CHM 320 Chapt 4 Lecture 5Hypothesis: More brown and less blue than other colorsConclusion: Less red, yellow, brown than other colors30 data points713115137024681012141618red orange yellow blue green brownM&M'save. total = 56st. dev. = 1.3Spring 2006CHM 320 Chapt 4 Lecture 5Spring ’05 results with 50 data points :Less Red M&M’s; other colors relatively equal50 data points10910101070246810121416red orange yellow blue green brownM&M colorNumber of M&M'sCHM 320 Chapt 4 Lecture 5Normal Error Curve00. 050. 10. 150. 20. 250. 30. 350. 40. 45-5-4-3-2-1012345ZyConfidence interval – provides a statistically valid range within which the true value occurs based on a group of measured valuesIn a population, a Gaussian curve can be used to determine confidence level.In any Gaussian Curve:Range confidenceμ±1σ 68.3 %μ±2σ 95.5 %μ±3σ 99.7 %These % are good when thinking about the population as compared to a sample.CHM 320 Chapt 4 Lecture 5Confidence Intervals for sample:Student’s t Confidence interval tells you that the true mean, μ, is likely to be within a certain value of the measured mean, x. The confidence interval can be expressed as:μ = x ± where s = measured standard deviationn = number of measurementst = value of Student’s tt values (Student’s t) are given in Table 4-2, pg. 67 of Harris. The values are arranged such that you must know the degrees of freedom (n –1), and the confidence level (50%, 90%, 95%, etc.)t s√nCHM 320 Chapt 4 Lecture 5CHM 320 Chapt 4 Lecture 5Common forms of statistical testing encountered by the analytical chemist• Comparing a result to a group of observations (the one-sample t-test)• Comparing two groups of objects: are the averages different? (the two-sample t-test)• Comparing two standard deviations: the F-test• Testing for outliers: the Q testStatistical Testing for SamplesCHM 320 Chapt 4 Lecture 5Example Calculation for Confidence IntervalWhat is the number of “Green M&M’s” in a bag of M&M’s?Average = 13 St. Dev. = 3.9 # of measurements = 30What is the 50% and 95% confidence levels for these measurements.Degrees of freedom = 29 t50%= 0.683 t95%= 2.042(use an estimate of 30)μ50%= 13 ± (0.683*3.9)/√30 = 13 ± 0.49 (range 12.51 to 13.49)μ95%= 13 ± (2.042*3.9)/√30 = 13 ± 1.45 (range 11.55 to 14.45)CHM 320 Chapt 4 Lecture 5Uncertainty can be reduced by increasing the number of measurementsSame example as before, but now you make 5 measurements instead of 30 measurements.Degrees of freedom = 4 t50%= 0.741 t95%= 2.776Degrees of freedom = 29 t50%= 0.683 t95%= 2.0425 measurements: μ50%= 13 ± 1.95 μ95%= 13 ± 4.8430 measurements: μ50%= 13 ± 0.49 μ95%= 13 ± 1.45factor of 4 improvement factor of 3.3 improvementSEE THE DIAGRAM IN THE LOWER RIGHT HAND CORNER OF PAGE 67 IN HARRIS.CHM 320 Chapt 4 Lecture 5Using the Student’s t to compare two sets of measurementsWe are testing the null hypothesis when comparing the mean values in two sets of data (hypothesis is that the two sets of measurements are statistically the same).Typically, we are interested in 95% possibility of being correct(1 out of 20 times is wrong).Cases to Consider:• Compare our measured values to a known value.• Compare two different samples measured with the same method (comparing replicate measurements)• Compare two different methods making measurements on same samples. (comparing methods)CHM 320 Chapt 4 Lecture 5Does my answer agree with the known value at a 95% confidence level?Calculate a t value, and compare to the t value from Table 4-2.Confidence equation: μ = x ±Rearrange: tcalc= (⏐x – known value⏐* √n) / sCompare tcalcto ttable. If tcalc> ttable, your answer does not agree.If tcalc< ttable, your answer does agree.t s√ntcalcttabletcalcttableCHM 320 Chapt 4 Lecture 5Example: How does the measured value of “blue M&M’s” compare to having a equal distribution of M&M’s in bag – i.e. 1/6 of M&M’s are blue or 9.3 M&M’s of the 56 total M&M’s?In order to answer the question, you must determine tcalc, and compare it to the ttableat 95% confidence.tcalc= (⏐x – known value⏐* √n) / s = (⏐ 11 – 9.3 ⏐ * √30) / 3.1= 3.00ttable= 2.042tcalc= 3.00 > 2.042 = ttableDifference is significant!Where:x = 11s = 3.1n = 30CHM 320 Chapt 4 Lecture 5Example: Is the chance of getting a red M&M from a bag of M&M’s the same as the chance of getting an yellow M&M?Red M&M: 7 ± 2.5 (n = 30)Yellow M&M: 5 ± 1.9 (n = 30)tcalc= (⏐xr–xo⏐/ spool) * (√(nr* no)/(nr+ no))spool= √[sr2(nr–1) + sy2(ny– 1)]/(nr+ ny–2)spool= √[2.5r2(30r– 1) + 1.9y2(30y– 1)]/(30r+ 30y–2)= 2.2tcalc= (⏐7 – 5⏐/2.2) * (√(30* 30)/(30+ 30))= 3.52ttable= 2.042tcalc= 3.52 > ttable= 2.042 DIFFERENCE IS SIGNIFICANTCHM 320 Chapt 4 Lecture 5F-test: Determines if two standard deviations are significantly different from one another.• The two samples are arranged so that S1>S2• The calculated value of F is compared to the critical value for the sample sizes and confidence level.CHM 320 Chapt 4 Lecture 5Example:Red M&M – standard deviation = 2.5 (n = 30)Orange M&M – standard deviation = 3.0 (n = 30)Fcalc= sO2/ sR2(use sOin numerator because it is greater than sR)= (3.0)2/ (2.5)2= 9.0 / 6.25 = 1.44Ftable= 1.30 (for degrees of freedom of 30)Fcalc= 1.30 < 1.84 = FtableNO significant differenceCHM 320 Chapt 4 Lecture 5• Used to evaluate the validity of outlying data. • An outlier is a data point that appears unusually different than most. Q-test: Determines if a data point is an outlier02468101202468xyCHM 320 Chapt 4 Lecture 5Example: Test scores (from last lecture)90, 89, 68, 95, 100Steps:• List data from low to high: 68, 89, 90, 95, 100• Determine the range: Difference in lowest and highest values (100 - 68 = 32)• Determine the gap: Difference from questionable data point and closest data point (89 – 68 = 21)• Qcalc= gap/range = 21 / 32 = 0.66• Compare calculated Qcalcto Qtable(from table pg. 75)Qcalc= 0.66 > 0.64 = Qtable• Qcalc> Qtablethe data point is considered an outlier and can be removed from the data