We've built a logistic regression model in RapidMiner, and would like to use it to make predictions for some new data points. Which operator do we need: Performance Apply Model O Cross Validation O Nominal to Numerical
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- You trained the regression model with 100 regressors and 1000 observations in the training and another 1000 in the test sample. You found that in-sample R2 over the training sample is 70% and the out-of-sample R2 over the test sample only - 30%. (select all that apply) a) Do you think there is any problem and how would you characterize it? Can adding more regressors (if you have them) help the model? b) Which approaches you may use to solve the problem? c) What would you expect the in-sample R2 to increase or decrease after that? What about the out-of-sample (test) R2?You build a model predicting blood pressure as a function of three variables: weight (numeric) age (numeric) income (categorical: low, medium, high) You first specify your model as: blood pressure ~ age * income + weight How many parameters (k) does your model have? (Remember, we do not count the grand mean in k) You change the above model specification to be: blood pressure ~ age * income + weight * income How many parameters does your model have now? You change your model to include the three-way interaction (which, remember, includes all two-way interactions and main effects, too!) Your model now looks like this: blood pressure ~ age * income * weight How many parameters does your model have now?This question tests your understanding of data transformation. We did an assignment on this process. If you test your data and find that it is not suitable for testing, which mathematical function can you put each score through to make it usable and suitable for testing? Select one: a. There is no way to make the data usable. b. square root each number c. subtract the df from each number d. use the variance instead of each number Clear my choice
- Analysts at a start-up company are analyzing 35 months of sales data. They partition the data (the first 20 observations are assigned to the training set; the most recent 15 months are in the test set). The only independent variable is T (month number, ranging from 1 to 35). Five models (polynomials of order 1 - 5) are fit to the data. The first order is just the linear model; the 2nd order polynomial is the quadratic model; order 3 is the cubic model, etc. In each case the model is fit on the training data, and scored on both the training and test data sets. The results are below. Based on this output, which is the best predictive model? Metrics AE RMSE MAE SSE Metrics AE RMSE MAE SSE Model 2 Model 3 Model 5 Model 4 Model 1 1 <0.000001 0.955978 0.792802 18.277907 1 -1.034550 1.424155 1.208991 30.423248 Training Data Scoring Models (Polynomial of order 1-5) 4 2 <0.000001 0.928791 0.759583 17.253086 3 <0.000001 <0.000001 0.928295 0.855568 0.761951 0.652212 17.234646 14.639962 Test Data…I need to plot the data from a simple linear regression with its values x and y with the fitted y values from fitting a logistic regression model. How do I plot the x values with two different y's, y values on the left side with fitted y values on the right? I want to generate the fitted/predicted y values using logistic regression and then create a plot using all those three variables? Here is an example of my SAS code that doesn't work. data coupons;input redeemed discount;cards;100 5147 9176 11211 13244 15277 17310 19343 21;run; ods graphics on;proc logistic data=coupons;model redeemed(event='1') = discount;effectplot;run; It give me this warnings and I do not understand…. NOTE: Option EVENT= is ignored since LINK=CLOGIT. NOTE: PROC LOGISTIC is fitting the cumulative logit model. The probabilities modeled are summed over the responses having the lower Ordered Values in the Response Profile table. Use the response variable option DESCENDING if you want to reverse the assignment of…Why use LASSO shrinkage methods for linear regression? Select ALL that are correct. They allow for greater model interpretability. They increase the prediction accuracy of linear regression on training data. They slightly decrease bias and can substantially decrease the variance of the model. They slightly increase bias, but can substantially decrease the variance of the model.
- consider the following model: y = b_0+ b_1*x what is the parameter b_0? O a. the slope coefficient. O b. O C. O d. The independent variable. the dependent variable. the intercept.You decide to run a simpler model to predict churn, using only the variables tenure (in months) and TotalCharges (in US$). The output is given below. The AIC of this model is 4727.6 (in contrast to the AIC of 4240 for the full model). On the basis of this which model would be expected to give superior predictive performance? Actual ## Coefficients: ## Estimate Std. Error z value Pr(>|z|) ## (Intercept) 2.471e-01 5.360e-02 4.611 4.01e-06 *** ## tenure < 2e-16 *** -1.124e-01 5.816e-03 -19.334 ## TotalCharges 8.236e-04 5.618e-05 14.660 < 2e-16 *** ## No --- ## Signif. codes: 0 ## Yes Yes ## Null deviance: 5701.5 on 4921 ## Residual deviance: 4721.6 on 4919 ## AIC: 4727.6 515 345 ## (Dispersion parameter for binomial family taken to be 1) ## Predicted ***** No 795 3267 0.001 Confusion Matrix (Training) **** Actual 0.01 Yes No degrees of freedom degrees of freedom Yes The simpler model (with just tenure and TotalCharges) The full model (with all variables) 0.05 0.1 220 145 Predicted No 339…You are working on a spam classification system using regularized logistic regression. "Spam" is a positive class (y = 1)and "not spam" is the negative class (y=0). You have trained your classifier and there are m= 1000 examples in the cross-validation set. The chart of predicted class vs. actual class is: Predicted class: 1 Predicted class: 0 Actual class: 1 85 15 For reference: Accuracy = (true positives + true negatives)/(total examples) Precision = (true positives)/(true positives + false positives) Recall = (true positives)/ (true positives + false negatives) F1 score = (2* precision * recall)/(precision + recall) What is the classifier's F1 score (as a value from 0 to 1)? Write all steps Use the editor to format your answer Actual class: 0 890 10
- A Ridge Linear Regression adds the sum of the squared values of the coefficients to the loss function to penalize large coefficients. Group of answer choices True FalseFor evaluation of regression models, typically, the higher the [ Select ] ["Adjusted R Squared", "Residual Standard Error"] the better, and the lower the [ Select ] ["Adjusted R Squared", "Residual Standard Error"] , the better.This is a coding question. Now that you have worked out the gradient descent and the update rules. "Try to progrum a Ridge regression. Please complete the coding. Note that here the data set we use has just one explanatory variable and the Ridge regression we try to create here has just one variable (or feature). Now that you have finished the program. What are the observations and the corresponding predictions using Ridge? Now, make a plot to showease how well your model predicts against the observations. Use spatter plot for observations, line plot for your model predictions. Observations are in color red. and predictions are in color green. Add appropriate labels to the x axis and y axis and a title to the plot You may also nood to fine tune hyperparameters such as leurning rate and the number of'aterations.