BAS ASSIGNMENT #1

BAS 320 - Assignment 1 - Working with R CARTER JONES Response: Analytics is the best major at UT! getwd () #This should output the path to your BAS folder; change with Session, Set Working Directory if not ## [1] "C:/Users/carte/Documents/320 DOCS" load ( "BAS320Assignment1.RData" ) library (regclass) ## Loading required package: bestglm ## Loading required package: leaps ## Loading required package: VGAM ## Loading required package: stats4 ## Loading required package: splines ## Loading required package: rpart ## Loading required package: randomForest ## randomForest 4.7-1.1 ## Type rfNews() to see new features/changes/bug fixes. ## Important regclass change from 1.3: ## All functions that had a . in the name now have an _ ## all.correlations -> all_correlations, cor.demo -> cor_demo, etc. Question 1: R as a calculator Translate into R syntax the following mathematical expressions. Have the answers be “printed to the screen” (i.e., don’t left -arrow them to anything) when it is knitted. Note: you can mouse over the equations to see them, or you can head to the Canvas page to see what they look like. If you only see half an equation, click the double up arrows on the far right of the equation to collapse it, then click again to expand it.

(8 − 2 1 + 4 ) (3 + 2 5 2 ) #about 23.4 ( 8 - ( 2 / ( 1 + 4 ))) * ( 3 + ( 2 / 5 ^ 2 )) ## [1] 23.408 8 + 3 8/5 − 2 1/3 2 3 + 3 2/3 + 4 5/7 #about 8.355 8 + (( 3 ^ ( 8 / 5 ) - ( 2 ^ 1 / 3 )) / ( 2 ^ 3 + 2 ^ ( 2 / 3 ) + 4 ^ ( 5 / 7 ))) ## [1] 8.418014 |√2 − 300𝑒 −2(8−6) | + log 10 294 − ln320 − 12 #about -11.22 abs ( sqrt ( 2 ) - 300 * ( exp ( - 2 * ( 8 - 6 )))) + log10 ( 294 ) - log ( 320 ) - 12 ## [1] -11.2195 "answer with help from chatgpt" ## [1] "answer with help from chatgpt" d. When translating this equation into R, write values in R’s version of scientific notation when possible 8 × 10 −2 + 8.34 × 10 2 + 5.32 × 10 3 #about 6154 ( 8 * 10 ^- 2 ) + ( 8.34 * 10 ^ 2 ) + ( 5.32 * 10 ^ 3 ) ## [1] 6154.08 Question 2: Left-arrow Step 1: Define a variable named heard to equal 10. Imagine this is the number of people that have heard of a new product on day 1 (additional people hear of the product by word of mouth). Step 2: Re-define heard to be 148% its current value, minus 2, then rounded to the nearest integer. Imagine this is the number of people that have heard of the product on day 2. You should find it will go from equaling 10 on day 1 to equaling 12.8, which rounds to 13, on day 2.

## [1] 132.3 130.8 129.3 127.8 126.3 124.8 123.3 121.8 120.3 118.8 117.3 115.8 ## [13] 114.3 112.8 111.3 109.8 108.3 106.8 105.3 103.8 102.3 100.8 99.3 97.8 ## [25] 96.3 94.8 93.3 91.8 90.3 88.8 87.3 d. Create a text vector named Q3d that consists of the nine words great, great, fair, fair, fair, average, average, average, average, repeated a total of four times. You’ll want to use the rep command, and your vector should contain a total of 36 elements. Print to the screen the contents of Q3d and show the result of running table(Q3d) to give a frequency table of its elements. Hint: although not required, try creating the text vector (the one that gets repeated 4 times) with three different rep commands. Q3d <- rep ( c ( "great" , "great" , "fair" , "fair" , "fair" , "average" , "average" , "average" , "ave rage" ), times= 4 ) Q3d ## [1] "great" "great" "fair" "fair" "fair" "average" "average" ## [8] "average" "average" "great" "great" "fair" "fair" "fair" ## [15] "average" "average" "average" "average" "great" "great" "fair" ## [22] "fair" "fair" "average" "average" "average" "average" "great" ## [29] "great" "fair" "fair" "fair" "average" "average" "average" ## [36] "average" table (Q3d) ## Q3d ## average fair great ## 16 12 8 e. Create a vector named items that contains the integer sequence from 43 down to 3. Create a vector named prices that contains the integer sequence from 10 up to 50. What’s the median of the vector produced when you multiply together the elements of items and prices ? items <- seq ( from= 43 , to= 3 , by= - 1 ) items <- 43 : 3 prices <- 10 : 50 median <- median (items * prices) median ## [1] 592 Question 4: Vectors (NFL data) Load in the NFL dataframe that is contained in the regclass library (you’ll need the library and data command to get this dataset into the environment). This data contains statistics for NFL teams from the 2002- 2012 seasons. Let’s look at the X13.OffPassYds column, which gives a team’s total number of yards gained when passing over the course of

a season. Save (left-arrow) the X13.OffPassYds column into a vector named pass , and use this vector in the following parts. For most parts, you’ll need to use the which function. library (regclass) data (NFL) pass <- NFL $ X13.OffPassYds a. Print to the screen the values in the 212th, 325th, and 326th positions of pass . pass <- NFL $ X13.OffPassYds pass[ c ( 212 , 325 , 326 )] ## [1] 4977 3683 2999 b. In what positions of pass will you find the number 2890 (there are three positions)? positions <- which (pass == 2890 , arr.ind = TRUE ) positions ## [1] 17 126 231 "chat gpt helped me on this one" ## [1] "chat gpt helped me on this one" c. Print to the screen all unique values of pass that are between 2880 and 2920 (inclusive). You’ll need to use the unique function (not discussed before; you can Google how to use it or ask ChatGPT, but it’s pretty self -explanatory). Note: there are eight elements of pass in this range, but only five are unique (i.e., only 5 values should be printed to the screen for this part). x <- pass[pass >= 2880 & pass <= 2920 ] x <- unique (x) x ## [1] 2890 2882 2916 2898 2891 d. What is the average value of all elements of pass that are less than 2320? You’ll need to use the mean function. av <- pass[pass < 2320 ] av <- mean (av) av ## [1] 2166.5 e. How many values of pass are at least 3995? You’ll need to use the length function. tnn <- pass[pass >- 3995 ] length (tnn) ## [1] 352 f. Some (in fact 75) of the elements of pass are multiple of 5. By using the %in% shortcut and seq , print to the screen all elements in pass that are multiples of 5,

## [1] "data.frame" "character" "factor" "numeric" "i used gpt to figure out how to use sapply function so i wouldnt have to write 4 lines" ## [1] "i used gpt to figure out how to use sapply function so i wouldnt have to write 4 lines" b. What is sum total of all values in the sections column of the data? ii <- SPOTHIT suuum <- sum (ii $ sections) suuum ## [1] 430613 c. What percentage (a number between 0-1) of songs have values of tempo that are strictly less than 80 (beats per minute)? Try using length and which to count up the number of entries in the tempo column that are strictly less than 80 then dividing by the number of rows in SPOTHIT . uu <- (ii $ tempo) uu <- uu[uu < 80 ] uu <- length (uu) OO <- nrow (SPOTHIT) uu / OO ## [1] 0.0714251 d. Print to the screen the song names in rows 29200, 35271, 37263, and 40969. SPOTHIT $ track[ c ( 29200 , 35271 , 37263 , 40969 )] ## [1] "You Belong With Me" "I Knew You Were Trouble." ## [3] "Bad Blood" "Shake It Off" e. Print to the screen the first 8 entries of the target column, which say “Yes” or “No” whether the song was considered a hit. head (SPOTHIT $ target) ## [1] Yes No No No No No ## Levels: No Yes f. Use the levels function to print to the screen the levels (possible values) of the time_signature column. Then, again using the levels function, rename them be Unknown , 1/4 , 3/4 , 4/4 and 5/4 . Then, use tail to show the values of time_signature in the last 15 rows of the data. levels (SPOTHIT $ time_signature) ## [1] "0" "1" "3" "4" "5" ts <- SPOTHIT $ time_signature levels (ts) <- c ( "Unknown" , "1/4" , "3/4" , "4/4" , "5/4" ) tail (ts, 15 )

## [1] 4/4 3/4 4/4 5/4 4/4 4/4 3/4 4/4 4/4 4/4 4/4 4/4 4/4 4/4 4/4 ## Levels: Unknown 1/4 3/4 4/4 5/4 g. Which rows (1-41106) contain the most energetic songs ( energy column), at least according to Spotify’s data? What are the song titles and artists (they are actually all by the same artist)? Hint: you can use the which command to find the row numbers that have the max value of energy , then find the track and artist in those positions. Try checking out these “songs” on Spotify; they probably aren’t what you’d expect (and any analysis that tries to determine if a song is a hit or flop should probably exclude these songs). ene <- which (SPOTHIT $ energy == max (SPOTHIT $ energy)) ene <- SPOTHIT[ene, c ( "track" , "artist" )] ene ## track artist ## 17457 Endless Rain Nataural ## 17707 Dripping Rain Nataural ## 22369 Under Shelter Rain Nataural ## 22667 Pouring Rain Nataural "chatgpt helped me with this one" ## [1] "chatgpt helped me with this one" h. What value is in the “cell” located in the 10th column and 8833rd row of SPOTHIT ? SPOTHIT[ 8833 , 10 ] ## [1] 0.264 i. Print out the entirety of the 41100th row to the screen. SPOTHIT[ 41100 ,] ## track artist uri ## 41100 Sweater Weather The Neighbourhood spotify:track:2QjOHCTQ1Jl3zawyYOpxh6 ## danceability energy key loudness mode speechiness acousticness ## 41100 0.612 0.807 10 -2.81 Major 0.0336 0.0495 ## instrumentalness liveness valence tempo duration_ms time_signature ## 41100 0.0177 0.101 0.398 124.053 240400 4 ## chorus_hit sections target ## 41100 91.20552 7 Yes j. Change the name of the column named target to Hit . Provide a frequency table of the elements in the Hit column. names (SPOTHIT)[ which ( names (SPOTHIT) == "target" ) ] <- "Hit" table (SPOTHIT $ Hit)

Note: there have previously been isolated reports of the Assignment1-OnlineSales.csv not being downloaded correctly (opening it up in Excel or Numbers shows “Unauthorized” in a single cell). We’re unsure of the root cause, but as a last resort, instead of reading in the data file from the computer (which won’t work), you can run: RETAILER <- read.csv("https://docs.google.com/spreadsheets/d/1tPaaeqn_8dJYrQ1mQwcnvfN1IIM BttXtXzJ0k6zUxp4/export?format=csv",stringsAsFactors=TRUE) Retailer <- read.csv ( "Assignment1-OnlineSales.csv" ) Retailer $ Product <- factor (Retailer $ Product) summary (Retailer) ## Product Variety Showings UnitCost ## Category10:7593 Min. : 1.00 Min. : 0.00 Min. : 0.00 ## Category9 :7064 1st Qu.: 1.00 1st Qu.: 0.00 1st Qu.: 10.70 ## Category6 :5262 Median : 3.00 Median : 0.00 Median : 19.99 ## Category8 :4976 Mean : 10.31 Mean : 11.68 Mean : 39.24 ## Category2 :3775 3rd Qu.: 7.00 3rd Qu.: 8.00 3rd Qu.: 39.48 ## Category3 :3381 Max. :763.00 Max. :1510.00 Max. :2200.00 ## (Other) :7447 ## Markup NetQuantity GrossQuantity Review ## Min. :-19.0000 Min. : -240.0 Min. : 0.0 Min. : 1.000 ## 1st Qu.: 0.5000 1st Qu.: 2.0 1st Qu.: 3.0 1st Qu.: 7.200 ## Median : 0.6100 Median : 9.0 Median : 11.0 Median : 8.500 ## Mean : 0.5385 Mean : 111.3 Mean : 137.4 Mean : 8.097 ## 3rd Qu.: 0.6600 3rd Qu.: 53.0 3rd Qu.: 66.0 3rd Qu.: 9.600 ## Max. : 1.0000 Max. :12112.0 Max. :14859.0 Max. :10.000 ## ## Text ## Length:39498 ## Class :character ## Mode :character ## ## ## ## b. Create a column in the RETAILER dataframe named Price that tabulates the typical selling price of items sold that week. Hint: a formula for the price is: 𝑃?𝑖𝑐𝑒 = 𝑈?𝑖?𝐶???(1 + 𝑀𝑎?𝑘??) Print to the screen the result of running summary on the Price column. Sanity check: you’ll find the Mean equals 58.29 from the summary. UC <- Retailer $ UnitCost Markup <- Retailer $ Markup Retailer $ Price <- UC * ( 1 + Markup) summary (Retailer $ Price)

## Min. 1st Qu. Median Mean 3rd Qu. Max. ## -552.50 16.70 30.32 58.29 60.27 3168.00 c. Sometimes items are offered on clearance to get rid of inventory. Unfortunately, when this happens, the record-keeping messes up the value of Markup , resulting in a Price that can be negative. Print to the screen the result of running subset on RETAILER , keeping only rows that have Price less than -100 and keeping only the Product, UnitCost, Markup, and Price columns. Sanity check: ensure that only 18 rows and 4 columns (5 including the row number) are printed to the screen. If this command is incorrect and the entirety of the dataframe is printed to the screen, that’ll be excessive output and the grader will stop! The first row number will be 10964 and the last row number will be 28608. th <- subset (Retailer,Price < - 100 , select = c ( "Product" , "UnitCost" , "Markup" , "Price" )) th ## Product UnitCost Markup Price ## 10964 Category6 20.80 -5.93 -102.544 ## 11126 Category8 79.97 -3.80 -223.916 ## 16653 Category6 19.50 -7.03 -117.585 ## 16952 Category10 85.00 -2.40 -119.000 ## 17533 Category1 20.00 -9.00 -160.000 ## 17583 Category6 19.50 -11.19 -198.705 ## 18308 Category6 19.50 -7.53 -127.335 ## 18457 Category6 19.50 -9.64 -168.480 ## 19338 Category10 85.00 -3.25 -191.250 ## 19354 Category1 20.00 -19.00 -360.000 ## 19472 Category10 85.00 -7.50 -552.500 ## 19527 Category1 20.00 -19.00 -360.000 ## 19945 Category1 20.00 -9.00 -160.000 ## 24615 Category9 65.00 -2.61 -104.650 ## 25094 Category2 27.65 -4.92 -108.388 ## 27664 Category9 35.00 -4.00 -105.000 ## 28523 Category9 65.00 -4.42 -222.300 ## 28608 Category9 35.00 -4.83 -134.050 "chatgpt helped me figure out how to keep only the indicated columns" ## [1] "chatgpt helped me figure out how to keep only the indicated columns" d. Let’s “fix” the issues with Price by throwing away any rows with values 0 or less and restricting our analysis to items in product categories 1, 3, and 5. Create a subset named SUB whose rows have strictly positive values of price and whose product categories are 1, 3, or 5. Once the subset has been created, remove the Text column. Then, remove unused levels of Product (since only 3 of the 10 possible categories are being examined) using a droplevels command.

you’ll see 10 bars (but the only ones with any height will be the ones for the three categories in the subset). • Make a boxplot of the unit costs • Bonus: Create a scatterplot showing the relationship between Price (horizontal axis) and Review (vertical axis) and add a trend line. You will need to Google / ChatGPT this one as it has not been covered yet. Looking up how to do stuff is a good skill to acquire! hist (SUB $ Price, breaks = seq ( from= 0 , to= 450 , by= 10 ),) hist ( log10 (SUB $ Price), breaks= seq ( from= - 2 , to= 3 , by= . 1 ))

barplot ( ( table (SUB $ Product)))