Precision & Accuracy Precision –How close are the measurements to each other? –Reproducibility Accuracy – How close is a measurement to the true value?

Precision & Accuracy

• Precision – How close are the measurements to each other?– Reproducibility

• Accuracy – How close is a measurement to the true value?

Could be affected by the following:Systematic errors – all higher or lower than actual value

(lack of accuracy)Random errors – some high and some low

(lack of precision)

Significant Figures

• Presenting measurements and calculated results with the appropriate significant figures and units is an indication of the precision of values.

Rules for Sig Figs• All nonzero digits are significant• Trapped zeros are significant• Trailing zeros are significant if there is a decimal point• Leading zeroes are NOT significant

300300.300.00.0030.00300.00300

# sig figs

1

3

4

1

2

3

How would one write 300 with 2 sig figs?

Scientific Notation3.0 x 10 2 has two sig figs

Written as a number between 1-10 x a power of ten

Unambiguously displays the precision of the value making it easier to make comparisons

300 3 x 102

300. 3 .00 x 102

300.0 3.000 x 102

0.003 3 x 10-3

0.0030 3.0 x 10-3

0.00300 3.00 x 10-3

Making Measurements - Thermometer

The number of significant figures in your measurement

depends on the measuring device.

The bottom of the meniscus is between 87 and 88 ° C.

This can be read to 1 digit more precision than indicated by the calibration.

The last estimated digit can vary from person to person, but each should record a value to the tenth’s place.

There are 3 sig figs and the last digit is the uncertain digit.

Generally, measurements are uncertain by ± 1 in that last digit unless otherwise indicated by your measuring device. Usually, 1/10 of an increment.

87.5±.1°C

Beaker vs. Graduated Cylinder

Each contains the same amount of water.

Beaker

10. ± 1mL

Graduated Cylinder

10.05 ± .05 mL

The Analytical Balance

All digits should be recorded as given, precision is to the 0.1 mg, & the accuracy is determined by the calibration.

Calculations & Sig Figs

• Multiplication & DivisionThe total number of sig figs in the answer is equal to the same number of sig figs in the measurement used in the calculation with the smallest number of sig figs.

Ex: 5.1 cm x 2.01 cm = 10.0701 cm2 = 10. cm2

• Round the final answer using the number to the right of the last sig fig.• Avoid round off errors by keeping extra digits beyond the last sig fig

when calculating intermediate values.

Calculations & Sig Figs

• Addition & SubtractionThe final answer should be rounded to the right-most filled column (according to the value with the biggest uncertain digit – the weakest link).

Ex: 6.5 cm

100.01 cm

+ .044 cm

106.554 cm

= 106.6 cm

“Scientific notation” can make it easier…..

• What is the sum of 4.5 x 10-6, 3.2 x10-5, and 15.2 x 10-7?

.45 x 10-5

3.2 x 10-5

.152 x 10-5

3.802 x 10-5

= 3.8 x 10-5

SI Prefixes

Prefix Symbol Meaning Power of 10Mega M 1,000,000 106

Kilo k 1,000 103

Deci d 0.1 10-1

Centi c 0.01 10-2

Milli m 0.001 10-3

Micro μ 0.000001 10-6

Nano n 0.000000001 10-9

Femto f 0.000000000000001 10-15

Atto a 0.000000000000000001 10-18

Fundamental SI Units

Physical Quantity UnitAbbreviation

Mass kilogram kg

Length meter m

Time second s

Temperature kelvin K

Dimensional Analysis• Use conversion factors (definitions, ratios) to convert

from one unit to another.

• Conversion factors are exact numbers that have no uncertainty.

• Ex. Convert 6.4 weeks to hours.6.4 weeks x 7 days x 24 hours = 1100 hrs

1 week 1 day

Group Problems• Convert 47 hours to weeks. 47 hours x 1 day x 1 week = 0.28 weeks

24 hours 7 daysThe same conversions were used as in the previous example. The top equals the bottom. Round off answers at the end. Keep additional sig figs for intermediate answers.• Calculate the sum of 2.5 + 3.5 + 4.5 +5.5. 2.0

3.54.55.5

14.5• The tread on a certain automobile tire wears 0.00100 inches per 2,600 miles driven. If

the car is driven 45 miles a day, how many months ( 1mo = 30 days) can a tire w/ 0.010 in of treat be used before it wears down and needs to be replaced?

.010in x 2,600 mi x 1 day x 1 month = 19.25 = 19 months 0.00100in 45 mi 30 days

• In a displacement of water by gas experiment the initial volume of water in a burette is 45.50 mL and the final volume is 37.50 mL. What is the total volume of water displaced? In mL? in L?

45.50 (4sf) - 37.50 mL (4sf)

8.00 mL (3sf) = 0.00800 L (still 3sf)

Statistical Analysis and Expression of Data

Reading: Lab Manual 29 - 40

Today: Some basics that will help you the entire year

n

xx i

i

The mean or average

= true value, measurements

≈ true value, finite # of measurements

Uncertainty given by standard deviation

1

2

n

xi

i

σ ≈ S = [(xi-x)2/(n-1)]1/2

For finite # of measurements Standard deviation: S

(Calculators can calculate & σ quite easily!!! Learn how to do this on your calculator.)

For small number of measurements σ ≈ S is very poor. Must use Student t value. σ ≈ tS; where t is Student t

Usually use 95% Confidence Interval

So, 95% confident that if we make ameasurement of x it will be in the range

x ± t95s

Uncertainty of a SINGLE MEASUREMENT

Usually interested in mean (average) and its uncertainty

Standard Deviation of the meann

SSm

Then average and uncertainty is expressed as x ± t95Sm

Often want to know how big uncertainty is compared to the mean:

Relative Confidence Interval (C.I.) = (Sm / x )(t95)

Expression of experimental results:

1. Statistical Uncertainties (S, Sm, t95S, Sm (t95) / x) always expressed to 2 significant figures

2. Mean (Average) expressed to most significant digit in Sm (the std. dev. of the mean)

Example Measure 3 masses: 10.5763, 10.7397, 10.4932 gramsAverage = 10.60307 gramsStd. Dev. S = .125411 = .13 grams

Then average = ? = 10.60 grams

Sm = .125411 / 3 = .072406 = .072 grams

95% C.I. = t95Sm = 4.303 * .072406 = .311563 = .31 gramsof the mean

Relative 95% C.I.of the mean = ?

= 95 % C.I. / Average = .311563/10.60307 = .029384 = .029 of the mean

Usually expressed at parts per thousands (ppt)

= .029 * 1000 parts per thousand = 29 ppt = Relative 95% C.I. of the mean

Now work problems.

What if measure 10.5766, 10.5766, 10.5767 grams? Ave. = 10.57663; Sm = .000033

Ave. = ? Ave. = 10.5766, not 10.57663 because limited by measurement to .0001 grams place

t95 for 3 measurments

Sally 5 times, average value of 15.71635% ; standard deviation of 0.02587%.

Janet 7 times, average value of 15.68134% ; standard deviation of 0.03034%.(different technique)

Express the averages and standard deviations to the correct number of significant figures.

Must use Sm. Sally: Sm = 0.02587/5 = 1.157 x 10-2 = 1.2 x 10-2%Janet: Sm = 0.03034/7 = 1.147 x 10-2 = 1.1 x 10-2%

Sally: 15.72%; S = 0.026% Janet: 15.68%; S = 0.030%

Using the proper statistical parameter, whose average value is more precise?

Must use Sm.Sm(Janet) < Sm(Sally) so Janet’s average value is more precise.

95% confidence intervals of the mean, relative 95% confidence intervals of the mean

Sally: 95% C.I. = ±t95Sm = ±2.776 (1.157 x 10-2) = ±3.21 x 10-2% = ±3.2 x 10-2% Range = 15.69 – 15.75% Relative 95% C.I. = 3.21 x 10-2 / 15.71635 * 1000 ppt = 2.04 = 2.0 ppt

Janet: 95% C.I. = ±t95Sm = ±2.447(1.147 x 10-2) = ±2.81 x 10-2% = ±2.8 x 10-2% Range = 15.65 – 15.71% Relative 95% C.I. = 2.81 x 10-2 / 15.68134 * 1000 ppt = 1.79 = 1.8 ppt

Are the two averages in agreement at this confidence level?

Because the 95% C.I. for both measurements overlap, the two averages are in agreement

If you owned a chemical company and had to choose between Sally’s and Janet’s technique, whose technique would you choose and why?

Choose Sally’s techniques because the uncertainty in a single measurement based on S is better than that using Janet’s technique.