lab 1 post lab NEW

# lab 1 post lab NEW - Exercise 1 How do scientists collect...

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Unformatted text preview: Exercise 1: How do scientists collect and analyze data? Post Lab Question 1 Weight of Water Measured by Various Volumetric Measuring Devices Trial 100 mL beaker (g) 5 mL pipet (g) 1 55.02 58.37 2.62 2.97 0.17 2 54.07 59.14 2.61 2.88 0.17 3 55.48 59.1 2.59 2.95 0.17 4 55.59 59.04 2.75 3.01 0.16 5 54.21 59.27 2.71 2.95 0.16 6 55.77 59.19 2.71 2.99 0.16 7 55.74 59.19 2.67 2.97 0.17 8 56.8 59.44 2.64 2.9 0.17 Mean 55.33 59.09 3.26 2.95 0.17 standard deviation 0.890065 g 0.31405g 1.78652g 0.04469g 0.00276g percent error of mean 7.78% 1.51% 8.55% 1.58% 2.50% 100 mL graduated cylinder (g) 10 mL graduated cylinder (g) 20-200 µL pipettor (g) Mean example calculation (55.018+54.069+55.482+55.585+54.211+55.770+55.736+56.802) ÷ 8 = 55.33413g Standard Deviation example calculation √[ (55.018²+54.069²+55.482²+55.585²+54.211²+55.770²+55.736²+56.802²) ÷ 7 ]= 0.890065g percent error of mean example calculation ( |55.33413-60|÷ 60 ) * 100 = 7.78% n1 = 8 Ttest value Calculation t = | 55.33413 – 59.09263 | / sqrt((0.793262696/8)+(0. 098628839/8))= 11.256 degree of freedom = (16-1) = 15 Hypothesis At value of P = 0.05 the critical t value at 15 degree is 2.13. Based on this null hypothesis is reject Sample Variance 0.79 0.1 0.41 Ttest value 11.26 1.34 6.31 degree freedom 15 15 15 p value 0.1 hypothesis Reject Accept Reject Question 2 (a) (b) ( c ) Sample Variance calculation S1^2 = {(55.018-55.33413)^2 + (54.069 – 55.33413)^2 + (55.482 – 55.33413)^2 + (55.58S1^2 = {(55....
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lab 1 post lab NEW - Exercise 1 How do scientists collect...

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