Laboratory Measurements

• Measurements consist of– Number: Tells amount measured– Units: Kind of measurement made– Uncertainty: Possible error

• No Measurement is Perfect due to– Skill of individual making measurement– Reliability or Sensitivity of instrument

Measurements…Continued

• Kinds of Uncertainty:• Human Error = carelessness in reading

instruments such as parallax (the apparent change in position due to a change in viewing location)

• Systematic = inaccuracies that occur each time and always in the same direction; such as balance not properly calibrated

• Random = fluctuations in measurements due to the precision limitations of the instrument

•How to detect and minimize these types of error:

• Human error is reduced by being careful in reading all instruments and recording all data;

• Systematic errors are very difficult to detect and can not be analyzed statistically – however, care in instrument calibration helps reduce them;

•Random errors can be evaluated statistically

• Statistics are a laboratory researchers best friend- not only in detecting errors but also in analyzing experimental data

•Descriptive Statistics consist of methods that will characterize your data

• Mean is the statisticians word for average

• Formula for the Mean

• Mean = ∑ x n where ∑, the Greek letter Sigma & is read “the sum

of” x = the individual measurements n = the number of measurementsSymbol for the mean is an “x” with a bar over it;

sorry I can’t make it with Microsoft software

• But the deficiency of the mean is that it does not tell you the about of variation in the data set

• What about the Range? It represents the amount of spread in the data but is only the difference between the maximum and minimum data values Let’s look at a couple of Data Sets of 10 measurements

• Trial 1: 8,10,8,12,16,14,16,12,10,14

• Trial 2: 12,8,11,12,13,16,12,11,13,12

• Calculate Both the Mean & Range for each

•Answer: Both Trials (data sets) have a Mean of 12 and a Range of 8

• Why a Range of 8? The maximum value in each is 16 and a minimum of 8

• Problem with the range is that it involves only two data points and does not reflect the distribution of the data values around the mean

• So what can we use to determine the amount of spread in the data? Look at both data sets again-

• Trial 1: (x = a measurement with that value)

x x x x x

x x x x x

8 9 10 11 12 13 14 15 16 • Trial 2: (x = a measurement with that value) x

x x x

x x x x x

8 9 10 11 12 13 14 15 16

• Definitely – Trial 2 has more data points surrounding the mean

• Statistically this spread is called variance

• There are formulas for calculating it and it is a straight forward process

• Step 1:You will calculate the difference between each measurement from the mean

• Step 2 is to square each of our calculated differences- Why? Because this gives us a positive value regardless of the sign of the difference- Try it

• Step 3: we add up all the squared values which we call the variance

• v = ∑ (x – mean)2

v = the variance x = the individual measurement

∑ = the sum of; you add them up

• Trial 2 as an example Measurement (x-mean) (x-mean)2

12 12-12=0 02= 08 8-12=-4 (-4)2=1611 11-12=-1 (-1)2=112 12-12=0 02= 013 13-12=1 (1)2=116 16-12=4 (4)2=1612 12-12=0 02= 011 11-12=-1 (-

1)2=1 13 13-12=1 (1)2=1

12 12-12=0 02= 0 v = 36

• Done? Nope- remember that all measurements have a number and a unit of measurement- so if, for example, we had measured a liquid in milliliters (mL), each measurement would have mL attached to it.

• So measurement 1 in Trial 2 would be 12mL; the second, 8mL; the third, 11mL, and so forth

• In our formula we squared the difference between each measurement and the mean, so our variance would be in reality, 36 mL2

• But 36 square milliliter (mL2) does not make any sense. So we need to take the square root of the variance

• This gives up the Standard Deviation after a little mathematical manipulation

• Standard Deviation, σ (lower case sigma)

• σ = √v/n-1• Standard deviation equals the square

root of the variance divided by the number of measurements minus 1

• Using Trial 2 data we have• σ = √36mL2/(10-1) • σ = √36mL2/9• σ = √4mL2

•σ = 2 mL• We combine the mean with the

standard deviation to report our summary as 12 +/- 2 mL

•Read as 12 plus or minus 2 mL

Sidebar- Why n-1?

• Statistics is based on UNBIASED data or treatment of the data

• n-1 is referred to as the Degrees of Freedom which means each number must have the “freedom” to be any number with in the range of numbers

Sidebar- Why n-1?...Continued

• For example, given 5 numbers that add up to 20 and we have

• Number 1 = 10• Number 2 = 3• Number 3 = 2• Number 4 – 1• Then Number 5 is fixed and can only

be 4 since the first four add up to 16

Back to the Treatment ofLaboratory Measurements

• We could perform the same statistical calculations on Trial Set 1 given earlier

• The result would be 12 +/- 3mL• Now compare that value to 12 +/- 2mL• So what does all this tell us since we

do not know the true value of our experimental data?

• Interpretation of our two statistical results

• We assume, with a high degree of confidence, that the true value is between the two extremes given using the mean and the standard deviation

• Trial 1: 12 +/- 3mL; so the true value is somewhere between 9mL (12-3) and 15mL (12 + 3)

• Trial 2: 12 +/- 2mL; so the true value is somewhere between 10mL (12-2) and 154mL (12 + 2)

Accuracy vs. Precision

• Some people would say that Trial 2 would be more accurate than Trial 1

• But Accuracy is how close our measurement is to the actual value and experiments would not be run if we already know the actual value--- So

• The correct term is Precision

• Precision is a measure of how close our measurement values are to one another

• Said another way, how repeatable are our measurements

• Thus, Trial 2 is more precise than Trial 1

Calculations involving Precision

• All measurements contain an estimated digit

• Example: If the meniscus (the bottom of the concave surface) of water is between interval gradations of a graduated cylinder, we can estimate its value

• So we could measure 15.8mL for a volume when the liquid’s meniscus is between 15 and 16mL

• We know for sure that the liquid is greater than 15 mL and less than 16 mL

• The .8 mL portion of our measurement is only an estimate

• If we could use a more sensitive instrument we could refine our measurement.

• Nevertheless, it will always contain an estimated digit or figure at the end

• The last digit in all measurements are therefore only an estimate

• So if we used a graduated cylinder divided into tenths of a milliliter we could refine our measurement

• We could determine it to be between 15.7 and 15.8 mL; we know that it is greater than 15.7 mL for sure according to our graduated cylinder BUT we can also estimate what the value is between the two intervals giving us 15.75 mL

• Is our new reading more precise than the first? Yes, but it still has an estimated digit – the .05mL value!

• What is the danger, if any, in performing calculations with measurements that have an estimated digit?

• Let’s see: What is the cylinder’s surface area if we measure its diameter to be 6.58cm?

• A = πr2 = π(3.29)2

• Our calculator value is 34.00491304cm2

• BUT was our original measurement carried out to the same number of places past the decimal?

Significant Figures/Digits to the Rescue

• Rules for Counting SigFigs; Count– All digits from 1 to 9– Zeros in between nonzero digits– All zeros following nonzero digits (these zeros

are called “trailing zeros”)

• Do NOT Count zeros in front of a value (these are called “leading zeros”) because they only serve to set the decimal point

• Exact Numbers – those by definition or counting numbers are infinite as to sigfigs

• 156 g has 3 sigfigs• 0.2608 m has 4 sigfigs• 120.50 L has 5 sigfigs• 0.05003 s has 4 sigfigs• 7.2 oC has 2 sigfigs• 12 students is infinite

• How can be tell the reader the number of sigfigs if a measurement like 2000 g?

• Is it accurate to only the thousands or the hundreds or the tens of grams?

• We use Scientific Notation to solve our dilemma – M x 10n

• What ever digits are used in the M portion of our notation are significant

• 2.0 x 103 g has 2 sigfigs• 2 x 103 g has 1 sigfig• 2.00 x 103 has 3 sigfigs• Got it?

Calculations involving SigFigs

• Addition and Subtraction- your final answer can not be more precise (places past the decimal) than the least precise measurement

• 12.23 cm 2.9 cm 10.28 cm 25.41 cm BUT we can only have 1 digit

past the decimal point- see rule aboveWhat do we now do- we round off our answer

Rounding Off

• Determine the number of sigfigs needed

• Examine the first digit following the last significant digit (viewing the number from left to right)

• If this digit is 4 or less, all sigfigs stay the same (don’t round up)

• If the digit is more than 5, the last significant digit is increased by 1 (round up)

Rounding Off………..Continued

• If this digit is 5 and the five is followed by other digits, increase the last significant digit by 1 (round up)

• If this digit is 5 and the five is followed by all zeros, the last significant digit must be even

• Examples- 3 sigfigs: 2.3466cm = 2.35cm; • 2.344cm = 2.34cm• 2,3451cm = 2.35cm• 2.34500cm = 2.34cm Think about the examples

Back to Calculations with SigFigs• Multiplication/Division• The Quotient or Product has to contain

the same number of sigfigs as the starting measurement with the lest number of sigfigs

• 2.34cm x 3.6cm = 8.424cm2 = 8.4cm2

There are only 2 sigfigs in “3.6”• 4.689cm/ 2.3 x 103 = 2038.695652cm =• 2.0 x 103 cm as we can only have 2

sigfig

One Last Concept

• Percent Error can be calculated when the actual or true value is known

• Formula is• %error = l your value-literature value l x 100 literature value

• Interpretation the absolute value of the difference between your value (observed value) and the literature value (accepted value) divided by the literature value

• Formula given another way• = l observed value – accepted value l x

100 accepted value• An example: You determine the atomic

mass for aluminum to be 28.9 amu whereas the literature cited value is 27.00amu.

• What is your percent error?• Filling in your formula you get• l 28.9 – 27.00 l x 100 = 7.04% (7.037037) 27.00

•That’s all Folks…