If we explicitly apply the cubic polynomial svm to the original $784-$dimensional raw pixel features, the resulting representation would be of massive dimensionality. Instead, we will apply the cubic polynomial svm to the $10-$dimensional PCA representation of our training data which we will have to calculate just as we calculated the $18-$dimensional representation in the previous problem. Use the sklearn package and build the SVM model on your local machine. Use random$\_$state $=0,$ kernel $=$ 'poly', degree $=3,$ and default values for other parameters. If you have done everything correctly, cubic polynomial svm should perform better $($on the test set$)$ using these features than either the $18-$dimensional principal components or raw pixels. The error on the test set using the cubic polynomial svm should only be around $0\mathrm{.}06,$ demonstrating the power of nonlinear classification models. Error rate for $10-$dimensional PCA features using cubic polynomial svm $=$Explicit Confidence Intervals for an Exponential Statistical Model

$6$ points possible $($graded$)$

Suppose you observe a sample data set consisting of $n=64$ inter$-$arrival times $x_1,$dots,$x_{64}$ for the subway,

measured in minutes. As before, we assume the statistical model that $x_1,$dots,$x_{64}{}^{\text{i}\text{i}\text{d}}$exp$()$ for some

unknown parameter $>0.$ In this data set, you observe that the sample mean is $\frac{1}{64}{\displaystyle \underset{i=1}{\overset{64}{}}}{x}_i=7\mathrm{.}8.$

Additional Instructions: For best results, please adhere to the following guidelines and reminders:

For the upcoming calculations, please truncate $q_{\frac{}{2}}$ at $2$ decimal places, instead of a more exact value.

For example, if $q_{\frac{}{2}}=3\mathrm{.}84941,$ use $3\mathrm{.}84$ instead of $3\mathrm{.}85$ or $3\mathrm{.}849.$

Input answers truncated at $4$ decimal places. For example, if your calculations yield $11\mathrm{.}327458,$ use

$11\mathrm{.}3274$ instead of $11\mathrm{.}3275$ or $11\mathrm{.}32745.$

You will be computing Cls at asymptotic level $90\%.$

Using the 'solve method' $($refer to the slide 'Three solutions'$),$ construct a confidence interval $I_{\text{s}\text{o}\text{l}\text{v}\text{e}}$ with

asymptotic level $90\%$ for the unknown parameter $.$

$,$

Using the 'plug$-$in method' $I_{\text{p}\text{l}\text{u}\text{g}-in}($refer to the slide 'Three solutions'$),$ construct a confidence interval with

asymptotic level $90\%$ for the unknown parameter $.$

Which interval is narrower?