class: center, middle, inverse, title-slide # Review ### Dr. D’Agostino McGowan --- layout: true <div class="my-footer"> <span> Dr. Lucy D'Agostino McGowan <i>adapted from slides by Hastie & Tibshirani</i> </span> </div> --- class: center, middle ## RStudio Cloud Update --- class: center, middle ## Make sure to bring a calculator to the exam on Thursday --- class: center, middle ## 📖 Canvas --- ## Office Hours this week: * Wednesday 10a * **Wednesday 2p** * Thursday 10a --- mpg | weight | automatic ----|--------|----- 21 | 2.6 | 1 22.8 | 2.32 | 1 18.7 | 3.4 | 0 14.3 | 3.5 | 0 24.4 | 3.1 | 0 You want to predict mpg using weight and automatic. Write out how you would calculate `\(\hat\beta\)` in matrix form using the data provided (you need not **solve** the matrices) --- mpg | weight | automatic -----|--------|----- 21 | 2.6 | 1 22.8 | 2.32 | 1 18.7 | 3.4 | 0 14.3 | 3.5 | 0 24.4 | 3.1 | 0 Solving the equation results in the following `\(\hat\beta\)`: `\(\begin{bmatrix} \hat{\beta}_0 \\ \hat{\beta}_1 \\ \hat{\beta}_2\end{bmatrix} = \begin{bmatrix}80.45 \\ -18.40 \\ -13.30\end{bmatrix}\)` Using the information provided, calculate the MSE (mean squared error) for this model. --- mpg | weight | automatic -----|--------|----- 21 | 2.6 | 1 22.8 | 2.32 | 1 18.7 | 3.4 | 0 14.3 | 3.5 | 0 24.4 | 3.1 | 0 `\(\begin{bmatrix} 0.66 & 0.34& 0.07& 0.19&-0.26 \\ 0.34& 0.66&-0.07&-0.19& 0.26 \\ 0.07&-0.07& 0.37& 0.42& 0.21 \\ 0.19&-0.19& 0.42& 0.55& 0.02 \\ -0.26& 0.26& 0.21& 0.02& 0.77 \\ \end{bmatrix}\)` Above is the **hat matrix** for the model. Use this to calculate LOOCV error for this model --- mpg | weight | automatic ----|--------|----- 15 | 3.8 | 0 30.3 | 1.9 | 1 40.9 | 1.5 | 1 20.1 | 3.4 | 0 26 | 2.1 | 1 You get a new test data set (above). Using the same model you fit to the training data, calculate the MSE in this test dataset --- class | x -------|--- 1 | -0.57 1 | -1.26 1 | 1.13 2 | 3.3 2 | 5.4 Using the data above, calculate `\(\pi_k\)`, `\(\mu_k\)`, and `\(\sigma^2\)` --- class | x -------|--- 1 | -0.57 1 | -1.26 1 | 1.13 2 | 3.3 2 | 5.4 Using LDA, what is are the discrimant scores for x = 3? Which class would you classify this point into? --- class | x -------|--- 1 | -0.57 1 | -1.26 1 | 1.13 2 | 3.3 2 | 5.4 What is the probability that an observation where x = 3 is in class 1? --- class | x -------|--- 1 | -0.57 1 | -1.26 1 | 1.13 2 | 3.3 2 | 5.4 What is the decision boundary for this LDA problem? --- ## P(+ | Drug User) = 0.8 ## P(Drug User) = 0.1 ## P(+ | Not a drug user) = 0.2 * What is the probability of being a drug user give you test positive? --- | Truth (+) | Truth (-) | total -|---------|-------| Predicted (+) | 10 | 20 | 30 Predicted (-) | 5 | 50 | 55 Total | 15 | 70 | 85 * What is the accuracy? * What is the false positive rate? * What is the false negative rate?