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Last week: Statistics I (average, standard deviation, standard deviation of the mean) This week: Statistics II (95% Confidence Interval) & Calibration of Volumetric Equipment & Thermometer

Last week: Statistics I (average, standard deviation, standard deviation of the mean) This week: Statistics II (95% Confidence Interval) & Calibration

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Last week: Statistics I (average, standard deviation, standard deviation of the mean)

This week: Statistics II (95% Confidence Interval)&

Calibration of Volumetric Equipment & Thermometer

Recall from last week…..

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

n

SSm Standard deviation of the mean

1. S , Sm - 2 significant figures

2. Mean (Average) expressed to most significant digit in Sm

For finite # of measurements σ ≈ S

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

± t95s

Uncertainty of a SINGLE MEASUREMENT

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

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)

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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.

CLEANING GLASSWARE

• Do NOT rinse at de-ionized water taps – take flask back to own sink to rinse.

• Rinsing with de-ionized water should leave no droplets.

• For open and easily rinsed containers, use a cleansing powder, rinse with tap water and finally with de-ionized water.

• For burets, pipets, volumetric flasks, etc., use liquid detergent, rinse with tap water and finally with de-ionized water.

NEVER use cleansing powder; it is too hard to get out and in the case of burets it will usually score the Teflon stopcocks.

Time After Delivery

Volume Change After Drainage with Different Buret Free Delivery Times

(aka time required to drain entire buret)

30 sec 100 sec

1 min 0.01 mL 0.00 mL

10 min 0.10 mL 0.02 mL

Buret Drainage Times: (Buret should drain within 100-120 seconds)

Buret

Check Duplicate Runs by Approximate Method

V W = V - W

Run l 25.15 ml 24.97 g 0.18Run 2 24.51 ml 24.36 g 0.15 Δ1 – Δ2 ≤ .03Run l 35.19 ml 35.03 g 0.16Run 2 34.47 ml 34.28 g 0.19

etc. Make sure to compare the same volume from DIFFERENT runs.

• Name__________________________________ Lab Section________________• Date Report Submitted___________________• CALIBRATION OF VOLUMETRIC EQUIPMENT• 1. Thermometer • Thermometric Standard, true reading (TT) ___________°C•• Desk Thermometer, observed reading (To) ___________°C•• Thermometer Correction (TT - To) ___________°C

• 2. Buret • Series A: Observed Mass of Corrected True Vol. Correction • T(H2O), °C Volume, Vo Water Mass Volume, VT VT – Vo • __________ ___________ ___________ ___________ ___________ ___________• (approximately 0-25mL)• ___________ ___________ ___________ ___________ ___________ • (approximately 0-35mL)• ___________ ___________ ___________ ___________ ___________ • (approximately 0-45mL)• Series B: • T(H2O), °C• __________ ___________ ___________ ___________ ___________ ___________•

____________ ____________ ____________ ____________ ____________ • • ___________ ___________ ___________ ___________ ___________ • Series C: • (if needed)• T(H2O), °C• __________ ___________ ___________ ___________ ___________ ___________• • ___________ ___________ ___________ ___________ ___________ • • ___________ ___________ ___________ ___________ ___________ • • • Nominal buret reading 0 – 25 0 - 35 0 - 45•• Average volume correction __________ __________ __________•

Buoyancy Correction:

B = buoyancy correctionda = 0.00115 g/cm3 (density of air at room temperature) dw = 8.0 g/cm3 (density of calibration weights) do = 1.00 g/cm3 (density of water to 3 sig. fig.) o = object (water) ww = mass read on balance (mass of weights) wo = actual mass of object

= ww + B

Archimedes’s Principle: an object immersed in a fluid is buoyed up by a force equal to the weight of the fluid displaced.

If an object is larger in volume than the weights that balance it, then it displaces more air and is buoyed up by a greater force.

ΔB = daww[1/do – 1/dw]

Mass of air displaced by object

Mass of air displaced by weights

Volumetric Pipet

Uncertainty of a SINGLE MEASUREMENT

Check the highest-lowest < 0.03g

***how is this quick check like the one for the buret?

Thermometer (on side bench)

Quartz thermometer

Hang your thermometer on a hook (count from left or right)

Should be within 0.5C

Stirrer

probe crossbar with hooks

Thermometer

Storage of buret and pipet:

• Fill with de-ionized H20

• Cork in top of buret (if old obtain a new cork)

• Rubber policeman on tips

• Loosen stopcock of buret slightly for storage. Remember to retighten when you use buret again.

• Store in buret cabinet

Things to remember...

• Wear goggles, lab coat, closed-toe shoes

• Backpacks on hooks

• No eating, drinking, using cell phones, etc.

• Make sure balances are calibrated, not moved (check the bubble)

• Work on the wet bench side (sinks)