your numerical answers to these three questions in the first line

below, in comma-delimited format.

Answer the same questions for the second time interval, and report in the second line,

using the same format as in the first.

Answer the same questions for the third time interval, and report in the third line,

using the same format as in the first.

Starting in the fourth line, explain how you obtained your results.

Then explain once more what the 'difference quotient' operation does to two columns of

numbers.

your response &&&&&&&&&&&&&&&&&&

(start in the next line):

10, 10, 1

10, 15, 1.5

10, 20, 2

The top row subtracted from the row below it: 10-0 = 10 and 10 - 0 = 10. 10/10 = 1

The 2nd row subtracted from the row below it: 20-10=10 and 25-10 = 15. 15/10=1.5

The third row subtracted from the row below it: 30-20 = 10 and 45-25 = 20. 20/10=2

The difference quotient operation will subtract the first set of data from the second

set of data, to obtain two differences, and then divide them to find the average

velocity (in this case).

#$&*

Select and analyze 5 random intervals from 30-interval data, using the data program to

find mean and standard deviation

Using a coin according to the following instructions, you will now select 5 intervals

randomly from your 3-interval data. You will do this by generating 5 numbers

corresponding to the numbers of your data point. The process should take only a couple

of minutes:

Using the coin you will generate a series of numbers between 0 and 31. Note that there

are 32 numbers between 0 and 31. This process can generate 32 possible numbers.

If you generate a number you have generated before you will discard it and generate an

alternative.

If you generate a number that does not correspond to one of your intervals (probably 1-

20 or 1-19) you will discard that number.

You will continue until you have generated 5 numbers that haven't been discarded.

To generate each number will require 5 flips of your coin. You will write down 5

numbers.

Your first flip is worth 1. Flip the coin. If you get Heads write down the number 1. If

you get tails write down 0. Whichever number you write down will be at the top of a

column.

Your second flip is worth 2. Flip the coin a second time. If you get Heads write down

the number 2. If you get tails write down 0. This number does in the column below the

previous.

The third, fourth, and fifth flips are respectively worth 4, 8 and 16 on Heads, 0 if you

get Tails.

You should now have five numbers in your column. Add them up.

The result will be not less than 0 + 0 + 0 + 0 + 0 and not more than 1 + 2 + 4 + 8 + 16

= 31.

Go ahead and generate your first number according to these instructions. If the number

is between 1 and the number of intervals you observed (e.g., between 1 and 30, or

between 1 and 29), circle the number.

Now generate another number, using the same procedure with 5 flips of the coin. If this

number is between 1 and your number of intervals (e.g., between 1 and 30), and if it

does not duplicate the first number you generated, circle it.

Continue this process, generating totals between 0 and 31 and circling those that lie in

the correct range and do not duplicate any your previous numbers. Stop when you have

generated 5 distinct numbers within the appropriate range.

Now select the time intervals corresponding to the numbers you have generated (e.g., if

you had a 30-interval set and your numbers were 23, 8, 11, 19, 5 and 22 you would select

the 23d, 8th, 11th, 19th, 5th and 22d time intervals).

Clear, then put these 5 time intervals into the textbox. Note that you will put time

intervals into the textbox, not the numbers you have generated between 0 and 31.

Click on the Mean and Standard Deviation button.

In the first line below, report the five random numbers you generated, in comma

delimited format.

In the second line below, report the five time intervals you put into the box, in comma

delimited format.

In the third line, report the mean and the standard deviation in comma-delimited format.

Starting in the fourth line give a brief explanation of what your numbers mean and they

were obtained. Optional comments may be added.

your response &&&&&&&&&&&&&&&&&&

(start in the next line):

15, 22, 29, 25, 8

1.113281, .953125, 1.097656, 1.199219, 1.054688

1.084, 0.18

The numbers are 5 random samples taken from the data. The 1.084 is their mean, or

average. The 0.08986 is the standard deviation. The standard deviation tells how far

away from the average a data point might be. A standard deviation of .18 means that one

standard deviation away from the average will be plus or minus .18. Two standard

deviations away from the average will be plus or minus .36.

#$&*

In three lines report the following numbers:

By how much does the mean of your 5-interval sample differ from the mean of the entire

data set of 30 intervals?

What is the standard deviation of the 30-interval set?

What is the first number you reported as a percent of the second. That is, what is the

difference between your sample and the entire data set, as a percent of the standard

deviation of the data set?

Starting in the fourth line give a brief explanation of what your numbers mean and how

you obtained them

your response &&&&&&&&&&&&&&&&&&

(start in the next line):

The mean of the 5-interval set is 1.084. The mean of the the data set of 30 intervals

was 1.061. This is a difference of .023.

The first number is 102% of the second number. The difference between the mean of the

30 data points and the 5-interval set is .024. This number is 13% of the standard

deviation of .18.

#$&*

Analyze a set of 'made-up' time intervals and look at their distribution

The set of numbers given below represents a set of 30 'made-up' quick-click time

intervals. You will answer a few questions about this data set, including the mean and

standard deviation of a 5-interval random sample. Later the results of all students will

be compiled and used to demonstrate the 'sample standard deviation', which is an

important statistical characteristic of sample and very relevant to interpretation of

experimental results.

.1752

.172

.1979

.1991

.176

.1711

.1664

.1665

.1858

.1764

.1765

.1885

.173

.1853

.1683

.1674

.1833

.1632

.1783

.1962

.1704

.1914

.1751

.1715

.1967

.1852

.1851

.1771

.1639

.1824

.1877

Copy these numbers into a cleared textbox, click on Mean and Standard Deviation, and

report their mean and standard deviation in comma-delimited format in the first line

below. Starting in the next line give a brief explanation of what your numbers mean and

how you obtained them.

your response &&&&&&&&&&&&&&&&&&

(start in the next line):

.1791, 0.01025.

The mean is the average of the numbers in the set. The standard deviation is obtained

by finding how far each data point is from the mean, squaring that absolute value,

adding those squared numbers, dividing them by the number of items in the data, and then

taking the square root of that quotient.

#$&*

below, enter the following numbers, one to a line, in the given order:

The number which is two standard deviations less than the mean.

The number which is one standard deviation less than the mean.

The number which is equal to the mean.

The number which is one standard deviation more than the mean.

The number which is two standard deviations more than the mean.

Starting in the next line give a brief explanation of what your numbers mean and how you

obtained them

your response &&&&&&&&&&&&&&&&&&

(start in the next line):

.1528

.16885

.1791

.18935

.1996

I figured out that 2 standard deviations would mean .0205, so I subtracted that from the

mean to get the first number and added it to get the last. I took .1791, which is the

mean, and subtracted .01025 from it for the second number. I added them for the fourth

number.

#$&*

below, report each of the following numbers, one number to each line:

The number of the 30 time intervals which are less than the number which is two standard

deviations less than the mean.

The number of the 30 time intervals which lie between two standard deviations less than

the mean and one standard deviation less than the mean.

The number of the 30 time intervals which lie between one standard deviation less than

the mean and the mean.

The number of the 30 time intervals which lie between the mean and one standard

deviation more than the mean.

The number of the 30 time intervals which lie between one standard deviation more than

the mean and two standard deviations more than the mean.

The number of the 30 time intervals which are greater than the number which is two

standard deviations more than the mean.

Starting in the 7th line give a brief explanation of what your numbers mean and how you

obtained them; as usual you may include optional comments.

your response &&&&&&&&&&&&&&&&&&

(start in the next line):

0

6

12

8

5

0

Two standard deviations is 0.0205. I took the mean of .1791 and subtracted 0.0205. I

looked for any numbers that were less than .1586. Then I subtracted .01025 from .1791

and looked for any numbers less than .16885 (that I had not already marked off). Next I

checked to see if any of the numbers were exactly .1791. Then I added .1791 and .01025

and looked for number less than .18935 that had not already been marked off. Next, I

added .1791 and .0205 and looked for any numbers less than .1996 that had not been used.

Finally, I double checked that I had accounted for all of the numbers in my data.

#$&*

below, report each of the numbers you reported above, but expressed as a percent of the

30 intervals (rounded to the nearest percent). For example, the number 10 would be 33%

of 30. Include a brief explanation of what your numbers mean and how you obtained them

your response &&&&&&&&&&&&&&&&&&

(start in the next line):

0%

19%

39%

26%

16%

0%

I divided the number of data points that fit each category above by 30 to find the

percent of data in each category. So, in the category of numbers that were between the

mean and 1 standard deviation less than the mean, there were 6 numbers. 6 divided by 31

(the number of time intervals given) is 19%. I then checked to be sure I had accounted

for 100% of the data.

#$&*

Perform a similar analysis with your 30-interval data

Return to your own 30 time intervals. Count the numbers in each range (less than mean -

2 std dev, between mean - 2 std dev and mean - 1 std dev, between mean - 1 std dev and

mean, etc.), using the mean and standard deviation of that data set.

Report each number as a percent of your total number of intervals, one number in each of

the first six lines below. Starting in the 7th line give a brief explanation of what

your numbers mean and how you obtained them

your response &&&&&&&&&&&&&&&&&&

(start in the next line):

0%

17%

30%

43%

7%

3%

I counted the number of data points in each category. I divided each number by 30 (the

total number of data points collected) to get the percents. I double checked that I had

accounted for 100% of the data.

#$&*

In a standard 'normal' distribution, we expect that the respective percents in the six

ranges will be about 2%, 14%, 34%, 34%, 14% and 2%. In a very large sample of data (say,

at least tens of thousands of data points), if the data are in fact distributed

normally, we expect actual results to very nearly reflect this distribution. If a large

distribution does not closely match the expected results, we suspect that something in

the system or in our observation process in fact deviates from the 'standard normal'

expectation. Not everything we observe does in fact follow the standard normal pattern.

You 3-interval results might or might not be expected to follow a standard normal

distribution.

If the data sample is not very large, there may of course be chance fluctuations in the

distribution and the percents may not be all that close to the expected distribution. In

a medium-sized sample of 30 or so, we definitely expect more observations to lie in the

middle two ranges and in either of the outer ranges, and we aren't too surprised if no

results at all appear in the outermost ranges (more than 2 standard deviations from the

mean).

Based on the percents you reported and the percents quoted above, by how much would you

say your actual 30-interval results deviated from the standard normal distribution? Did

your results deviate enough to make you suspect that your clicks were not normally

distributed about their mean?

your response &&&&&&&&&&&&&&&&&&

(start in the next line):

I was close on 4 of the 6 intervals. My second 34% was actully 43%, and the second 14%

was 7% in mine, but for this small sample, I think that it was very close to what I

should have seen.

#$&*

Answer the same question for the 30 made-up time intervals given earlier.

your response &&&&&&&&&&&&&&&&&&

(start in the next line):

The made up was also somewhat close. The normal had 2% at each end, and mine did not.

The middles were off somewhat, but the two inner amounts were more than the their

adjacent intervals.

#$&*

Compare your distribution with the standard normal distribution

We will in a subsequent exercise learn to sketch a standard normal curve, and to

represent our information using this sketch.

For the present, simply copy this figure below and label it as indicated below:

There are five vertical lines on the graph, representing respectively

mean - 2 * std dev, also labeled z = -2

mean - 1 * std dev, also labeled z = -1

mean - std dev, also labeled z = 0

mean + 1 * std dev, also labeled z = 1

mean + 2 * std dev, also labeled z = 2.

Label the x axis with the z numbers -2, -1, 0, 1 and 2.

Below these labels, place the respective numbers you obtained earlier for your 30-

interval results, the numbers corresponding to mean - 2 * std dev, mean - 1 * std dev,

etc..

The five lines divide the region between the curve and the x axis into six smaller

regions. Each of these regions will include either 2%, 14% or 34% of the total area

between the curve and the x axis.

Within each region, write the percent that indicates its area as a percent of the total.

below:

Indicate in the first line in comma-delimited format the percents you placed in the

regions, from left to right.

Indicate in the second line the x-axis labels corresponding to z = -2, -1, 0, 1 and 2.

Starting in the third line give a brief explanation of what your numbers mean and how

you obtained them.

your response &&&&&&&&&&&&&&&&&&

(start in the next line):

0, 17, 30, 43, 7, 3

0, 5, 9, 3, 2, 1

The first line has the percentages for each section of the graph, from left to right.

These represent the percentages of my data for each interval: more than 2 standard

deviations less than the mean, between 1 & 2 standard deviations less than the mean,

between the mean and one standard deviation less than the mean, between the mean and one

standard deviation more than the mean, between 1 and 2 standard deviations above the

mean, and more than 2 standard deviations above the mean. The second line is the actual

number of data point in each category.

#$&*

self-critique #$&*

#$&* self-critique

self-critique rating

rating #$&*:

&#Your work on this lab exercise looks very good. Let me know if you have any questions. &#