Statistical Analysis of Experimental Data: Hypothesis Testing and Confidence Intervals - P, Exams of Chemistry

Download Statistical Analysis of Experimental Data: Hypothesis Testing and Confidence Intervals - P and more Exams Chemistry in PDF only on Docsity! CHEN 3010 Applied Data Analysis Fall 2009 Final Examination Solutions 1) (30 pts) Eight independent measurements are taken of the diameter of a piston. The measurements (in inches) are: 3.236 3.223 3.242 3.244 3.228 3.253 3.253 3.230 The average and standard deviation are 3.239 and 0.0113 inches, respectively, and the true variance of piston diameter is unknown. a. (8 pts) Is there evidence that the true mean piston diameter is less than 3.24 inches? Use a risk factor of 0.05. Solution: We wish to test the hypothesis : 3.24 vs. : 3.24. First, we need to compute the test statistic (based on t-distribution since variance is not known) corresponding to our sample average: √⁄ 3.239 3.24 0.0113 √8⁄ 0.250 We compare this to the T distribution with 7 degrees of freedom: . , 1.895 Since the test statistic is NOT less than the critical t-value for this lower-tailed test, we cannot accept the alternate hypothesis and there isn’t evidence that the true mean piston diameter is less than 3.24 inches. b. (7 pts) Compute a 95% two-sided confidence interval on the variance of piston diameter. Solution: The 95% confidence interval can be given by the following equation: 1 Χ . , 1 Χ . , 8 1 · 0.0113 16.01 8 1 · 0.0113 1.69 . . CHEN 3010 Final Exam Page 2 c. (15 pts) Eight independent measurements are taken of the diameter of another piston. The measurements on this second piston are: 3.295 3.232 3.261 3.248 3.289 3.245 3.576 3.201 The average and sample standard deviation are 3.293 and 0.118 inches, respectively. Is there evidence that the diameter of the two pistons significantly differ from one another? You must prove any assumptions you make about the variance. Use an alpha value of 0.05. Solution: First, we must determine if the variances for the two populations are different or if they can’t be considered different. We wish to test the hypothesis : vs. : (lower-tailed because our data suggests that the variance from the first piston is lower than variance of the second piston. Our test statistic is: 0.0113 0.118 0.0091 We compare this test statistic to the F distribution with 7 degrees of freedom in the numerator and 7 degrees of freedom in the numerator, using the statistic for the lower-tailed test: . , , 1 . , , 1 3.79 0.264 Since our test statistic (0.0091) is much less than the critical f-value of 0.264, we can accept our alternate hypothesis. The variances are unequal. The next step is to compute the test statistic for our comparison of means. For equal sample size when variance is unknown and unequal, we use the following test statistic: 3.239 3.293 0.0113 0.118 8 1.29 We compare this to the T distribution with degrees of freedom, where: ⁄ 1 ⁄ 1 7 CHEN 3010 Final Exam Page 5 Since this is a resolution IV design, this means that binaries will be aliased with binaries and main effects will be aliased with trinaries. 5) (20 pts) In a landmark 2002 publication in the New England Journal of Medicine, researchers reported that of 136 cardiac arrest patients that had been subjected to induced hypothermia, 75 emerged from the hospital with healthy brain function. In a control group of 137 patients – patients who had not been cooled – only 53 patients emerged from the hospital with healthy brain function. a. (10 pts) Is there a significant difference between inducing hypothermia and withholding hypothermia (test using α = 0.05)? Solution: We wish to compare the binomial proportions of the two populations. We can set up the following hypotheses: : vs. : .  A test statistic for this comparison of binomial proportions is based upon the normal distribution: ̂ ̂ ̂ 1 ̂ 1 1 0.551 0.387 0.469 1 0.469 1136 1 137 2.73 where ̂ 0.551, ̂ 0.387, and ̂ 0.469. We compare the test statistic to the normal distribution: . 1.645 CHEN 3010 Final Exam Page 6 Since the test statistic exceeds the critical z value for this upper-tailed test, we can accept the alternate hypothesis and claim that inducing hypothermia leads to a better resonse than withholding hypothermia. b. (5 pts) What is the P-value for this test? Solution: The P-value is the area to the right (in the upper tail) corresponding to our test statistic of 2.73. We can look in the z tables to see what the area is corresponding to the right of a z-value of 2.73. This is equal to 1- 0.9968 = 0.0032. c. (5 pts) Find a 90% confidence interval on the difference between percentage of patients emerging from the hospital with healthy brain function for the two groups. Solution: The confidence interval for a binomial proportion difference is based on the normal distribution (normal approximation to the binomial distribution): ̂ ̂ / ̂ 1 ̂ ̂ 1 ̂ ̂ ̂ / ̂ 1 ̂ ̂ 1 ̂ 0.551 0.389 1.645 0.551 1 0.551 136 0.389 1 0.389 137 0.551 0.389 1.645 0.551 1 0.551 136 0.389 1 0.389 137 . . 6) NEW. (25 pts) The data below represent the %-yield for batches of a chemical product. Statistics for the data set are shown below the table. CHEN 3010 Final Exam Page 7 Average = 88.9 Stdev = 4.07 n = 60 a. (7 pts) Based on the parameters of a box plot, do any of these measurements appear to be outliers? Solution: First, we need to determine the interquartile range (IQR). For the IQR, we need to determine what the 1st and 3rd quartiles are. If we take a look at the sorted data, the 1st quartile will be (since there are an even number of data points in each half) the average of the 15th and 16th data points, which is 86.1. The 3rd quartile will be the average of the 45th and 46th data points, which is 91.9. Therefore, the IQR = 91.9 – 86.1 = 5.8. We also need the median, which is the average of the 30th and 31st data points, which is 88.95. Outliers are greater than 1.5IQR on either side of the median. For this case, outliers would occur below (88.95 - 1.5*5.8) = 80.25 or above (88.95 + 1.5*5.8) = 97.65. There is one point below 80.25, so 74.3 is an outlier. b. (15 pts) With any outliers removed, determine the number of bins, compute the bin boundaries, and determine frequencies for each bin for a proper histogram. Basically, determine everything for a proper histogram but you don’t have to draw it. Solution: With the single outlier removed, we are left with 59 data points. We can determine the number of bins for a histogram using the following two formulas (the first is an upper number, the second is a lower number): #   √ √59 7 sorted 94.1 94.1 84.6 74.6 87.3 90.6 93.2 92.1 83.6 83.0 87.3 90.6 90.6 96.4 85.4 83.6 87.3 90.6 91.4 88.2 89.7 84.0 87.5 91.4 88.2 86.4 87.6 84.1 87.6 91.7 86.1 74.6 85.1 84.1 87.7 92.1 95.1 84.9 89.6 84.2 88.2 92.4 90.0 87.3 90.0 84.6 88.2 92.4 92.4 89.6 90.1 84.9 88.3 93.2 87.3 90.3 94.3 85.1 88.8 93.2 87.3 92.4 85.4 85.1 89.1 93.7 84.1 90.6 86.6 85.1 89.6 94.1 90.1 89.1 91.7 85.4 89.6 94.1 95.2 88.8 87.5 85.4 89.7 94.3 86.1 86.4 84.2 86.1 90.0 94.3 94.3 85.1 85.1 86.1 90.0 95.1 93.2 84.0 90.5 86.4 90.1 95.2 86.7 93.7 95.6 86.4 90.1 95.3 83.0 87.7 88.3 86.6 90.3 95.6 95.3 90.6 84.1 86.7 90.5 96.4 CHEN 3010 Final Exam Page 10 The responses (in absorbance units) were algal population after 7 days’ incubation and three responses were recorded for each treatment combination. a. (5 pts) In the chart on the following page, what is the value “ContrastAB” equal to? The contrast of AB is just the dot product of the AB column with the responses (sum of responses): 1 1.35 1 1.023 1 1.327 1 0.939 1 1.974 1 1.862 1 2.073 1 2.381 . b. (30 pts) Complete a full ANOVA table for these results and suggest which effects should be retained (use α=0.05). You may write directly on the following sheet but remember to turn in with your exam. Solution: See the attached worked out sheet. · 2 3 · 2 12 · 2 3 · 2 24 There is 1 degree of freedom for each of the 7 effects, 23 total degrees of freedom, leaving (23 – 7) = 16 degrees of freedom for error. We can calculate the following columns for our ANOVA table: F_critical can be found in the back of the book for an alpha of 0.05, 1 degree of freedom for the numerator and 16 degrees of freedom for the denominator: F_critical = 4.49. Conclusion: C, BC, and AC are significant. c. (5 pts) Determine a suitable regression equation relating the coded variables to the algal population after 7 days’ incubation. CHEN 3010 Final Exam Page 11 Solution: Since C, BC, and AC are significant, we can set up the following regression equation: 0.539 0.152 0.030 0.038 where the coefficients are just the effects of the significant terms divided by 2. d. (10 pts) If the phosphorous concentration were 0.24 mg/L, the nitrogen concentration were 10.00 mg/L, and carbon concentration were 96 mg/L, what is an estimate for the algal population after 7 days’ incubation? Solution: The coded variables corresponding to the above concentrations are: A = 0.5, B = 1.0, and C = 0.9. If we plug these coded variables into the model above, we get that the algal population after 7 days’ incubation should be about 0.720. Treatment A B C AB BC AC ABC y1 y2 y3 sum (1) ‐1 ‐1 ‐1 1 1 1 ‐1 0.448 0.576 0.326 1.35 a 1 ‐1 ‐1 ‐1 1 ‐1 1 0.391 0.309 0.323 1.023 b ‐1 1 ‐1 ‐1 ‐1 1 1 0.412 0.434 0.481 1.327 ab 1 1 ‐1 1 ‐1 ‐1 ‐1 0.376 0.251 0.312 0.939 ACBCCy 038.0030.0152.0539.0 +++= c ‐1 ‐1 1 1 ‐1 ‐1 1 0.639 0.656 0.679 1.974 ac 1 ‐1 1 ‐1 ‐1 1 ‐1 0.583 0.631 0.648 1.862 bc ‐1 1 1 ‐1 1 ‐1 ‐1 0.657 0.736 0.68 2.073 abc 1 1 1 1 1 1 1 0.768 0.814 0.799 2.381 Contrast ‐0.519 0.511 3.651 ContrastAB 0.725 0.911 0.481 Effect ‐0.04325 0.042583 0.30425 0.0299167 0.060417 0.075917 0.040083 12.93 Grand Total SSEffect 0.011223375 0.01088 0.555408 0.00537 0.021901 0.03458 0.00964 0.539 Grand Average Coeff 0.152125 0.030208 0.037958 0.702 SST ANOVA (Part B)    Source SS dof MS F0 Fcrit A 0.011223375 1 0.011223 3.396 4.494 B 0.010880042 1 0.01088 3.292 4.494 C 0.555408375 1 0.555408 168.038 4.494 Part A: ContrastAB 0.359 AB 0 005370042 1 0 00537 1 625 4 494. . . . BC 0.021901042 1 0.021901 6.626 4.494 AC 0.034580042 1 0.03458 10.462 4.494 Part C: ABC 0.009640042 1 0.00964 2.917 4.494 Model: Error 0.053 16 0.003305 T t l 0 702 23 P t D d d i bl A 0 5o a . ar   : co e  var a es . B 1 C 0.9 y 0.71989