Question 1.

> x  <- c(5, 10, 15, 20, 25, 30)
> y  <- c(-1, NA, 75,  3,  5,  8)
> z  <- c(5)

Question 2

> a <- x*z
> b <- y*z
> print(a)
[1]  25  50  75 100 125 150
> print(b)
[1]  -5  NA 375  15  25  40

Question 3

> library(haven)
> library(readr)
> setwd("C:/Users/drayr/OneDrive/Desktop/DEM Fall 2020/STATS 7273")
> stata <- read_dta("Data/stata_PSID_w1.dta")
> ##view the data*
> View(stata)
> ##select variables into a new data set*
> assignment1<-subset(x=stata,select=c("id","age","marpi","adjwlth2","educ","pubhs","h_race_ethnic_new","race5"))
> dim(assignment1)
[1] 131361      8

Question 3.1

> dim(assignment1)
[1] 131361      8
> str(assignment1)
tibble [131,361 x 8] (S3: tbl_df/tbl/data.frame)
 $ id               : num [1:131361] 4003 4003 4003 4003 4003 ...
  ..- attr(*, "format.stata")= chr "%9.0g"
 $ age              : num [1:131361] 49 51 53 55 57 59 47 49 51 53 ...
  ..- attr(*, "label")= chr "Age of respondent"
  ..- attr(*, "format.stata")= chr "%8.0g"
 $ marpi            : num [1:131361] 1 1 1 1 1 1 0 0 0 0 ...
  ..- attr(*, "label")= chr "Marital pairs indicator"
  ..- attr(*, "format.stata")= chr "%8.0g"
 $ adjwlth2         : num [1:131361] 113 119 116 129 112 ...
  ..- attr(*, "label")= chr "Wealth (including home equity) in 1000s of yr 2000 "
  ..- attr(*, "format.stata")= chr "%9.0g"
 $ educ             : num [1:131361] 9 9 9 9 9 10 12 12 12 12 ...
  ..- attr(*, "label")= chr "Years completed education"
  ..- attr(*, "format.stata")= chr "%9.0g"
 $ pubhs            : num [1:131361] 0 0 0 0 0 0 0 0 0 0 ...
  ..- attr(*, "label")= chr "1 = lives in public housing"
  ..- attr(*, "format.stata")= chr "%8.0g"
 $ h_race_ethnic_new: chr [1:131361] "NL White" "NL White" "NL White" "NL White" ...
  ..- attr(*, "label")= chr "Race/ethnicity updated codes (5/26/14)"
  ..- attr(*, "format.stata")= chr "%16s"
 $ race5            : dbl+lbl [1:131361] 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, ...
   ..@ label       : chr "Race/ethnicity updated codes (5/26/14)"
   ..@ format.stata: chr "%16.0g"
   ..@ labels      : Named num [1:5] 1 2 3 4 5
   .. ..- attr(*, "names")= chr [1:5] "Latino- Any Race" "NL Asian" "NL Black" "NL Other" ...
> print(assignment1)
# A tibble: 131,361 x 8
      id   age marpi adjwlth2  educ pubhs h_race_ethnic_new        race5
   <dbl> <dbl> <dbl>    <dbl> <dbl> <dbl> <chr>                <dbl+lbl>
 1  4003    49     1     113.     9     0 NL White          5 [NL White]
 2  4003    51     1     119.     9     0 NL White          5 [NL White]
 3  4003    53     1     116.     9     0 NL White          5 [NL White]
 4  4003    55     1     129.     9     0 NL White          5 [NL White]
 5  4003    57     1     112.     9     0 NL White          5 [NL White]
 6  4003    59     1     104.    10     0 NL White          5 [NL White]
 7  4004    47     0     493     12     0 NL White          5 [NL White]
 8  4004    49     0     447.    12     0 NL White          5 [NL White]
 9  4004    51     0     386.    12     0 NL White          5 [NL White]
10  4004    53     0    2493.    12     0 NL White          5 [NL White]
# ... with 131,351 more rows

Question 3.2

> hist(assignment1$race5)

Question 3.3

> mean(assignment1$adjwlth2,na.rm = T)
[1] 187.1656
> median(assignment1$adjwlth2,na.rm = T)
[1] 32.804

Question 3.4

> min(assignment1$age)
[1] 1
> max(assignment1$age)
[1] 999
> IQR(assignment1$age)
[1] 33
> mean(assignment1$age)
[1] 32.02676
> median(assignment1$age)
[1] 29

Question 3.5

> table(assignment1$pubhs)

     0      1 
124366   6961 
> # 6961 people received public assistance in the form of public housing.
> newdata<- subset(x=assignment1,pubhs==1, select = all())
> table(newdata$race5)

   1    2    3    4    5 
 366   27 5472  114  982 
> table(newdata$h_race_ethnic_new)

Latino- Any Race         NL Asian         NL Black         NL Other 
             366               27             5472              114 
        NL White 
             982 
> # 366 Latinos received public assistance in the form of public housing.
> 
> newdata$race5<-factor(newdata$race5,
+                       levels = c(1,2,3,4,5),
+                       labels=c("Latino","Asian","Black","Other","White"))
> prop.table(table(newdata$race5))

     Latino       Asian       Black       Other       White 
0.052578652 0.003878753 0.786093952 0.016376957 0.141071685 
> table(newdata$race5)

Latino  Asian  Black  Other  White 
   366     27   5472    114    982 

Question 3.6

> # I would be interested to know information about individuals' parents characteristics such as, Parent education attainment and Parent level of income and Geographic information to compare demographic structures across geographies.