Analytical Chem istry - DePauw University

Chapter 4 Evaluating Analytical Data111To compare the means for the two analysts we use the following null andalternative hypotheses.H : µ = µ H : µ ≠ µ0 A B A A BBecause we cannot pool the standard deviations, we calculate t exp usingequation 4.20 instead of equation 4.2186. 83−82.71t exp=( 032 . ) ( 216 . )+6 62 2= 462 .and calculate the degrees of freedom using equation 4.23.⎡( 032 . ) ( 216 . ) ⎤+⎢⎥ν= ⎣ 6 6 ⎦2⎛ 2 ⎞( 032 . ) ⎛ 2 ( 216 . ) ⎞2⎝⎜6 ⎠⎟⎝⎜6 ⎠⎟+6+1 6+12 2 2− 2= 5.3≈5From Appendix 4, the critical value for t(0.05,5) is 2.57. Because t exp isgreater than t(0.05,5) we reject the null hypothesis and accept the alternativehypothesis that the means for the two analysts are significantly differentat a = 0.05.Practice Exercise 4.10To compare two production lots of aspirin tablets, you collect samplesfrom each and analyze them, obtaining the following results (in mg aspirin/tablet).Lot 1: 256 248 245 245 244 248 261Lot 2: 241 258 241 244 256 254Is there any evidence at a = 0.05 that there is a significant difference inthe variance between the results for these two samples? This is the samedata from Practice Exercise 4.9.Click here to review your answer to this exercise.Pa i r e d Da t aSuppose we are evaluating a new method for monitoring blood glucoseconcentrations in patients. An important part of evaluating a new methodis to compare it to an established method. What is the best way to gatherdata for this study? Because the variation in the blood glucose levelsamongst patients is large we may be unable to detect a small, but significantTypical blood glucose levels for mostnon-diabetic individuals ranges between80–120 mg/dL (4.4–6.7 mM), rising to ashigh as 140 mg/dL (7.8 mM) shortly aftereating. Higher levels are common for individualswho are pre-diabetic or diabetic.