Final Assignment

Suggested answers

Take a random sample of size 25, with replacement, from the original sample. Calculate the proportion of students in this simulated sample who work 5 or more hours. Repeat this process 1000 times to build the bootstrap distribution. Take the middle 95% of this distribution to construct a 95% confidence interval for the true proportion of statistics majors who work 5 or more hours.
The exact 95% CI is (40%, 80%). Answers reasonably close to the upper and lower bounds would be accepted.
(e) None of the above. The correct interpretation is “We are 95% confident that 40% to 80% of statistics majors work at least 5 hours per week.”
(c) For every additional $1,000 of annual salary, the model predicts the raise to be higher, on average, by 0.016%.
$R^2$ of raise_2_fit is higher than $R^2$ of raise_1_fit since raise_2_fit has one more predictor and $R^2$ always
The reference level of performance_rating is High, since it’s the first level alphabetically. Therefore, the coefficient -2.40% is the predicted difference in raise comparing High to Successful. In this context a negative coefficient makes sense since we would expect those with High performance rating to get higher raises than those with Successful performance.
(a) “Poor”, “Successful”, “High”, “Top”.
Option 3. It’s a linear model with no interaction effect, so parallel lines. And since the slope for salary_typeSalaried is positive, its intercept is higher. The equations of the lines are as follows:
- Hourly:
  
  \[ \begin{align*} \widehat{percent\_incr} &= 1.24 + 0.000014 \times annual\_salary + 0.913 salary\_typeSalaried \\ &= 1.24 + 0.000014 \times annual\_salary + 0.913 \times 0 \\ &= 1.24 + 0.000014 \times annual\_salary \end{align*} \]
- Salaried:
  
  \[ \begin{align*} \widehat{percent\_incr} &= 1.24 + 0.0000137 \times annual\_salary + 0.913 salary\_typeSalaried \\ &= 1.24 + 0.0000137 \times annual\_salary + 0.913 \times 1 \\ &= 2.153 + 0.0000137 \times annual\_salary \end{align*} \]
A parsimonious model is the simplest model with the best predictive performance.
(c) The exponentiated coefficient (6.502427) represents the factor by which the percentage increase is higher for Successful ratings compared to Poor ratings.\/(a) and (d).
(a) and (d).
Let $u(x) = \sin(x^2) + \cos(ax)$. Then, $g(x) = [u(x)]^k$.

Using the chain rule, we get:

$g'(x) = k [u(x)]^{k-1} \cdot u'(x)$

Now, we need to compute $u'(x)$:

$u(x) = \sin(x^2) + \cos(ax)$

Using the chain rule for each term:

$\frac{d}{dx} \sin(x^2) = \cos(x^2) \cdot 2x ] [ \frac{d}{dx} \cos(ax) = -\sin(ax) \cdot a$

Thus,

$u'(x) = 2x \cos(x^2) - a \sin(ax)$

Combining these results:

$g'(x) = k \left( \sin(x^2) + \cos(ax) \right)^{k-1} \left( 2x \cos(x^2) - a \sin(ax) \right)$
We can split the integral into two separate integrals:

$\int{a}^{b} e^{cx} , dx + \int{a}^{b} \frac{1}{x^n} , dx$
1. Integral of $e^{cx}$:
$\int e^{cx} , dx = \frac{1}{c} e^{cx}$

Thus,

$\int{a}^{b} e^{cx} , dx = \frac{1}{c} \left[ e^{cx} \right]{a}^{b} = \frac{1}{c} \left( e^{cb} - e^{ca} \right)$
1. Integral of $\frac{1}{x^n}$:
For $n \neq 1$,

$\int \frac{1}{x^n} , dx = \int x^{-n} , dx = \frac{x^{-n+1}}{-n+1} = \frac{1}{1-n} x^{1-n}$

Thus,

$\int{a}^{b} \frac{1}{x^n} , dx = \frac{1}{1-n} \left[ x^{1-n} \right]{a}^{b} = \frac{1}{1-n} \left( b^{1-n} - a^{1-n} \right)$

Combining these results:

$\int_{a}^{b} \left( e^{cx} + \frac{1}{x^n} \right) dx = \frac{1}{c} \left( e^{cb} - e^{ca} \right) + \frac{1}{1-n} \left( b^{1-n} - a^{1-n} \right)$
The transpose of the vector $y$ is:

$x^\top = \begin{bmatrix} x_1 & x_2 & x_3 & x_4\end{bmatrix}$
The transpose of the matrix $N$ is: $N^\top = \begin{bmatrix} n_{11} & n_{21} & n_{31} & n_{41} \\ n_{12} & n_{22} & n_{32} & n_{42} \end{bmatrix}$
Solution parts:
1. The dimensions of $C$ are $3 \times 2$.
2. The dimensions of $D$ are $2 \times 3$.
3. For the matrix product $CD$:
  1. The product is valid because the number of columns in $C$ (which is 2) is equal to the number of rows in $D$ (which is 2).
  2. The dimensions of the resulting matrix $CD$ will be $3 \times 3$ (the number of rows of $C$ by the number of columns of $D$).
Solutions:
1. The dimensions of $E$ are $3 \times 2$.
2. The dimensions of $F$ are $2 \times 1$.
3. For the matrix product $EF$:
  1. The product is valid because the number of columns in $E$ (which is 2) is equal to the number of rows in $F$ (which is 2).
  2. The resulting matrix $EF$ is computed as follows:
  \[ EF =\begin{bmatrix} e_{11} & e_{12} \\ e_{21} & e_{22} \\ e_{31} & e_{32} \end{bmatrix} \begin{bmatrix} f_{11} \\ f_{21} \end{bmatrix}=\begin{bmatrix} e_{11}f_{11} + e_{12}f_{21} \\ e_{21}f_{11} + e_{22}f_{21} \\ e_{31}f_{11} + e_{32}f_{21} \end{bmatrix} \]
The resulting matrix $EF$ has dimensions $3 \times 1$.