Assignment 1

Due date: Friday, September 11, 2026
Total: 30 points

NoteSubmission

For this assignment, please use R Markdown or Quarto.

Submit both:

1. your source file (.Rmd or .qmd), and
2. the corresponding rendered PDF.

For multiple-choice questions, clearly indicate your selected answer.
For computational questions, include the R code used to obtain your answer.

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Question 1: R Programming and Statistical Computing [10]

Select the best answer. Each question is worth 1 point.

1a)

Which of the following creates a numeric vector in R?

A. x <- list(1, 2, 3)
B. x <- c(1, 2, 3)
C. x <- data.frame(1, 2, 3)
D. x <- matrix(1, 2, 3)

TipAnswer

B. The function c() combines values into a vector.

1b)

Which expression correctly returns the mean of x after removing missing values?

A. mean(x, remove = TRUE)
B. mean(x, missing = FALSE)
C. mean(x, na.rm = TRUE)
D. mean(na.omit = x)

TipAnswer

C. Use the argument na.rm = TRUE.

1c)

Suppose

\[ x=(10,20,30,40). \]

What does the following R expression return?

x[c(1, 3)]

A. 10 20
B. 10 30
C. 20 40
D. 30 40

TipAnswer

B. R uses 1-based indexing, so positions 1 and 3 are selected.

1d)

Which statement about R indexing is correct?

A. R indexing starts from 0.
B. R indexing starts from 1.
C. R indexing depends on the data type.
D. R indexing starts from \(-1\).

TipAnswer

B. R uses 1-based indexing.

1e)

Which function is most appropriate for applying a function to the columns of a numeric matrix?

A. apply()
B. lm()
C. filter()
D. rbind()

TipAnswer

A.

For example,

apply(X, MARGIN = 2, FUN = mean)

applies mean() to the columns of X.

1f)

What is the main advantage of using vapply() instead of sapply()?

A. It always runs in parallel.
B. It requires a prespecified output type.
C. It works only on matrices.
D. It automatically removes missing values.

TipAnswer

B. vapply() requires the expected output type, which makes the result more predictable.

1g)

Which of the following is the native R pipe operator?

A. %>%
B. ->
C. |>
D. %*%

TipAnswer

C. The native R pipe operator is |>.

1h)

Which dplyr function is used to retain rows satisfying a condition?

A. select()
B. filter()
C. mutate()
D. summarise()

TipAnswer

B. filter() retains rows satisfying logical conditions.

1i)

Which statement concerning apply() and a for loop is most accurate?

A. apply() is always faster than a for loop.
B. A for loop is always faster than apply().
C. Neither is always faster; performance depends on the computation.
D. apply() performs computations in parallel by default.

TipAnswer

C. Neither approach is universally faster. Performance depends on the computation and implementation.

1j)

Why is reproducibility important in statistical computing?

A. It guarantees that every statistical model is correct.
B. It allows an analysis to be rerun and verified.
C. It eliminates sampling variability.
D. It eliminates numerical error.

TipAnswer

B. Reproducibility allows an analysis to be rerun, checked, and verified.

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Question 2: Computational Results Using the Iris Data [10]

Use the built-in iris dataset in R.

For each computational question, include the R code used to obtain your answer.

2a) [2]

Report the sample mean of Sepal.Length, rounded to three decimal places.

\[ \boxed{\phantom{5.843}} \]

TipAnswer

mean(iris$Sepal.Length)

The result is

\[ \boxed{5.843}. \]

2b) [2]

Report the sample standard deviation of Sepal.Width, rounded to three decimal places.

\[ \boxed{\phantom{0.436}} \]

TipAnswer

sd(iris$Sepal.Width)

The result is

\[ \boxed{0.436}. \]

2c) [2]

How many observations in the iris dataset belong to the species virginica?

A. 25
B. 40
C. 50
D. 75

TipAnswer

C. There are 50 observations from each species.

For example,

table(iris$Species)

2d) [2]

Fit the linear regression model

\[ \texttt{Sepal.Length} = \beta_0 + \beta_1\texttt{Sepal.Width} + \varepsilon. \]

Report the estimated slope coefficient \(\hat{\beta}_1\), rounded to three decimal places.

\[ \boxed{\phantom{-0.223}} \]

TipAnswer

fit <- lm(Sepal.Length ~ Sepal.Width, data = iris)
coef(fit)

The estimated coefficients are approximately

\[ \hat{\beta}_0=6.526, \qquad \hat{\beta}_1=-0.223. \]

Therefore,

\[ \boxed{\hat{\beta}_1=-0.223}. \]

2e) [2]

Based on the fitted model in part (d), which statement is correct?

A. Sepal length tends to increase as sepal width increases.
B. Sepal length tends to decrease as sepal width increases.
C. The estimated slope is exactly zero.
D. The model cannot be fitted because both variables are continuous.

TipAnswer

B.

The estimated slope is negative:

\[ \hat{\beta}_1=-0.223<0. \]

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Question 3: R Coding and Visualization [6]

Use the iris dataset for both parts.

3a) [3]

Using ggplot2, construct a scatter plot with:

• Sepal.Width on the \(x\)-axis,
• Sepal.Length on the \(y\)-axis,
• points coloured according to Species.

Your figure must contain:

1. appropriate axis labels,
2. an informative title, and
3. a readable legend.

Include both your R code and the resulting figure.

TipAnswer

One possible solution is:

library(ggplot2)

ggplot(
  iris,
  aes(
    x = Sepal.Width,
    y = Sepal.Length,
    colour = Species
  )
) +
  geom_point() +
  labs(
    x = "Sepal Width",
    y = "Sepal Length",
    colour = "Species",
    title = "Sepal Length versus Sepal Width"
  ) +
  theme_minimal()

Grading

• 1 point: Correct variables on the \(x\) and \(y\) axes.
• 1 point: Points correctly coloured by Species.
• 1 point: Appropriate title, labels, and readable figure.

3b) [3]

Add a linear regression line to the plot from part (a).

Use

geom_smooth(method = "lm", se = FALSE)

and include the final plot in your report.

TipAnswer

One possible solution is:

ggplot(
  iris,
  aes(
    x = Sepal.Width,
    y = Sepal.Length,
    colour = Species
  )
) +
  geom_point() +
  geom_smooth(
    aes(group = 1),
    method = "lm",
    se = FALSE,
    colour = "black"
  ) +
  labs(
    x = "Sepal Width",
    y = "Sepal Length",
    colour = "Species",
    title = "Sepal Length versus Sepal Width"
  ) +
  theme_minimal()

Grading

• 1 point: Correct regression line is added.
• 1 point: Uses method = "lm" and se = FALSE.
• 1 point: Final figure and code are both included.

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Question 4: Numerical Linear Algebra [4]

4a) [2]

Suppose \(\Sigma\) is a positive-definite covariance matrix with Cholesky decomposition

\[ \Sigma=R^\mathsf{T}R, \]

where \(R\) is upper triangular.

Which expression correctly computes \(\log\det(\Sigma)\)?

\[ \sum_i \log(R_{ii}) \]

\[ 2\sum_i \log(R_{ii}) \]

\[ \sum_i R_{ii}^2 \]

\[ 2\log\left(\sum_i R_{ii}\right) \]

TipAnswer

B.

Since

\[ \Sigma=R^\mathsf{T}R, \]

we have

\[ \det(\Sigma) = \det(R^\mathsf{T})\det(R) = \det(R)^2. \]

Because \(R\) is triangular,

\[ \det(R)=\prod_iR_{ii}. \]

Therefore,

\[ \log\det(\Sigma) = 2\log\left(\prod_iR_{ii}\right) = 2\sum_i\log(R_{ii}). \]

4b) [2]

Let

\[ R= \begin{pmatrix} 2 & 1\\ 0 & 3 \end{pmatrix}, \qquad \Sigma=R^\mathsf{T}R. \]

Compute \(\log\det(\Sigma)\).

A. \(\log(6)\)
B. \(\log(12)\)
C. \(2\log(6)\)
D. \(6\)

TipAnswer

C.

Since

\[ \det(R)=2(3)=6, \]

we have

\[ \det(\Sigma)=\det(R)^2=36. \]

Hence,

\[ \log\det(\Sigma) = \log(36) = 2\log(6). \]

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Quick Answer Key

ImportantInstructor Answer Key

Question 1

Part	Answer
1a	B
1b	C
1c	B
1d	B
1e	A
1f	B
1g	C
1h	B
1i	C
1j	B

Question 2

Part	Answer
2a	5.843
2b	0.436
2c	C
2d	-0.223
2e	B

Question 4

Part	Answer
4a	B
4b	C