Q1

Find the current packages in the environment; list out the objects in each package; count them

lapply(lapply(search(), ls), length)

Q2

# r : the rate
# L : loan
# Y : Years

Y <- seq(10, 30, 5)
L <- c(5000000, 10000000, 15000000)
r <- c(0.02, 0.05, 0.07)

#make all combinations into data.frame
df <- expand.grid(Year = Y, Loan = L , rate = r)
df$pay <- apply(df, 1, function(x) { x[2]*(x[3]/(1-(1+x[3])^(-x[1]*12))) })
df
   Year    Loan rate     pay
1    10 5.0e+06 0.02  110240
2    15 5.0e+06 0.02  102914
3    20 5.0e+06 0.02  100870
4    25 5.0e+06 0.02  100264
5    30 5.0e+06 0.02  100080
6    10 1.0e+07 0.02  220481
7    15 1.0e+07 0.02  205827
8    20 1.0e+07 0.02  201741
9    25 1.0e+07 0.02  200527
10   30 1.0e+07 0.02  200160
11   10 1.5e+07 0.02  330721
12   15 1.5e+07 0.02  308741
13   20 1.5e+07 0.02  302611
14   25 1.5e+07 0.02  300791
15   30 1.5e+07 0.02  300241
16   10 5.0e+06 0.05  250719
17   15 5.0e+06 0.05  250038
18   20 5.0e+06 0.05  250002
19   25 5.0e+06 0.05  250000
20   30 5.0e+06 0.05  250000
21   10 1.0e+07 0.05  501437
22   15 1.0e+07 0.05  500077
23   20 1.0e+07 0.05  500004
24   25 1.0e+07 0.05  500000
25   30 1.0e+07 0.05  500000
26   10 1.5e+07 0.05  752156
27   15 1.5e+07 0.05  750115
28   20 1.5e+07 0.05  750006
29   25 1.5e+07 0.05  750000
30   30 1.5e+07 0.05  750000
31   10 5.0e+06 0.07  350104
32   15 5.0e+06 0.07  350002
33   20 5.0e+06 0.07  350000
34   25 5.0e+06 0.07  350000
35   30 5.0e+06 0.07  350000
36   10 1.0e+07 0.07  700209
37   15 1.0e+07 0.07  700004
38   20 1.0e+07 0.07  700000
39   25 1.0e+07 0.07  700000
40   30 1.0e+07 0.07  700000
41   10 1.5e+07 0.07 1050313
42   15 1.5e+07 0.07 1050005
43   20 1.5e+07 0.07 1050000
44   25 1.5e+07 0.07 1050000
45   30 1.5e+07 0.07 1050000

Q3

(1)

dta3 <- read.table("hs0.txt", h = T)
mapply(function(i,j) { t.test(dta3[,i], dta3[,j])$p.value },
       rep(7:11, each = 5),7:11) %>% matrix(5,5)
       [,1]   [,2]   [,3]   [,4]   [,5]
[1,] 1.0000 0.5813 0.6728 0.7572 0.8677
[2,] 0.5813 1.0000 0.8903 0.3776 0.7150
[3,] 0.6728 0.8903 1.0000 0.4516 0.8118
[4,] 0.7572 0.3776 0.4516 1.0000 0.6378
[5,] 0.8677 0.7150 0.8118 0.6378 1.0000

若p值大於0.05，則無顯著差異。

(2)

yvar <- names(dta3)[7:11]

lapply(yvar, function(y) {
  oneway.test(substitute(i ~ race, list(i = as.name(y))), data = dta3)$p.value})
[[1]]
[1] 0.0008795

[[2]]
[1] 0.0003045

[[3]]
[1] 6.491e-05

[[4]]
[1] 1.218e-05

[[5]]
[1] 0.04209

p值均小於0.05，各種族平均分數有差異。

(3)

tidy(lm(math ~ ses - 1, dta3), conf.int = TRUE) %>%
  mutate(term = ifelse(term == "seshigh", 
                       "High", ifelse(term == "sesmiddle", "Middle", "Low"))) %>%
  ggplot(aes(term, estimate)) +
  geom_pointrange(aes(ymin = conf.low, ymax = conf.high)) +
  coord_flip() +
  labs(x = "Ses", y = "Estimated mean") +
  theme_bw()

Q4

circle <- function(x) {sqrt(1^2 - x^2)}
N <- 100000; x <- runif(N, 0, 1); y <- runif(N, 0, 1)

# the area of the green part
A <- sum(circle(runif(N, 0, 1)) > runif(N, 0, 1))/N*(1*1)

# circular area = πr^2; π = area/r^2
A*4/1^2
[1] 3.15
curve(circle, 0, 1, ylim = c(0, 1), ylab = "f(x)")
points(x, y, col = ifelse(circle(x) > y, 3, 2), pch = ".")