BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: For a regression model of executive compensation, you are given:
i) The following statistics:

\begin{tabular}{|l|l|l|l|l|}
\hline \multicolumn{5}{|c|}{Executive Compensation} \\
\hline Coefficients: & Estimate & Std. Error & \(t\)-statistic & \(p\)-value \\
\hline (INTERCEPT) & -28,595.5 & 220.5 & -129.7 & <0.001 \\
\hline AGEMINUS35 & 7,366.3 & 12.5 & 588.1 & <0.001 \\
\hline TOPSCHOOL & 50.0 & 119.7 & 0.4 & 0.676 \\
\hline LARGECITY & 147.9 & 119.7 & 1.2 & 0.217 \\
\hline MBA & 2,490.9 & 119.7 & 20.8 & <0.001 \\
\hline YEARSEXP & 15,286.6 & 7.2 & 2132.8 & <0.001 \\
\hline
\end{tabular}
ii) The acceptable significance level is \(\alpha=0.10\).

Determine which predictor or predictors should be removed first prior to rerunning the model.

Options:
A
AGEMINUS35
B
AGEMINUS35, MBA, and YEARSEXP
C
TOPSCHOOL
D
TOPSCHOOL and LARGECITY
E
YEARSEXP

Correct answer: C

Solution: The variables TOPSCHOOL and LARGECITY both lack significance and are candidates for removal. However, only one variable should be removed at a time, and it should be the one with the highest \(p\)-value, which is TOPSCHOOL. After removing TOPSCHOOL and rerunning the model it is possible that LARGECITY will become significant.
Question: You are fitting a boosted regression tree with shrinkage parameter \(\lambda=0.08\). There is a single numeric predictor variable, \(x\), and the target variable is \(y\). Each tree is to have a single split. The first tree, built using all the training data has the following split:
- If \(\mathrm{x}<17.5\), then the predicted value of y is 125.4 .
- If \(x \geq 17.5\), then the predicted value of \(y\) is 350.5 .

The following are three observations from the training set:

\begin{tabular}{|l|c|c|}
\hline Observation & x & y \\
\hline 1 & 15.2 & 139 \\
\hline 2 & 12.9 & 156 \\
\hline 3 & 23.6 & 289 \\
\hline
\end{tabular}

Calculate the updated residuals for these three observations ( \(r_{1}, r_{2}, r_{3}\) ).

Options:
A
\(\quad(-406.1,-389.1,-31.1)\)
B
\(\quad(-211.4,-194.4,163.6)\)
C
\(\quad(10.0,10.0,28.0)\)
D
\(\quad(111.0,128.0,279.0)\)
E
\(\quad(129.0,146.0,261.0)\)

Correct answer: E

Solution: The initial residuals are the observations (139, 156, 289). They are updated by subtracting \(\lambda\) times the prediction from the tree. For the three observations the predictions are 125.4, 125.4, and 350.5. The updated residuals are \(139-0.08(125.4)=129.0,156-0.08(125.4) =146.0\), and \(289-0.08(350.5)=261.0\).
Question: Determine which of the following statements is applicable to \(K\)-means clustering and is not applicable to hierarchical clustering.

Options:
A
If two different people are given the same data and perform one iteration of the algorithm, their results at that point will be the same.
B
At each iteration of the algorithm, the number of clusters will be greater than the number of clusters in the previous iteration of the algorithm.
C
The algorithm needs to be run only once, regardless of how many clusters are ultimately decided to use.
D
The algorithm must be initialized with an assignment of the data points to a cluster.
E
None of (A), (B), (C), or (D) meet the meet the stated criterion.

Correct answer: D

Solution: (A) For \(K\)-means the initial cluster assignments are random. Thus different people can have different clusters, so the statement is not true for \(K\)-means clustering.
(B) For \(K\)-means the number of clusters is set in advance and does not change as the algorithm is run. For hierarchical clustering the number of clusters is determined after the algorithm is completed.
(C) For \(K\)-means the algorithm needs to be re-run if the number of clusters is changed. This is not the case for hierarchical clustering.
(D) This is true for \(K\)-means clustering. Agglomerative hierarchical clustering starts with each data point being its own cluster.
Question: Determine which of the following statements about clustering methods is NOT true.

Options:
A
Clustering is used to discover structure within a data set.
B
Clustering is used to find homogeneous subgroups among the observations within a data set.
C
Clustering is an unsupervised learning method.
D
Clustering is used to reduce the dimensionality of a dataset while retaining explanation for a good fraction of the variance.
E
In \(K\)-means clustering, it is necessary to pre-specify the number of clusters.

Correct answer: D

Solution: Item D is a statement about principal components analysis, not clustering.
Question: Sarah performs a regression of the return on a mutual fund \((y)\) on four predictors plus an intercept. She uses monthly returns over 105 months.
Her software calculates the \(F\) statistic for the regression as \(F=20.0\), but then it quits working before it calculates the value of \(R^{2}\). While she waits on hold with the help desk, she tries to calculate \(R^{2}\) from the \(F\)-statistic.

Determine which of the following statements about the attempted calculation is true.

Options:
A
There is insufficient information, but it could be calculated if she had the value of the residual sum of squares (RSS).
B
There is insufficient information, but it could be calculated if she had the value of the total sum of squares (TSS) and RSS.
C
\(R^{2}=0.44\)
D
\(R^{2}=0.56\)
E
\(R^{2}=0.80\)

Correct answer: C

Solution: Even though the formula for \(R^{2}\) involves RSS and TSS, she just needs their ratio, which can be obtained from \(F\).
\(F=\frac{(T S S-R S S) / 4}{R S S /(105-4-1)}=20\)
\(\frac{T S S-R S S}{R S S}=20(4) / 100=0.80\)
\(\frac{T S S}{R S S}=1.80\)
\(\frac{R S S}{T S S}=\frac{1}{1.80}\)
\(R^{2}=1-\frac{R S S}{T S S}=1-\frac{1}{1.80}=\frac{0.80}{1.80}=0.44\)
Question: You are given a stationary AR(1) model, \(y_{t}=\beta_{0}+\beta_{1} y_{t-1}+\varepsilon_{t}, \quad t=2, \ldots, T\). Determine which or the following is always true.

Options:
A
\(\quad \beta_{0} \neq 0\)
B
\(\quad \beta_{0}=1\)
C
\(\quad \beta_{1}=0\)
D
\(\quad \beta_{1}=1\)
E
\(\quad\left|\beta_{1}\right|<1\)

Correct answer: E

Solution: From the definition of a stationary AR(1) process on Page 254 of Frees, the parameter \(\beta_{0}\) can be any fixed constant, ruling out answers (A) and (B). To be stationary it is necessary that \(-1<\beta_{1}<1\), which makes answer (E) correct.
Question: An actuary is investigating the probability of a policyholder incurring positive medical expenditures. Using a sample of data from the company's claims files, a logistic regression model is fit and produces the following table of estimates:

\begin{tabular}{|l|l|l|l|}
\hline Predictor Variable & Coefficient Estimate & Standard error & \(p\)-value \\
\hline Age & 0.257 & 0.102 & 0.0117 \\
\hline Sex & -0.955 & 0.542 & 0.0781 \\
\hline Education & 0.103 & 0.069 & 0.1355 \\
\hline Log(Income) & -0.516 & 0.157 & 0.0010 \\
\hline
\end{tabular}

Determine which of the following predictor variables are significant, using a 5\% level of significance.

Options:
A
Log(Income) only
B
Education only
C
Age and Log(Income) only
D
Sex and Education only
E
All four are significant

Correct answer: C

Solution: Any variable with a \(p\)-value less than 0.05 is significant. Hence, Age and Log(Income) are the only significant variables.
Question: The regression model \(Y=\beta_{0}+\beta_{1} X_{1}+\beta_{2} X_{2}+\beta_{3} X_{1} X_{2}+\varepsilon\) is being investigated. The following maximized log-likelihoods are obtained:
- Using only the intercept term: -1126.91
- Using only the intercept term, \(X_{1}\), and \(X_{2}:-1122.41\)
- Using all four terms: -1121.91

The null hypothesis \(H_{0}: \beta_{1}=\beta_{2}=\beta_{3}=0\) is being tested at the \(5 \%\) significance level using the likelihood ratio test.

Determine which of the following is true.

Options:
A
The test statistic is equal to 1 and the hypothesis cannot be rejected.
B
The test statistic is equal to 9 and the hypothesis cannot be rejected
C
The test statistic is equal to 10 and the hypothesis cannot be rejected.
D
The test statistic is equal to 9 and the hypothesis should be rejected.
E
The test statistic is equal to 10 and the hypothesis should be rejected.

Correct answer: E

Solution: The only two models that need to be considered are the full model with all four coefficients and the null model with only the intercept term. The test statistic is twice the difference of the log-likelihoods, which is 10 .
The number of degrees of freedom is the difference in the number of coefficients in the two models, which is three.
At \(5 \%\) significance with three degrees of freedom, the test statistic of 10 exceeds the 7.81 threshold, so the null hypothesis should be rejected.
Question: Toby observes the following coffee prices in his company cafeteria:
- 12 ounces for 1.00
- 16 ounces for 1.20
- 20 ounces for 1.40

The cafeteria announces that they will begin to sell any amount of coffee for a price that is the value predicted by a simple linear regression using least squares of the current prices on size.
Toby and his co-worker Karen want to determine how much they would save each day, using the new pricing, if, instead of each buying a 24-ounce coffee, they bought a 48ounce coffee and shared it.

Calculate the amount they would save.

Options:
A
It would cost them 0.40 more.
B
It would cost the same.
C
They would save 0.40 .
D
They would save 0.80 .
E
They would save 1.20.

Correct answer: C

Solution: The regression line is \(y=0.40+0.05 x\). This can be obtained by either using the formula for the regression coefficients or by observing that the three points lie on a straight line and hence that line must be the solution. A 24 -ounce cup costs 1.60 . Two of them cost 3.20. A 48 -ounce cup costs \(\$ 2.80\). So the savings is 0.40 .
Question: Trevor is modeling monthly incurred dental claims. Trevor has 48 monthly claims observations and three potential predictors:
- Number of weekdays in the month
- Number of weekend days in the month
- Average number of insured members during the month

Trevor obtained the following results from a linear regression:

\begin{tabular}{|l|l|l|l|l|}
\hline & Coefficient & Standard Error & \(t\) Stat & \(p\)-value \\
\hline Intercept & -45,765,767.76 & 20,441,816.55 & -2.24 & 0.0303 \\
\hline Number of weekdays & 513,280.76 & 233,143.23 & 2.20 & 0.0330 \\
\hline Number of weekend days & 280,148.46 & 483,001.55 & 0.58 & 0.5649 \\
\hline Average number of members & 38.64 & 6.42 & 6.01 & 0.0000 \\
\hline
\end{tabular}

Determine which of the following predictors should be dropped, using a 5\% significance level.

Options:
A
None
B
Number of weekdays only
C
Number of weekend days only
D
Average number of members
E
All three predictors should be dropped

Correct answer: C

Solution: Only variables with a \(p\)-value greater than 0.05 should be considered. Because only one of the variables (Number of weekend days) meets this criterion, it should be dropped.
Question: In a simple linear regression model based on over 100 observations, you are given the following estimates.
i) The estimated slope is -1.03 .
ii) The standard error of the estimated slope is 0.06 .

Calculate the \(95 \%\) confidence interval for the slope.

Options:
A
\(\quad(-1.15,-0.91)\)
B
\(\quad(-1.13,-0.93)\)
C
\((-1.11,-0.95)\)
D
\(\quad(-1.09,-0.97)\)
E
\(\quad(-1.07,-0.99)\)

Correct answer: A

Solution: Using the rule of thumb \(t\)-value of 2 produces and interval of \(-1.03 \pm 2(0.06)\) for an interval of \((-1.15,-0.91)\). Note that if the more precise value of 1.96 is used, the same answer (to two decimal places) is obtained.
Question: Determine which of the following statements about prediction is true.

Options:
A
Each of several candidate regression models must produce the same prediction.
B
When making predictions, it is assumed that the new observation follows the same model as the one used in the sample.
C
A point prediction is more reliable than an interval prediction.
D
A wider prediction interval is more informative than a narrower prediction interval.
E
A prediction interval should not contain the single point prediction.

Correct answer: B

Solution: A is false because different models are likely to produce different results.
B is true because using the fitted model implies that this model continues to apply.
C is false as there is no easy way to compare reliability of the two approaches.
D is false in that a narrower interval provides more useful information about the true value.

E is false in that the interval contains the most likely values with the point prediction being the single most likely point.
Question: The actuarial student committee of a large firm has collected data on exam scores. A generalized linear model where the target is the exam score on a \(0-10\) scale is constructed using a log link, resulting in the following estimated coefficients

\begin{tabular}{|l|c|}
\hline Predictor Variables & Coefficient \\
\hline Intercept & -0.1 \\
\hline Study Time (in units of 100 hours) & 0.5 \\
\hline Attempt (1 for first attempt, else 0) & 0.5 \\
\hline Master's degree (1 for Yes, 0 for No) & -0.1 \\
\hline Interaction of Attempt and Master's degree & 0.2 \\
\hline
\end{tabular}

The company is about to offer a job to an applicant who has a Master's degree and for whom the exam would be a first attempt. It would like to offer half of the study time that will result in an expected exam score of 6.0 .

Calculate the amount of study time that the company should offer.

Options:
A
123 hours
B
126 hours
C
129 hours
D
132 hours
E
135 hours

Correct answer: C

Solution: Let \(T\) be the study time offered. The equation to solve is
\(6.0=\exp [-0.1+0.5(2 T)+0.5-0.1+0.2]=\exp (0.5+T)\)
\(1.79176=0.5+T\)
\(T=1.29176\) or 129 hours.