Assignment-1

2B03 Assignment 1 Descriptive Statistics (Chapters 1 & 2) Matthew Musulin 400329990 Due Thursday September 23 2021 Instructions: You are to use R Markdown for generating your assignment output file. You begin with the R Markdown script downloaded from A2L, and need to pay attention to information provided via introductory material posted to A2L on working with R and R Markdown. Having downloaded all necessary files, placed them in the same folder/directory, and added your answers to the R Markdown script, you then are to generate your output file using “Knit to PDF” and, when complete, upload both your R Markdown file and your PDF file to the appropriate folder on A2L. 1. Define the following terms in a sentence (or short paragraph) and state a formula if appropriate (this question is worth 5 marks). i. Categorical Data Categorical Data represents types of data that can be divided into categories or groups. ii. Frequency Distribution Frequency Distribution is a function that displays the number of observations within a givin interval. iii. Sturgess’s Rule Sturgess’s Rule is a rule used for determining the desirable number of classes. Where K is the interger rounded to the closest whole number. 1 + 3 . 3 log 10 n iv. Cross Tabulations Cross Tabulations are tabular summaries for two variables. v. Sample Median The Sample Median is the middle value after putting all observations in ascending order. X ( n/ 2) + X ( n/ 2+1) 2 2. Consider the following dataset on the final grade received in a particular course ( grade ) and attendance ( attend , number of times present when work was handed back during the semester out of a maximum of six times). Note that R has the ability to read datafiles directly from a URL, so here (unlike the odesi data that you manually retrieve) you do not have to manually download the data providing you are connected to the internet (this question is worth 8 marks). course <- read.table( "https://socialsciences.mcmaster.ca/racinej/files/attend.RData" ) attach(course) 1

i. Create a scatterplot of the data with attend on the horizontal axis and grades on the vertical axis via the command plot(attend,grade) 0 1 2 3 4 5 6 40 60 80 100 attend grade Do you see any pattern present in the data? If so describe it in your own words. In the scatterplot, students that attended more classes generally had a higher final grade, inversely, students that attended less classes had a lower final grade. ii. Construct the average grades for persons attending 0 times, and then repeat for those attending 1 time, 2 times, and so on through 6 times using something like mean(grade[attend== 0 ]) ## [1] 43.5 mean(grade[attend== 1 ]) ## [1] 56.83333 mean(grade[attend== 2 ]) ## [1] 51.14286 mean(grade[attend== 3 ]) ## [1] 66 mean(grade[attend== 4 ]) ## [1] 74.88889 mean(grade[attend== 5 ]) 2

fivenum(hs) ## [1] 3.21 17.31 22.00 29.86 108.13 fivenum(ba) ## [1] 3.30 24.53 34.62 45.99 107.39 ii. What can you say about relative wages of high school and university graduates? University Graduates are generally paid more than High School Graduates according to the median. HS=22.00 and UG=34.62 iii. Using Sturges’ rule, how many classes would you construct for the hs and ba wage data (hint - length() gives you the length of the vector, log10() may also be useful, so something like round(1+3.3*log10(length(hs))) might do the trick for the hs data at least)? round( 1+3.3 *log10(length(hs))) ## [1] 14 round( 1+3.3 *log10(length(ba))) ## [1] 14 iv. Plot histograms for the hs and ba data on separate graphs (hint: hist() ). hist(hs) Histogram of hs hs Frequency 0 20 40 60 80 100 0 500 1500 2500 hist(ba) 4

Histogram of ba ba Frequency 0 20 40 60 80 100 0 500 1000 1500 2000 v. Do the number of classes correspond to Sturges’ rule? No, neither the histogram of hs or ba corresponds to the Sturges’ rule which calculated 14 classes for both hs and ba. vi. Plot density curves for the hs and ba data on the same graph and add a legend (hint: first use something like plot(density(...),col="blue",lty=1) (you need to fill in (...) parts with the name of your data object, e.g., hs etc.) then lines(density(...),col="red",lty=2) , then see the help page by typing ?legend in the console pane. Note that you can add a legend using something like plot(density(hs), col= "blue" , lty= 1 ) lines(density(ba), col= "red" , lty= 1 ) legend( "topright" ,c( "High School" , "University" ), lty= c( 1 , 1 ), col= c( "blue" , "red" ), bty= "n" ) 5

iii. How many classes would you create if you used Sturges’ rule? n = 12 k = round( 1+3.3 *log10( 12 )) k ## [1] 5 iv. What are the class widths and class boundaries based on your answers to the previous two questions, using Sturges’ rule, the sample minimum as the first lower class boundary, and the sample maximum as the last upper class boundary? Width = (max(profits)-min(profits))/k Width ## [1] 4.171 v. Complete the table below showing the absolute frequency, relative frequency, cumulative frequency, and cumulative relative frequency for the above data. For this question you will need to do some manual data entry in the table skeleton provided below after you have figured out what the counts are based on your answers to the previous set of questions. In particular, you are to use Sturges’ rule (above) to obtain the desired number of classes, and use the range of the data (above) when constructing your class boundaries (note that you need to have a blank line between each new row that you add to the table, and the last class must be closed at the right - this question is worth 8 marks). Class Absolute Frequency Relative Frequency Cumulative Absolute Frequency Cumulative Relative Frequency [0.145,4.316) 2 0.1666 2 0.1666 [4.316,8.487) 4 0.3333 6 0.5 [8.487,12.658) 2 0.1666 8 0.6666 [12.658,16.829) 3 0.25 11 0.9166 [16.829,21] 1 0.0833 12 1 5. Since we use the summation operator ( Σ n i =1 ) often in class, let’s make sure we understand how to calculate objects that can be expressed succinctly using this operator. i. Care must be exercised when expanding certain sums and quantities. Let the sample size be n = 3 , and let X 1 = 1 , X 2 = − 1 , and X 3 = 3 . Demonstrate in R that it is generally not true that QQQQQQQ n i =1 X 2 i = ( QQQQQQQ n i =1 X i ) 2 (this question is worth 2 marks). data <- c( 1 , - 1 , 3 ) # Create a vector to hold data. data ## [1] 1 -1 3 sum = 0.0 # Calculate the sum of Xi ' s. sum2 = 0.0 for (i in 1 : 3 ) { sum = sum + data[i] } cat(sprintf( "Sum of Xi ' s is %.2f \n " , sum)) ## Sum of Xi ' s is 3.00 7

for (i in 1 : 3 ) # Calculate the sum of Xi ' s squared. { sum2 = sum2 + data[i]*data[i] } cat(sprintf( "Sum of Xi ' s squared is %.2f \n " , sum2)) ## Sum of Xi ' s squared is 11.00 sq_sumx = sum*sum # Calculate the square of the sum of Xi ' s. cat(sprintf( "The square of the sum of Xi ' s %.2f \n " , sq_sumx)) ## The square of the sum of Xi ' s 9.00 cat( "From the math above you can see that the sum of Xi ' s squared is not equal to the square of ## From the math above you can see that the sum of Xi ' s squared is not equal to the square of t ii. Using the same data as in the previous question, compute the sample mean ¯ X = QQQQQQQ n i =1 X k /n then compute the sample standard deviation ˆ σ = rrrrrrr QQQQQQQ n i =1 ( X i − ¯ X ) 2 / ( n − 1) in two ways: longhand (you can use R and use longhand notation, e.g., X[1], X[2], and X[3] or 1, -1, and 3, whichever you prefer), then using R functions such as mean() and sd() (this question is worth 2 marks). mean = sum/ 3 # Calculate the mean longhand. cat(sprintf( "Mean: %.2f \n " , mean)) ## Mean: 1.00 sd = 0.0 var = 0.0 for (i in 1 : 3 ) { var = var + ((data[i] - mean)ˆ 2 )/ 2 } sd = sqrt(var) # Calculate the standard deviation longhand. cat(sprintf( "Standard Deviation: %.2f \n " , sd)) ## Standard Deviation: 2.00 mean2 = mean(data) # Calculate the mean using R function. cat(sprintf( "Mean using mean() %.2f \n " , mean2)) ## Mean using mean() 1.00 sd2 = sd(data) # Calculate the standard deviation using R function. cat(sprintf( "Standard Deviation using sd() %.2f \n " , sd2)) ## Standard Deviation using sd() 2.00 iii. Express QQQQQQQ n i =1 K , where K is a constant (i.e., a number that does not change hence has no subscript i ), in terms of n and K only (Hint - a constant does not have a subscript as it does not change with i , but it is being added/summed, so type out a string of n constants etc.). Then for K = 3 and n = 5 determine QQQQQQQ n i =1 K using your result purely using n and K (i.e., without a summation sign - this question is worth 2 bonus marks, and you do not use R, rather use your powerful sense of logic and type out your answer with an explanation). 8