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University of Pennsylvania *

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405

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Statistics

Date

Feb 20, 2024

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#### STAT 405/705-402 Homework 0, Spring 2022 #### Name: #### Please list all external sources you have consulted for completing this assignment: #### Instructions: #### 1. If a question refers to a function not seen in class, #### use the help facility in R to learn about it. #### 2. Insert your answers and the R code you used to generate them, beneath each question. #### Even though for some questions you could find the answer "by inspection", for this homework you need #### to write code to get the answers, unless it is explicitly stated that no code is required. #### 3. When you are asked to print an answer, it is NOT ENOUGH to simply write a print statement, #### for example print(2*3) as an answer, you also need to #### cut and paste into this document the value(s) that is actually printed (i.e.,6). #### 4. When submitting to Canvas, please change the file extension to .txt, otherwise it cannot be submitted. ## ------------------------------------------------------------------------ #### Q1. R as a calculator. #a Store your birthday number, in YYYYMMDD format (example: 20010831), into a numeric variable called bday. bday <- 20040520 bday #> bday #[1] 20040520 # Now write code to find and print the answer for all parts: #b The cube root of your birthday number. x <- bday^(1/3) x # > x <- bday^(1/3) # > x # [1] 271.625 #c The log base 16 of your birthday number. y <- log (bday, base = exp(16)) y # > y <- log (bday, base = exp(16)) # > y # [1] 1.050829 #d Your birthday number, modulo 7. (That's the remainder, after dividing by 7. Use the "%%" operator.) z <- (bday %% 7) z # > z <- (bday %% 7) # > z # [1] 3 ## ------------------------------------------------------------------------ #### Q2. Some shortcuts for creating vectors. # In this question, you will learn a few shortcuts for creating vectors of special shapes.

#a The code below creates a vector of all integers from 1 to 10: vec0 <- c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) # This approach would be unwieldy if we want a vector of integers from, say, 1 to 10000. # Use the help facility of R to study the colon (":") operator and understand how it can help with this task. # Then, create a vector called vec1 consisting of all integers from your birthday number to 20210831. vec1 <- c(20040520:20210831) vec1 #b When you read the help page for ":", you may have noticed a function called "seq". # You can understand its behavior by running the code below: seq(9.1, 11.1, by = 0.2) seq(0, 10, length = 30) # Now use the "seq" function to create a vector called vec2 consisting of all integers divisible by 7 # between 0 and 100 (including 0), in DESCENDING order. vec2 <- seq(98,0,-7) vec2 # > vec2 <- seq(98,0,-7) # > vec2 # [1] 98 91 84 77 70 63 56 49 42 35 28 21 14 7 0 #c Try the following code to learn what the "rep" function does: rep(c(1, 2), times = 4) rep(c(0, 1), times = c(3, 5)) # You can also look up the help page for "rep" for additional info. # Now create a vector called vec3 consisting of all even numbers between 100 and 200 in ascending order, # with the K-th number repeated K times: (100, 102, 102, 104, 104, 104, ...). # HARD. Hint: Use seq, rep and : within a single line. vec3 <- rep(seq(100,200,2),seq(1,51,1)) vec3 # > vec3 <- rep(seq(100,200,2),seq(1,51,1)) # > vec3 # [1] 100 102 102 104 104 104 106 106 106 106 108 # [12] 108 108 108 108 110 110 110 110 110 110 112 # [23] 112 112 112 112 112 112 114 114 114 114 114 #list goes on... ## ------------------------------------------------------------------------ #### Q3. Applying functions and operators to vectors. #a Create the two vectors of numbers alpha = (200, 225, 250,..., 1000) # beta = (600, 590, 580, ..., 280) vec1 <- seq(200,1000,25) vec2 <- seq(600, 280,-10) # > vec1 <- seq(200,1000,25) # > vec2 <- seq(600, 280,-10) vec1 vec2 # > vec1 # [1] 200 225 250 275 300 325 350 375 400

# [10] 425 450 475 500 525 550 575 600 625 # [19] 650 675 700 725 750 775 800 825 850 # [28] 875 900 925 950 975 1000 # > vec2 # [1] 600 590 580 570 560 550 540 530 520 510 500 # [12] 490 480 470 460 450 440 430 420 410 400 390 # [23] 380 370 360 350 340 330 320 310 300 290 280 #b What is the sum of the element-wise product of the two vectors? vec3 <- vec1 + vec2 vec3 # > vec3 <- vec1 + vec2 # > vec3#c What is the maximum value in the vector containing the element-wise products? # [1] 800 815 830 845 860 875 890 905 920 # [10] 935 950 965 980 995 1010 1025 1040 1055 # [19] 1070 1085 1100 1115 1130 1145 1160 1175 1190 # [28] 1205 1220 1235 1250 1265 1280 #c What is the maximum value in the vector containing the element-wise products? max(vec3) # > max(vec3) # [1] 1280 ## ------------------------------------------------------------------------ #### Q4. Applying functions and operators to vectors, continued. #a Set the random number seed as 20210831. set.seed(20210831) #b Create two vectors, named x and y of standard normal random variables, each of length 10000000. x <- rnorm(10000000,0,1) y <- rnorm(10000000,0,1) #c Create a third vector, called z, containing the elementwise square root of the sum # of the squares of the elements. # That is, if you call the first elements x1 and y1, etc., you need to calculate # # x1,y1 --->>>> sqrt(x1^2 + y1^2) # x2,y2 --->>>> sqrt(x2^2 + y2^2) # x3,y3 --->>>> sqrt(x3^2 + y3^2) # and so on. z <- sqrt((x^2)+(y^2)) # > z <- sqrt((x^2)+(y^2)) # > z #d Find the average of the elements in the z vector. Print the result. (sum(z))/10000000 # > (sum(z))/10000000 # [1] 1.253316 f <- (sum(z))/10000000 #e Square the average and double it. Print the result. (f^2)*2 # > (f^2)*2 # [1] 3.141604 #f The probability that a standard normal random variable has absolute value

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greater than 1.96 is about 0.05. # Let's verify it by experiment: find the proportion of elements in x with absolute value greater than 1.96. # You may only use the functions and operators covered in Class 1 and 2. Print the result. # Hint: For a vector x, the command x>1.96 will give you a list of TRUE and FALSE corresponding # to each element. g<- x>1.96 g sum(g) # > g<- x>1.96 # > sum(g) # [1] 249191 prob <- 249191/10000000 prob # > prob <- 249191/10000000 # > prob # [1] 0.0249191

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