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

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Presentation on theme: "Last week: Statistics I (average, standard deviation, standard deviation of the mean) This week: Statistics II (95% Confidence Interval) & Calibration."— Presentation transcript:

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

2 Recall from last week….. The mean or average =  true value,  measurements ≈  true value, finite # of measurements Uncertainty given by standard deviation Standard deviation of the mean 1. S, Sm - 2 significant figures 2. Mean (Average) expressed to most significant digit in S m For finite # of measurements σ ≈ S

3 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 a measurement of x it will be in the range  ± t 95 s Uncertainty of a SINGLE MEASUREMENT

4 Example Measure 3 masses: 10.5763, 10.7397, 10.4932 grams Average = 10.60307 grams Std. Dev. S =.125411 =.13 grams Then average = ?= 10.60 grams S m =.125411 /  3 =.072406 =.072 grams 95% C.I. = t 95 S m = 4.303 *.072406 =.311563 =.31 grams of 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; S m =.000033 Ave. = ? Ave. = 10.5766, not 10.57663 because limited by measurement to.0001 grams place t 95 for 3 measurments

5 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 S m. Sally: S m = 0.02587/  5 = 1.157 x 10 -2 = 1.2 x 10 -2 % Janet: S m = 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 S m. S m (Janet) < S m (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. = ±t 95 S m = ±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. = ±t 95 S m = ±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

6 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) **************************************** 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.

7 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.

8 Time After Delivery Volume Change After Drainage with Different Buret Free Delivery Times (aka time required to drain entire buret) 30 sec100 sec 1 min0.01 mL0.00 mL 10 min0.10 mL0.02 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 0.18 Run 224.51 ml 24.36 g 0.15 Δ 1 – Δ 2 ≤.03 Run l 35.19 ml 35.03 g 0.16 Run 234.47 ml 34.28 g 0.19 etc. Make sure to compare the same volume from DIFFERENT runs.

9 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 __________ __________ __________

10 Buoyancy Correction:  B= buoyancy correction d a =0.00115 g/cm 3 (density of air at room temperature) d w = 8.0 g/cm 3 (density of calibration weights) d o = 1.00 g/cm 3 (density of water to 3 sig. fig.) o =object (water) w w = mass read on balance (mass of weights) w o = actual mass of object =w w +  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 = d a w w [1/d o – 1/d w ] Mass of air displaced by object Mass of air displaced by weights

11 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 0.5  C

12 Storage of buret and pipet: Fill with de-ionized H 2 0 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

13 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)


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