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CategoryProgramming
SubjectR | R Studio
DifficultyUndergraduate
StatusSolved
More InfoProbability Homework Help
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Short Assignment Requirements

A 10 problem homework assignment on basic R programming.

Assignment Description

1

Stat 471/671, Spring 2019, Hwk 1, due Feb 14, 11:00am

Q1. Use the rep() function (once for each part) to generate the following vectors:

1)   1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3 3 3

2)   1 1 1 1 1 1 2 2 2 2 2 3 3 3 3 4 4 4 5 5 6

3)   1 1 1 1 1 2 3 3 3 4 4 4 4 5 5

Q2. Let x = y = (0,−3,NA,13)0. Without help of the function identical(), check whether or not x = y.

Q3. Use vectorization to evaluate for x = 5 in a single R expression without using loops or writing out the summation one term at a time. You may use the R function factorial().

Q4. Write single R expressions to do the following. For the numbers

7.5, 6.9, 0.06, 0.006, 12, 2.7, 17, 8, 0.8, 6.9,

compute

(a)    the square roots of the absolute deviations of the numbers from their mean.

(b)    the largest of the squared deviations of the numbers from their median.

(c)    the absolute errors due to rounding: Round the square roots of the numbers totwo decimal places and square them. Find the absolute differences between these and the original numbers.

Q5. Write a single R expression to calculate the sample correlation coefficient r between the two variables: Age and Circumference from the Orange data set using the formula

You may not use the R functions cov() or var() for writing this expression. Use the cor() function to check your answer.

Q6. Use a single R expression and the the sample variance formula to calculate the sample variance of the weight in the chickwts data set.

Q7. Using a single R expression, calculate the sample median absolute deviation (mad)

1.4826 × med|xi − med(xi)|

of the weight in the chickwts data set, where med(xi) is the sample median and med|xi− med(xi)| is the sample median of the absolute deviations of xi from med(xi).

Q8. Create an R matrix object named A using the following data:

2

4.5 -2.1 6.3 2.7 1.6 -2.6 3.6 8.4 5.2 8.3

2.5 1.7 3.7 -4.4 3.7

3.8 5.8 5.2 1.9 -4.2

Do this by using the scan() function to read in the data from screen. Perform the following operations and report the result of each operation:

(a)    extract the middle 3 elements of row 3 of A to a vector y;

(b)    create the following 3 × 3 matrix B by subsetting matrix A:

3.6 5.2 8.3

1.7 -4.4 3.7

5.8 1.9 -4.2

(c)    compute vector d = By;

(d)   append the data vector (9.4,10.5,−11.5)0 to the left hand side of B to form W;

(e)    compute matrix P = WW0;

(f)     compute the inverse of P;

(g)    solve the equation Pa = d for a using a = P−1d;

(h)    solve the equation Pa = d for a using function solve()

(i)     factorize P so that P = R0R, where R is an upper triangular matrix.

Q9. Use the matrix A created in Q8 to write a single expression to extract

(a)    the row of A that contains the largest value of the matrix;

(b)    the column of A that contains the smallest value of the matrix.

Q10. Elements of an R data frame can also be accessed the same way elements of a matrix are accessed using subscripts. Subsets of a data frame extracted this way will themselves be data frames, except when they are single rows, columns, or scalars. In general, a single row will be a data frame and a single column, an R vector.

Load the package named MASS and the Cars93 data set available in the package using data(Cars93, package="MASS"). Use subscripting operations on Cars93 to find the vehicles with highway mileage of less than 25 miles per gallon (variable MPG.highway) and weight (variable Weight) over 3500 lbs. Report the model name, the price range (low, high), highway mileage, and the weight of the cars that satisfy these conditions. This can be done in a single R expression; however, you may simplify the extraction to several steps.

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