Chapter 16.2, Problem 16.5E

A classification study involving several classifiers was carried out. After training and the usual validation step, the following table shows results for classifiers tried. In below, classifiers are identified as M1, M2 and so on. Sensitivity Specificity M1 0.82 0.82 M2 0.92 0.72 M3 0.72 0.92 M4 0.46 0.47 M5 0.03 0.16 M6 0.13 0.02 M7 0.33 0.64 M8 0.72 0.24 M9 0.47 0.82 M10 0.06 0.84 The following list has statements about the classifiers. In the list, there is a single incorrect statement. Please identify the incorrect statement. Hint 1: recall that a classifier dominates another if both performance measures are better. Hint 2: to help you visualize and compare classifiers, you may want to plot the classifier data in a ROC graph (which will not be submitted). Select one: Qa. O b. Oc. d. By swapping zeroes and ones, M5 can be worsened M8 does not dominate M9 M1 dominates all of M4, M5 and M6 M1 is a good classifier ? e. By swapping zeroes and ones, the performance of M4 does not change…

Consider the following training data, shown below before centering. XY 1 0 1 1 1 1 1 1 00 1 1 1 0 0 1 1 1 0 1 1 This data set will be analysed after centering all columns (not scaling). In what follows, the centered data columns are referred to as X and Y. Using these centered columns, we have the following quantities: XTX = 24/11 = 2.1818; XTY = 13/11 = and YTY = 24/11 = 2.1818. Ridge regression Q1 For 2 = R AR = 1.1818 0.56, compute and write in the provided space the ridge estimate ẞ (0.56). Use decimal numbers, not fractions. Q2 Using the ridge estimate ẞ (0.56) you just computed, determine the percentage of shrinkage achieved with respect to the squared L2 norm. That is, compute the shrinkage using || (0.56)||||||with the OLS estimate. In the provided space, write the shrinkage as percentage between 0 and 100 with decimal values. Lasso AR Q3 The following are several expressions for the lasso estimate: (2) = 0.5833 * (1 - 0.84622); L L (a) = 0.5833 * (1 -0.78572); (A) = 0.5417 *…

Calculate the​ 95% confidence intervals for the proportion of children​ surviving, and the proportion of​ non-crew adult passengers surviving. We want to use the given data to make inferences about the general population of all large boat​ crashes, so the data set should be treated as a random sample for this purpose. Part 2 The​ 95% confidence interval for survival rate amongst​ non-crew adults runs from    enter your response here​% to    enter your response here​%. ​(Round to one decimal place as needed. Use ascending​ order.) Part 3 The​ 95% confidence interval for survival rate amongst children runs from    enter your response here​% to    enter your response here​%. ​(Round to one decimal place as needed. Use ascending​ order.) Part 4 Test the alternative hypothesis that the proportion of children surviving does not equal​ 35%, and​ next, test the alternative hypothesis that the proportion of​ non-crew adult passengers surviving does not equal​ 35%. Again, the data set should…