FS1 - Chapter 4 onwards.Rmd

---
title: 'FS1 - Chapter 4 onwards'
date: "`r Sys.Date()`"
output:
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---
```{r}
rm(list=ls())
library(magrittr, quietly = T)
library(tidyverse, quietly = T)
library(data.table)
library(dplyr, quietly=T)
library(ggplot2)
theme_simple <- theme_bw() + theme(legend.position="none")
dnbinomw<- function(trials, size, prob){
    # Negative binomial distribution
    # How many trials (x) needed for a fixed number of successes (size)
    # Where probability of success is constant (p)
    return(prob*choose(trials-1, size-1)*(prob)^(size-1)*(1-prob)^(trials-size))
}

dgeomw <- function(trials, prob){
    return(prob*(1-prob)^(trials-1))}
```

# Chapter 4 


### Preliminary questions

```{r}
dpois(x=0:3, lambda=4) %>% sum()
dpois(x=7:10, lambda=4) %>% sum()

library(data.table)
library(tidyverse)
options(scipen = 999)
dt <- data.table(x=c(0:25),
           p=dbinom(x=0:25, size=25, prob=0.2),
           cp=dbinom(0:25, size=25, prob=0.2) %>% cumsum())

dt[cp>0.9, x][1]
```

### Section 4.1

```{r}
dpois(x=0:1, lambda=6) %>% sum() # less than 0.05 so accept at 5% level.
1 - dpois(x=0:5, lambda=2.5) %>% sum() # this is greater than 2%, but two sided so 1% level. 
1 - dpois(x=0:16, lambda=10) %>% sum() # 0.027, less than 0.05
```
```{r}
# Question 7
1- sum(dpois(0:2, lambda=1)) # 5% answer is no
# Question 8
sum(dpois(0:6, lambda=9)) # 0.206, so no
# Question 9
sum(dpois(0:4, lambda=6*1.25)) # 0.132, so no
# Question 10
sum(dpois(0:3, lambda=6.9)) # 0.0871296 > 0.025 so nope. 

# Question 11
# day = 0.3; 20 = 6
dpois(x=5, lambda=6) # 0.161
dpois(x=0:8, lambda=6) %>% sum() # 0.8472
# new lambda; 30 days = 5. Old lambda would be 9 breakdowns.
sum(dpois(0:5, lambda=9)) # insufficient evidence

# Question 12 
# Old lambda is 2.25 per week, or 9 every 4 weeks
# H0: lambda = 9, H1: lambda < 9
sum(dpois(x=0:4, lambda=9)) # 0.055

# Question 13
# lambda = 1.5 a week
# 6 week period there are 13 breakdowns; two sided test
# H0: lambda = 9, H1: lambda != 9
1-sum(dpois(0:12, lambda=9)) # 0.1242266 < 0.025

# Question 14
# 1% defect rate, 1000 in a box. X ~ B(1000, 0.01) ~ Po(10)
dpois(x=9, lambda=10) # 0.12511
dpois(x=0:7, lambda=10) %>% sum() # 0.22
# Approximation OK as n large, p small, np>=10. Also CAR. 
# H0: Lambda = 10, H1: Lambda < 10
dpois(x=0:5, lambda=10) %>% sum() # 0.067
dbinom(x=0:5, size = 1000, prob = 0.01) %>% sum() # 0.066
```
### Section 4.2
```{r}
# Example 4
options(scipen = 999)
dt <- data.table(x = 0:9, dpois(x=0:9, lambda=9), cumsum(dpois(x=0:9, lambda=9)))
dt[V3<0.05, max(x)] # for a one sided test, sufficient evidence if 3 or fewer homes. 
dt[V3<0.05] # P[X\le 3] = 0.0212

# Example 5 
# H0: Lambda = 7, H1: Lambda > 7 
dt <- data.table(x=0:20, y = dpois(0:20, lambda=7), z = cumsum(dpois(0:20, lambda=7)))
dt # 12 goals or more is sufficient to reject the Null! Incorrect; 13 or more because P[X\ge c] = 1 - P[X \le c-1]
dt[x==11]

 # Example 6
# 0.325 per minute, therefore 20 minutes has 6.5
dt <- data.table(x=0:12, y = dpois(0:12, lambda=6.5), z = cumsum(dpois(0:12, lambda=6.5)))
# You'll reject the Null if one or zero calls comes in. 
# You'll reject the null if 12 or more calls come in, but not for reasons you'd think. For upper tails you have to add one! Why? 1 - prob.
dt[x<2, sum(y)] + 1- dt[x==11,z] # Notice y, not z for the first.

# Question 1
dt <- data.table(x = 0:4, y=dpois(0:4, lambda=5.5), z=cumsum(dpois(0:4, lambda=5.5)))
dt[z<0.05] # Critical region is with x \in [0,1]
dt <- data.table(x=0:15, y=dpois(0:15, lambda=8), z=dpois(0:15,lambda=8) %>% cumsum())
dt[z>0.99] # X greater than 16 
dt <- data.table(x=0:15, y=dpois(0:15, lambda=4), z=cumsum(dpois(0:15, lambda=4)))
dt[z>0.95] # X greater than 8+1 = 9. 

# Question 2
dt <- data.table(x=0:20, y=dpois(0:20, lambda=8), z=cumsum(dpois(0:20, lambda=8)))
dt[z>0.95] # 13+1 days, so 14. 

# Question 7
dt <- data.table(x=0:13, y=dpois(0:13, lambda=4), z=cumsum(dpois(0:13, lambda=4)))
dt # 0 and 9

dt <- data.table(x=0:30, y = dpois(0:30, lambda=9.5), z = cumsum(dpois(0:30, lambda=9.5)))
dt[, a:=z-0.975]
dt # c1 = 3, c2 = 16 + 1 = 17

# Question 8 
dt <- data.table(x=0:20, y=dpois(0:20, lambda=7.5), z = cumsum(dpois(0:20, lambda=7.5)), 
                 a = cumsum(dpois(0:20, lambda=7.5))-0.025, 
                 b = cumsum(dpois(0:20, lambda=7.5)) - 0.975)
dt # c1 = 2, c2 -1 = 13, so c2 = 14

dpois(x=0:2, lambda=7.5) %>% sum() + (1 - dpois(x=0:13, lambda=7.5) %>% sum())
1 - dpois(3:13, lambda=7.5) %>% sum()  
# 11 is within the acceptance region

# Question 9 
dt <- data.table(x=0:15, z = dpois(0:15, lambda=5) %>% cumsum())
dt %>% mutate(a = z - 0.05, b = z-0.95) %>% tail(10)
# c1 = 1, c2
1 - sum(dpois(x=2:8, lambda=5))

# Question 10
dt <- data.table(x=0:20, z = cumsum(dpois(0:20, lambda=9)))
dt %>% mutate(a = z-0.025, b = z - 0.975) %>% as.data.table() # c1 = 3 , c2 -1 = 15
1 - sum(dpois(4:15, lambda=9)) # 0.0433


# Question 11
temp <- NULL
for(i in seq(3, 20, by=0.5)){
temp <-rbind(temp, c(i,1 - sum(dpois(3:14, lambda= i))))
}
temp # lambda = 6.5?

opt <- function(a){abs(sum(dpois(x=0:2, lambda=a))+ 1-sum(dpois(x=0:14, lambda=a))- 0.05)}
# opt <- function(a){(sum(dpois(x=0:2, lambda=a))+ 1-sum(dpois(x=0:14, lambda=a)- 0.05)^2)}
optimise(f = opt, interval = c(0,20)) 
dpois(0:2, lambda=6.4) %>% sum() + 1-dpois(0:14, lambda=6.4) %>% sum()
dpois(0:2, lambda=8) %>% sum() + 1 - dpois(0:14, lambda=8) %>% sum()
```
Challenge:
```{r}
library(data.table)
library(magrittr)
library(dplyr)
options(scipen = 999)
# Question 12
dt <- data.table(x=0:20, z = cumsum(dbinom(0:20, size=30, prob=0.5)))
dt[z<0.02] # 8 

dt <- data.table(x=1:10, z=cumsum(dgeomw(trials=1:10, prob=0.5)))
dt[z>0.975] # 6 + 1 = 7
# Challenge
dt <- data.table(x=3:20, y=cumsum(dnbinomw(trials=3:20, size = 3, prob=0.18)))
dt[y<0.05] # x<= 5

dnbinomw(trials=3:5, size=3, prob=0.18) %>% sum()
```
# Chapter 5
```{r}
# Prior knowledge
1-pnorm(q=115, mean=120, sd=8) # 0.7340145
pnorm(q=130, mean=120, sd=8) - pnorm(q=120, mean=120, sd=8) # 0.3944
qnorm(p=0.25, mean=120, sd=8) # 114.6

1-dnbinomw(trials = 5:11, size = 5, prob=0.5) %>% sum() #27.4%


1-pnorm(q=1.5) # 0.0668072

z <- (40.5 - 40)/(2.25/4)^0.5
1-pnorm(z) #0.252
z <- (40.5 - 40)/(2.25/49)^0.5
1 - pnorm(z) # 0.009815329

# Question 5
1-pnorm(0.5/(1/12)^0.5) # 0.04163226
pnorm((20/7 - 3.5)/(1/12)^0.5) # 0.01297623

# Question 6
pnorm(1/(35/(12*750))^0.5)
pnorm(1/(35/(12*30))^0.5)
pnorm(-0.5/(1/6)^0.5)

# Question 7
1-pnorm(3, mean=2.85, sd=(1011/40000)^0.5) # 0.1727; forgot the zero
# Question 8 
dt <- data.table(x=seq(1,3000, by=1), 
                 z = pnorm(q=3.6, mean=3.5, sd=sqrt(35/(12*seq(1, 3000, by=1)))) - 
                     pnorm(q=3.4, mean=3.5, sd=sqrt(35/(12*seq(1, 3000, by=1)))))
dt[z>0.99][1,] # 1936 or more. 

# Question 9 
pnorm(q=30000, mean=28500, sd=sqrt((6800^2)/15)) - 
pnorm(q=25000, mean=28500, sd=sqrt((6800^2)/15))

# Question 10
dt <- data.table(x=1:100, z= pnorm(q=0.5*sqrt(1:100)/2.5)-pnorm(q=-0.5*sqrt(1:100)/2.5))
dt[z>=0.94]
```
## Chapter 5.2
```{r}
# Example 4
dnbinomw(trials=12, size=10, prob=2/3)
pnorm(0.5/sqrt(0.3))
# Question 1 
sum(dpois(0:25, lambda=30))# 0.2083574
pnorm(2.5, mean=3, sd=sqrt(0.3)) # 0.1806552
# Question 2 
1-pnorm(1)
# Question 3 
pnorm(2.4, mean=2, sd=sqrt(0.08))
# Question 4
pnorm(-sqrt(2))
# Question 5
dpois(4, lambda=3)
# Question 6 
pnorm(-1)
#Question 6
0.8/(0.2)^2
1-pnorm(0.5)
# Question 7
dnbinomw(trials=35, size=10, prob=1/3)
pnorm(5/sqrt(3))
# Question 8 
1-sum(dpois(0:350, lambda=300))
# Question 9 
pnorm(-sqrt(3))
```
Mixed exercises:
```{r}
#Question 1
1-pnorm(2)
# Question 2
(c(1,2,4,5,7,8) * 1/6 )%>% sum()
(c(1,2,4,5,7,8)^2 * 1/6) %>% sum() - ((c(1,2,4,5,7,8) * 1/6 )%>% sum())^2
pnorm(-0.5/sqrt(6.25/20)) # 0.1855467
# Question 3
options(scipen = 999)
dt <- data.table(x=1:100, z=pnorm(-sqrt(1:100)))
dt[z<0.05]
# Question 4
1-pnorm(2, mean=5/3, sd = sqrt(25/(18*20)))
# Question 5
dpois(3, lambda=2)
1-pnorm(25/24, mean=1, sd=sqrt(1/24)) 
# Question 6 
1-dnbinomw(trials=10:24, size=10, prob=1/2) %>% sum()
1-pnorm(2.4, mean=2, sd=sqrt(0.2))
# Question 7
1-pnorm(59, mean=60, sd=sqrt(25/48))
pnorm(5, mean=4.8, sd=sqrt(4.32/30))
# Question 8 
pnorm(5, mean=4.9, sd=sqrt(0.8^2/100))
# Question 9 
dt <- data.table(x=1:100, z=1-pnorm(42, mean=40, sd=sqrt(9/1:100)))
dt[z<0.05]
# Question 10
1-pnorm(37, mean=35, sd=sqrt(0.45))
# Question 11
pnorm(0.57, mean=0.6, sd=sqrt(0.24/500)) + (1-pnorm(0.63, mean=0.6, sd=sqrt(0.24/500))) # these are the same!
dt <- data.table(x=1000:2000, z=pnorm(0.57, mean=0.6, sd=sqrt(0.24/1000:2000)))
dt[z<0.025]
```

```{r}
dt <- data.table(x=4:33, z=dnbinomw(size=4, prob=0.12, trials=4:33) %>% cumsum())
dt[z<0.025]
```
# Chapter 6
```{r}
# Prior knowledge
# Question 1 
6*exp(-5)
sum(dpois(0:1, lambda=5))
# Question 2
(2/3)^5
1-dgeomw(trials=1:5, prob=1/3) %>% sum()
# Question 3 
dbinom(x=70:100, size = 100, prob=0.6) %>% sum() # Actual probability
1-pnorm(q=69.5, mean=60, sd=sqrt(24)) # Normal approximation
```
```{r}
# Example 1

dt <- data.table(n=2:8, 
                 OB=c(12,15,22,41,33,21,16), 
                 OM=c(6,12,21,37,35,29,20), 
                 E =c(10,20,30,40,30,20,10))

names <- c("OB","OM")
dt[,paste0(names, "E"):=(.SD - E)^2/E, 
   .SDcols=names]
dt[,apply(X = .SD, FUN = function(X){sum(X)}, MARGIN=2),
   .SDcols=paste0(names,"E")]

dt[,paste0(names,"E2"):=(.SD)^2/E, .SDcols=names]
dt[,.(sum(OBE2) - sum(OB), sum(OME2) - sum(OM))]

# Mel's goodness of fit is higher, so more likely to be biased. 
```

```{r}
# Question 6.1.2
library(data.table)
dt <- data.table(n=1:6, O=c(27,33,31,28,34,27))
dt[,E:=sum(O)/6]
dt[,sum(((O - E)^2)/E)] # 1.6
dt[,sum((O^2)/E)-sum(O)] # 1.6

# Question 6.1.3
library(magrittr)
dt <- data.table(year=c(7:11), O=c(190, 145, 145, 140, 130)) %>% dplyr::mutate(E=sum(O)/length(7:11)) %>% as.data.table()
dt[,sum(((O-E)^2)/E)]
dt[,sum((O^2)/E) - sum(O)]

# 6.1.4
dt <- data.table(O=c(117,43), E=c(0.75*160 , 0.25*160))
dt[,sum(((O-E)^2)/E)] # 0.3

# 6 .1.5
dt <- data.table(O=c(28,22), E1=c(25,25), E2=c(30,20))
dt[,.(sum((O-E1)^2/E1),sum((O-E2)^2/E2))] # Smaller for 60%
```
```{r}
qchisq(p=0.95, df=5)
qchisq(p=0.05, df=3)
qchisq(p=0.90, df=4)

qchisq(p=0.05, df=2) # Pr[Chi2 > q] = 0.95
qchisq(p=0.01, df=4) # Pr[Chi2>q] = 0.99

# 6.2.1
# 6 degrees of freedom
# 6.2.2 
qchisq(p=0.95, df=5) # 11.0705
# 6.2.3
qchisq(p=0.95, df=5)
qchisq(p=0.99, df=8) # 20.09
qchisq(p=0.9, df=10) # 15.98718
# 6.2.4
qchisq(p=0.95, df=10) # 18.30704
# 6.2.5
qchisq(p=0.9, df=8)
# 6.2.6
qchisq(p=0.01, df=8) # 1.646
# 6.2.7
qchisq(p=0.05, df=5) 
# 6.2.8
qchisq(p=0.05, df=12)
qchisq(p=0.95, df=12)
# 6.2.9
dt <- data.table(x=1:5, O=c(24,12,6,6,2), E=50*dgeomw(trials = 1:5, prob = 0.5))
dt
# dt[x>=3] <- dt[x>=3] %>% mutate(O=sum(O), E=sum(E)) %>% as.data.table()
# dt %>% slice(1:3)
dt # We see that the bottom two cells need combining. 
dt <- data.table(x=1:3, O=c(24,12,14), E=50*c(dgeomw(trials = 1:2, prob = 0.5), sum(dgeomw(trials=3:5, prob=0.5))))
dt[,sum((O-E)^2/E)] < qchisq(p=0.99, df=2) 
```

```{r}
# 6.3.1
dt <- data.table(a=1:6, b=c(16,11,13,15,8,9), c=rep(12,6))
#H0: b ~ U_d(1,6)
dt[,sum((b-c)^2/c)] # 4.3333
qchisq(p=0.95, df=5) # 11.0705 
# Insufficient evidence to reject the Null of uniform. 

# 6.3.2
dt <- data.table(a=1:0, O=c(15,105), E=120*c(0.2, 0.8))
dt[,sum((O-E)^2/E)] # 4.21875
qchisq(p=0.95, df=1) # 3.841459

# 6.3.3
dt <- data.table(a=0:2, O=c(4,73, 23), E=c(0.1, 0.6, 0.3)*100)
dt[,sum((O-E)^2/E)] < qchisq(p=0.95, df=0.975)

#6.3.4
dt <- data.table(a=0:4, O=c(45,19,11,8,17), E=c(55,20,10,7,8))
dt[,sum((O-E)^2/E)]
qchisq(p=0.95, df=4)

# 6.3.5
dt <- data.table(O=c(7286, 9304, 32121, 112535, 244472, 281942), E=c(0.013, 0.015, 0.05, 0.165, 0.357, 0.4)*687660)
dt[,sum((O-E)^2/E)]
qchisq(p=0.95, df=5) # made an error here, it is the number of cells!
```
## Chapter 6.4
```{r}
# Example 8
dt <- data.table(O=c(11,8,8,7,8,9,12,9,13,15))
dt <- dt %>% mutate(E=sum(O)/length(O))
dt[,sum((O-E)^2/E)] # 6.2
qchisq(p=0.95, df=9) # 16.9

# Example 9

dt <- data.table(a=0:8, O=c(12,28,28,17,7,4,2,2,0))
dt <- dt %>% mutate(E=dbinom(0:8, size=10, prob=0.2)*sum(O))

dt <- dt[a<=3] %>% as.matrix() %>% rbind(colSums(filter(dt, a>3))) %>% as.data.table()


dt[,sum((O-E)^2/E)] 
qchisq(p=0.95, df=4) # 9.487729


# 6.4.1
# It is 4+
dt <- data.table(a=0:4, 
                 O=c(12,23,24,24,17), 
                 E=100*c(dpois(0:3,lambda=2),
                         1-sum(dpois(0:3, lambda=2))))
dt[,sum(a*O)/sum(O)]
dt[,sum((O-E)^2/E)]

# 6.4.2
dt <- data.table(O=c(15,23,19,20,14,11)) %>% mutate(E=sum(O)/length(O))
dt[,sum((O-E)^2/E)] # 5.764706
qchisq(p=0.95, df=5) # 11.0705

# 6.4.3
dt <- data.table(a=0:3, O=c(15,13,9,13), E=50*c(dpois(0:2, lambda=1.4), 1-sum(dpois(0:2, lambda=1.4))))
dt[,sum(a*O)/sum(O)] # 1.4; minus one degree of freedom
dt[,sum((O-E)^2/E)] # 5.04233
qchisq(p=0.90, df=2) # 4.60517
# We have sufficient evidence to reject the Null that this data is from a Poisson distribution with coefficient 1.4

# 6.4.4
dt <- data.table(a=0:6, O=c(0,26,36,20,10,6,2))
# Finding P; 
dt[,sum(O*a)/(sum(O)*6)] # p = 0.4
dbinom(x=0:6, size=6, prob=0.4)*100 # have to combine the last three cells and the first cell
dt <- data.table(a=1:4, 
                 O=c(26,36,20,18), 
                 E=100*c(sum(dbinom(0:1, size=6, prob=0.4)), 
                     dbinom(2:3, size=6, prob=0.4), 
                     sum(dbinom(4:6, size=6, prob=0.4))))
dt[, sum((O-E)^2/E)] # 3.19
qchisq(p=0.95, df=2) # 5.99 
# Insufficient evidence to reject the NULL that this data came from a Binomial distribution with p=0.4. 

# 6.4.5 - Wrong
dt <- data.table(Emply=c(4,3,5,1,2), 
                 Acc = c(22,14,25,8,12))
dt <- dt %>% mutate(O=Acc/Emply, E = sum(O)/length(O), E2 = E*Emply)
dt
dt[,sum((O-E)^2/E)] # 1.18
qchisq(p=0.95, df=4) # 9.49

# 6.4.6
dt <- data.table(a=0:8, 
                 O=c(2,8,15,18,14,13,7,3,0))
# Poisson coefficient is the mean
dt[,sum(a*O)/sum(O)] # 3.45
dt[,sum(O)] # 80
dpois(0:8, lambda=3.45) * 80 # collapse back three and fronst two cells
dt <- data.table(a=1:6, O=c(10, 15,18,14,13, 10), E=80*c(sum(dpois(0:1, lambda=3.45)), dpois(2:5, lambda=3.45), 1-sum(dpois(0:5, lambda=3.45))))
# <=1, 2, 3, 4, 5, 6+
dt[,sum((O-E)^2/E)] # 0.9896
qchisq(p=0.95, df=4) # 9.487729

# 6.4.7
# Independent of each other, constant average rate
dt <- data.table(a=0:5, O=c(50, 24, 12, 9, 5, 0))
dt[,sum(a*O)/sum(O)] # 0.95
dt[,sum(O)] # 100
dpois(0:5, lambda=0.95) * 100 # merge the last three cells
dt <- data.table(a=0:3, 
                 O=c(50,24,12,14), 
                 E=100*c(dpois(0:2, lambda=0.95), 1-sum(dpois(0:2, lambda=0.95))))
dt[,sum((O-E)^2/E)] # 16.04645
qchisq(p=0.95, df=2) # 5.99

# 6.4.9
dbinom(3:5, size=8, prob=0.5)*200

```


```{r}
# 6.6.4
library(TREX)
dt <- data.table(a=1:5, O=c(76, 17, 4, 2, 1))
1/dt[,sum(a*O)/sum(O)] #0.741

dt[,E:=c(dgeomw(trials=1:4, prob=0.741), 1-sum(dgeomw(1:4, prob=0.741)))]
dt
dt[,E:=sum(O)*E]
dt <- dt %>% as.matrix()
dt <- rbind(dt[1:2,], t(colSums(dt[3:5,])))
dt[,sum((O-E)^2/E)] # 0.311756
qchisq(p=0.975, df=1) # 5.02

# 6.6.5
dt <- data.table(a=1:8, O=c(12,14,11,10,5,9,7,7), 
                 E=75*c(dgeomw(trials = 1:7, prob=5/26), 
    1-sum(dgeomw(trials=1:7, prob=5/26))))
#X~geom(5/26)
dt <- data.table(a=1:6, O=c(12,14,11,10,5,23), 
                 E=75*c(dgeomw(trials = 1:5, prob=5/26), 
    1-sum(dgeomw(trials=1:5, prob=5/26))))
dt[,sum((O-E)^2/E)] # 2.419
qchisq(p=0.95, df=5) ## 11.0705

# Challenge
dt <- data.table(a=10:15, 
                 O=c(10,25,29,15,15,10), 
                 E = c(dnbinomw(trials=10:14, size=10, prob=0.8), 
  1-sum(dnbinomw(trials=10:14, size=10, prob=0.8))) * 104)
# MLE by hand, p=0.8
# Negative binomial mean is r/p, so 10/\bar{x} is the one
10/dt[,sum(a*O)/sum(O)] # 0.8137715

dt
dt[,sum((O-E)^2/E)] # 3.638
qchisq(p=0.95, df=4) # 9.487729
```
Mixed exercises 6
```{r}
# 1
qchisq(p=0.99, df=10) # 23.20925
# 2
qchisq(p=0.95, df=8) # 15.50731
# 3
# 8 degrees of freedom
qchisq(p=0.95, df=8) # If X2>15.50731 then reject
# 4
# 6 df
qchisq(p=0.95, df=6) # 12.59159
# 5 
# H0: Drug taking is statistically independent of catching the cold
O <- matrix(c(34, 66,
              45, 55), nrow=2, byrow=T)
E <- outer(rowSums(O), colSums(O))/sum(O)
sum((O-E)^2/E) # 2.531646
qchisq(p=0.95, df=1) # 3.84159

# 6
dt <- data.table(a=0:3, O=c(38, 32, 10, 0))
# Find P; MLE is mean of O
dt[,sum(O*a)/sum(O)] # lambda = 0.65

dpois(0:3, lambda=0.65) * 80 # Have to collapse last cell

dt <- data.table(a=0:2, O=c(38, 32, 10),
                 E = 80*c(dpois(0:1, lambda=0.65), 1-sum(dpois(0:1, lambda=0.65))))

dt[,sum((O-E)^2/E)] # 1.314099
qchisq(p=0.95, df=1) # 3.841459

# 7 
O <- matrix(c(23, 27, 
              32, 18), nrow=2, byrow=T)
E <- outer(rowSums(O), colSums(O))/sum(O)
sum((O-E)^2/E) #3.2727
qchisq(p=0.90, df=1) #2.71

#8 
dt <- data.table(O=c(20,16,25,18,21), E=rep(20,5))
dt[,sum((O-E)^2/E)] #2.3
qchisq(p=0.95, df=4)

#9
# MLE implies p=1/\bar{x}
dt <- data.table(a=0:5, O=c(1,1,5,11,24,8))
dt[,sum(a*O)/(sum(O)*5)] # 0.72

dbinom(0:5, size=5, prob=0.72)*50
dt <- data.table(a=2:5, O=c(7,11,24,8), 
                 E=50*c(sum(dbinom(0:2, size=5, prob=0.72)),
                     dbinom(3:5, size=5, prob=0.72)))
dt[,sum((O-E)^2/E)] #2.624061
qchisq(p=0.95, df=4) # 9.487729

# 10
dt <- data.table(a=0:3, O=c(112, 56, 40, 0))
dt[,sum(a*O)/sum(O)] # 0.6538462
dpois(0:3, lambda=0.6538462) * 208
dt[,E:=dpois(0:3, lambda=0.6538462) * 208]
dt[,sum((O-E)^2/E)] #20.65154
qchisq(p=0.95, df=2) #5.99
# Reject the null

# 11 - Wrong?
dt <- data.table(O=c(10,35,15), E=c(30,55,15))
dt[,sum((O-E)^2/E)] # 20.61
qchisq(p=0.95, df=2) # 5.99

# 12
library(TREX)
dt <- data.table(a=0:9, O=c(0,4,7,8,10,6,7,4,4, 0))
dt[,sum(a*O)/sum(O)] # 4.28
dpois(0:9, lambda=4.28)*50 # collapse first three, last three
dt <- data.table(a=2:7, O=c(11, 8, 10, 6, 7, 8), 
                 E=50*c(sum(dpois(0:2, lambda=4.28)),
                     dpois(3:6, lambda=4.28), 
                     1-sum(dpois(0:6, lambda=4.28))))

dt[,sum((O-E)^2/E)] # 1.18
qchisq(p=0.95, df=4) #9.49

# 13
O <- matrix(c(100, 600, 
              80, 800), nrow=2, byrow=T)
E <- outer(rowSums(O), colSums(O))/sum(O)
sum((O-E)^2/E) # 10.42259
qchisq(p=0.95, df=1) # 3.84

# 14
O <- matrix(c(74, 28, 68, 
              45, 40, 45), nrow=2, byrow=T)
E <- outer(rowSums(O), colSums(O))/sum(O)
sum((O-E)^2/E) # 8.687399
qchisq(p=c(0.95,0.99), df=2) # 5.99, 9.21034

# 15
exp(-2.15)*2.15 # dpois(1, lambda=2.15)
1-dpois(0:1, lambda=2.15) %>% sum() # 0.633
dt <- data.table(a=0:6, O=c(10, 12, 14, 12, 8, 3, 1))
dt[,sum(a*O)/sum(O)] # 2.15 as required

dt <- data.table(O = c(10,12,14,12, 12),
                 E=c(dpois(0:3, lambda=2.15), 
                     1-sum(dpois(0:3, lambda=2.15))) * 60)
dt[,sum((O-E)^2/E)] #2.507
qchisq(p=0.95, df=3) #7.815

# 16
dt <- data.table(a=1:7, O=c(130,54,24,28,13,5,1))
1/dt[,sum(a*O)/sum(O)] # 0.4866412
dgeomw(1:7, prob=0.4866412) * 255 # collapse last 2
dt <- data.table(a=1:6, O=c(130,54,24,28,13,6), 
                 E=255*c(dgeomw(1:5, prob = 0.4866412), 1-sum(dgeomw(1:5, prob=0.4866412))))
dt[,sum((O-E)^2/E)] #14.84
qchisq(p=0.95, df=4) # 9.487729

# 17 

# Challenge
dt <- data.table(a=seq(2.5, 22.5, by=5), 
                 O = c(7,63,221,177,32))
M <- dt[,sum(a*O)/sum(O)]
V <- dt[,sum(a^2*O)/sum(O)] - dt[,sum(a*O)/sum(O)]^2
# V <- c(rep(2.5, 7), rep(7.5, 63), rep(12.5, 221), rep(17.5, 177), rep(22.5, 32)) %>% var()
dt <- data.table(a=seq(2.5, 22.5, by=5), 
                 O = c(7,63,221,177,32), 
                 E = 500*c(pnorm(5, mean=M, sd=sqrt(V)),

                           pnorm(10, mean=M, sd=sqrt(V)) - pnorm(5, mean=M, sd=sqrt(V)),

                           pnorm(15, mean=M, sd=sqrt(V)) - pnorm(10, mean=M, sd=sqrt(V)),

                           pnorm(20, mean=M, sd=sqrt(V)) - pnorm(15, mean=M, sd=sqrt(V)),

                           1- pnorm(20, mean=M, sd=sqrt(V))))
dt
dt[,sum((O-E)^2/E)] # 3.24



```

Chapter 7 - Probability Generating Functions
```{r}
# Prior knowledge
# Q1
dpois(x=5, lambda=4.5)  # 0.1708269
exp(-4.5)*(4.5^5)/factorial(5)

sum(dpois(0:3, lambda=4.5)) # 0.342296
# 4.5
# 4.5

# Q2
dgeomw(trials=5, prob=0.3) # 0.3*0.7^4 = 0.7203
(1-0.3)^2 # 0.49
1-sum(dgeomw(trials=1:2, prob=0.3))
# E = 10/3, V = 0.7/0.3^2 = 7.777778

# Q3
dnbinomw(trials=13, size=5, prob=0.35) # 0.08284232
0.35 * choose(n=12, k=4) * (0.35^4) * (0.65^8) # 0.08284232
sum(dnbinomw(trials=5:12, size=5, prob=0.35)) # 0.4166549
# E = 5/0.35 = 14.28571; V = r(1-p)/p^2 = 5(0.65)/0.35^2 = 26.53061

```
# Chapter 8
```{r}
# Prior knowledge test
# X ~ N(\mu, 2.3^2), test whether \hat{mu} = 11.1 from 20 observations is sufficient evidence to conclude that \mu is greater than 10. 
# HO: \mu = 10, H1: \mu > 10
pnorm(q=11.1, mean=10, sd=2.3) # Insufficient evidence?
1-pnorm(q=11.1, mean=10, sd=sqrt(0.2645)) # 0.01622401, so fail to reject at 1% level. 

# MS2 - Chapter 3, Example 15
# Random sample of 50 bolts from an assumed normal distribution of N(0.580, 0.015) - what's the rejection region for a 1% test?
qnorm(p=0.005, mean=0.580, sd=0.015/sqrt(50)) # 0.5745358; lower bound
qnorm(p=0.995, mean=0.580, sd=0.015/sqrt(50)) # 0.5854642 ; upper bound

# Question 2 - Poisson at 10%
dt <- data.table(a=0:15, b=cumsum(dpois(0:15, lambda=5)), c=ppois(0:15, lambda=5))
dt <- data.table(x=0:15, z = cumsum(dpois(0:15, lambda=5)))
dt %>% mutate(a = z-0.05, b = z - 0.95) %>% as.data.table() # c1 = 3 , c2 -1 = 15
1 - sum(dpois(4:15, lambda=9)) # 0.0433
dt[b<0.05] # 1 and below
dt[b>0.95] # 9 and above
sum(dpois(0:1, lambda=5)) # 0.04042768
1-sum(dpois(0:8, lambda=5)) #  0.06809363

dt <- data.table(a=0:10, b=cumsum(dgeomw(0:10, prob = 0.15)))
dt
dt[b>0.95]
```


```{r}
1-pnorm(q=11.1, mean=10, sd=sqrt(0.2645)) # 0.01622401
# Question 2
dt <- data.table(a=0:10, z=cumsum(dpois(0:10, lambda=5)))
dt[z<0.05]
dt[z>0.95] # 9 therefore c2 -1 = 9, or c2 = 10. Region is c1=1, c2=10

log(0.95)/log(0.85)
1-(1-0.15)^(0:20)


# Example 4
options(scipen = 9)
sum(dbinom(0:7, size = 20, prob = 0.5)) # 0.131588; fail to reject at conventional levels
dt <- data.table(a=0:20, z=cumsum(dbinom(x = 0:20, size = 20, prob = 1/2)))
dt[z<0.025] # 5
dt[z>0.975] #14 -> 15

# Actual significance level?
dt[z<0.025] %>% tail(1)[2] 
dt[z<0.025][6,2] + 1- dt[z>0.975][1,2] # 0.04138947	

# Power given tails three times more likely than heads?
1-sum(dbinom(c(0:5, 15:20), size=20, prob=0.25)) # 0.3828235
sum(dbinom(c(6:14), size=20, prob=0.25)) # 0.3828235

# Example 5
1+log(0.05)/log(0.9) # 29.43 # Therefore critical region is X>=30
1-sum(dgeomw(trials=1:29, prob=0.1)) # 0.047  == 0.9^29
sum(dgeomw(1:29, prob=0.01)) # 0.2528279

```
Exercise 8a
```{r}
# Binomial with sample size of 10
options(scipen = 9)
dt <- data.table(a=0:10, b=dbinom(0:10, prob=0.25, size=10) %>% cumsum()) # K>=6
1-dt[b>0.95][1,2] # 0.01972771	
sum(dbinom(0:5, size=10, prob=0.3)) # 0.952651

# Binomial with sample size of 20
dt <- data.table(a=0:20, z= cumsum(dbinom(x=0:20, size=20, prob=0.3)))
dt[z<0.01] # c=1, T1 = 0.0076372598
1-sum(dbinom(0:1, size=20, prob=0.25)) # 0.9756874

# Two sided test
dt <- data.table(a=0:10, z=cumsum(dbinom(x=0:10, size=10, prob=0.45)))
dt[z<0.025] # cl = 1; 0.023
dt[z>0.975] # cu = 8+1; 0.995

# Question 10 
dt <- data.table(a=1:2000, z=cumsum(dgeomw(trials=1:2000, prob = 0.004)))
dt[z<0.05] # cl = 12
dt[z>0.949] # cu = 747+1 

# Question 11 
dt <- data.table(a=0:40, z=cumsum(dbinom(0:40, prob=0.05, size=40)))
dt[z>0.95] # k=4+1=5
1-dt[z>0.95][1,2] # 0.04802826	

dt2 <- data.table(a=1:50, z=cumsum(dgeomw(trials=1:50, prob=0.05)))
dt2[z<=0.05] # c1=1; 0.05 chance of T1

1-0.0588 # 0.9412
sum(dbinom(x=0:4, size=40, p=0.0588)) # 0.9161818
```
## 8.2
```{r}
# Example 6
qnorm(p=0.025, mean=1, sd=0.04/sqrt(10)) # 0.9752082
qnorm(p=0.975, mean=1, sd=0.04/sqrt(10)) # 1.024792
a <- pnorm(q=c(0.9752082,1.024792), mean=1, sd=0.04/sqrt(10))
a[1] + 1-a[2] # 0.05

# If mean weight is actually 1.02, what's the probability of a type 2 error?
pnorm(q=1.024792, mean=1.02, sd=0.04/sqrt(10)) - pnorm(q=0.9752082, mean=1.02, sd=0.04/sqrt(10)) # 0.6473978

# Example 7
qnorm(p=0.05, mean=150, sd=6/5) # 148.0262
# qnorm(p=0.975, mean=150, sd=6/5) # 152.352

qnorm(p=0.01, mean=150, sd=6/5) # 147.2084
# qnorm(p=0.995, mean=150, sd=6/5) # 153.091

# True new mu is 147, then implies the type 2 probability is 
1-pnorm(148.0262, mean=147, sd=6/5) # 0.1962294
1-pnorm(147.2084, mean=147, sd=6/5) # 0.4310637
```
Exercise 8.b
```{r}
qnorm(p=0.99, mean=50, sd=3/sqrt(20)) # 51.56056; only a 1% chance more than that given H0 is correct. 
#0.01
# True mean is 53
pnorm(51.56056, mean=53, sd=3/sqrt(20)) # 0.01594485

# Question 2
qnorm(p=0.05, mean=30, sd=2/sqrt(16)) # 29.17757
# 0.05
1-pnorm(q=29.17757, mean=28.5, sd=2/sqrt(16)) # 0.08768648; reverse in mind

# Question 3
qnorm(p=0.005, mean=40, sd=4/5) # 37.93934
qnorm(p=0.995, mean=40, sd=4/5) # 42.06066
# 0.05
# True is 42
pnorm(42.06066, mean=42, sd=4/5) - pnorm(37.93934, mean=42, sd=4/5) # 0.5302206

# Question 4
qnorm(p=0.025, mean=15, sd=1/5) # 14.60801
qnorm(p=0.975, mean=15, sd=1/5) # 15.39199
pnorm(15.39199, mean=15.6, sd=1/5) - pnorm(14.60801, mean=15.6, sd=1/5) # 0.149158

# Question 5
qnorm(p=0.95, mean=40, sd=8/sqrt(30)) # 42.40246
pnorm(q=42.40246, mean=42, sd=8/sqrt(30)) # 0.6085514
# Increase the sample size only way of keeping significance the same and lowering the P2

```
8.3
```{r}
# Example 8
qbinom(p=0.1, size=25, prob=0.3)
dt <- data.table(x=0:25, z=cumsum(dbinom(x=0:25, size=25, prob=0.3)))
dt[z<=0.1] # 4!
dt[x==4][,2] # 0.09047192
# p =0.2, what is the power of this test? Given H0 is false, correctly rejecting H0. So probability that X is in the critical region given p=0.2. 
dt <- data.table(x=0:25, z=cumsum(dbinom(x=0:25, size=25, prob=0.2)))
dt[x==4] # 0.4206743	

# Example 9, if less than 106 then regulatory action
qnorm(p=0.05, mean=106, sd=5/sqrt(30)) # 104.4985
pnorm(q=104.4985, mean=102, sd=5/sqrt(30)) # 0.9968996
# Do the diagrams in your head


# Example 10
dgeomw(trials=1:5, prob=0.02) %>% sum() # 0.0960792
1-dgeomw(trials=1:100, prob=0.02) %>% sum() # 0.1326196
0.98^100 # 0.1326196
1 - 0.0960792 - 0.98^100 # 0.7713012 inbetween
# Then have a 0.0960792 chance of rejecting, and a 0.9039208 chance of accepting
# 1-dgeomw(trials=1:5, prob=0.02) %>% sum()
# Size is the probability of incorrectly accepting H1; so rejecting given p=0.02 is true
0.0960792 + (1 - 0.0960792 - 0.98^100)*0.0960792 # 0.1701852
# First term is if first time X\le 5, second term is conditional on the first go being between 6 and 100, the probability of rejecting on the second term. 

# If p=0.015, power is the probability of correctly rejecting the Null given the cutoffs we have.
sum(dgeomw(trials=1:5, prob=0.015)) + sum(dgeomw(trials=6:100, prob=0.015))*sum(dgeomw(trials=1:5, prob=0.015)) # 0.1242129 

```
Exercise 8.3
```{r}
# Question 1 - Sample mean from normal distribution with H0 mean = 20, sd=3^2, n=25. ONe sided upper.
qnorm(p=0.95, mean=20, sd=3/5) # 20.98691
1-pnorm(q=qnorm(p=0.95, mean=20, sd=3/5), mean = 20.8, sd=3/5) # 0.3777026
# Question 2 - Binomial one sided p =0.35
dt <- data.table(a=0:20, z=cumsum(dbinom(x=0:20, size=20, prob=0.35)))
1-dt[z>0.95] %>% slice(1) %>% select(z) # 12 or higher
# Size = 0.01957936	
dt <- data.table(a=0:20, z=cumsum(dbinom(x=0:20, size=20, prob=0.36)))
1-dt[a==11][,2] # 0.02469464	

# Question 3 - Poisson; lambda < 4.5 - size of the test, power given 4.1
dt <- data.table(a=0:20, z=cumsum(dpois(x=0:20, lambda = 4.5)))
dt[z<0.05] # 0.01110900
dpois(x=0, lambda=4.1) # 0.01657268 = exp(-4.1)

# Question 4
0.004 %>% sqrt()
qnorm(p=0.025, mean=2, sd=sqrt(0.004/25)) # 1.975208
qnorm(p=0.975, mean=2, sd=sqrt(0.004/25)) # 2.024792

1 - (pnorm(qnorm(p=0.975, mean=2.02, sd=sqrt(0.004/25)), mean=2, sd=sqrt(0.004/25)) - pnorm(qnorm(p=0.025, mean=2.02, sd=sqrt(0.004/25)), mean=2, sd=sqrt(0.004/25))) # 0.3526081

# Question 5
dt <- data.table(a=0:10, z=cumsum(dbinom(x=0:10, size=10, prob=0.4)))
1-dt[a==6] # 0.05476188
dt <- data.table(a=0:10, z=cumsum(dbinom(x=0:10, size=10, prob=0.8)))
1-dt[a==6] # 0.8791261
# Power is higher for values of p further from 0.3. 

# Question 6
# Type one is incorrectly accepting the alternative hypothesis. 
# The size is the probability of committing a type 1 error. 
1-pnorm(q=25, mean=20, sd=sqrt(10)) # 0.05692315

# Question 7
# p>0.01 for geometric with alpha = 0.05
dt <- data.table(a=1:20, z=cumsum(dgeomw(trials=1:20, prob=0.01)))
dt[z<0.05] # lower limit is 5

dt <- data.table(a=1:5, cumsum(dgeomw(trials=1:5, prob=0.2)))
dt # 0.67232

# Question 8
dt <- data.table(a=1:400, z=cumsum(dgeomw(trials=1:400, prob=0.01)))
dt[z<0.025] # 2? 
dt[z>0.9749] # 367 + 1 = 368
sum(dgeomw(trials=1:2, prob=0.02)) + 1 - sum(dgeomw(trials=1:367, prob=0.02)) # 0.04020252

# Question 9
options(scipen = 9)
1-sum(dpois(x=0:11, lambda=8)) + sum(dpois(x=10:11, lambda=8))*(1-sum(dpois(0:7, lambda=8))) # .2057148
1-sum(dpois(x=0:11, lambda=1)) + sum(dpois(x=10:11, lambda=1))*(1-sum(dpois(0:7, lambda=1))) # 0.00000000083274425

# Question 10
library(TREX)
pnorm(q=79, mean=80, sd=sqrt(25/20)) # 0.1855467
dgeomw(trials=1:10, prob=pnorm(q=79, mean=80, sd=sqrt(25/20))) %>% sum() #  0.8715713

dgeomw(trials=1:10, prob=pnorm(q=79, mean=81, sd=sqrt(25/20))) %>% sum() # 0.3128074

# Challenge
dt <- data.table(a=0:20, z=cumsum(dbinom(x=0:20, size=20, prob=0.08)))
dt[,z-0.95] # 3 or 4; larger error. More conservative so critical region is X>=5
1-sum(dbinom(x=0:4, size=20, prob=0.08)) # probability of getting it wrong each time
dt <- data.table(a=1:20, z=cumsum(dgeomw(trials=1:20, prob =1-sum(dbinom(x=0:4, size=20, prob=0.08)))))
dt # maximum five boxes can be checked before the probability of a type one error is less than 0.1

# If the first four boxes pass then no recalibration
# Say first two boxes are fine at 0.08, and the second two have 0.2. 4 boxes were checked; what's the power of this test? What's the probability of rejecting the null of p=0.08 given the last two boxes were p=0.2?

p1 <-  1-sum(dbinom(x=0:4, size=20, prob=0.08)) # rejecting a box R12
p2 <- 1-sum(dbinom(x=0:4, size=20, prob=0.2)) # rejecting a box R34
# 4 boxes checked, total probability we reject the null correctly?
p1 + p1*(1-p1) + (1-p1)^2 * p2 + (1-p1)^2 * (1-p2) * p2 # 0.6179552

```
Exercise 8.4
```{r}
# Example 11
options(scipen = 9)
1-sum(dpois(0:6, lambda=3.5)) # 0.0652881

dt <- data.table(a=seq(0.5, 10.5, by=0.5), 
                 b=1-sum(dpois(0:6, lambda=seq(0.5, 10.5, by=0.5))))
# ppois(6, lambda=3.5) = sum(dpois(0:6, lambda=3.5)) = Pr[X\le6]
dt <- data.table(a=seq(0.5, 10.5, by=0.5), 
                 b=1-ppois(6, lambda=seq(0.5, 10.5, by=0.5)))
# To colour the points arbitrarily
dt <- dt %>% mutate(c=if_else(b>0.5, 1,0))

library(ggplot2)
library(hrbrthemes)
ggplot(data=dt, mapping=aes(x=a, y=b, colour=c)) + 
    geom_point() + 
    geom_line(colour="black") + theme_simple

theme_simple <- theme_bw() + theme(legend.position="none")

# Example 12
sum(dbinom(x=0:1, size = 10, prob=0.2)) # 0.3758096
p <- 0.4
(1-p)^5 + (1-(1-p)^5)*(1-p)^5 # 0.1494734

dt <- data.table(a=c(0.1, 0.2, 0.25, 0.3, 0.35))
dt <- dt %>% mutate(A = (1-a)^10 + 10*a*(1-a)^9, 
              B=(1-a)^5 + (1-(1-a)^5)*(1-a)^5)
dt # test B strictly dominates for all plausable levels of A considered for a one-sided test, but also requires less time. 

# Example 3
1-dgeomw(1:30, prob=0.1) %>% sum() # 0.04239116
# Power function is (1-p)^30

```
Exercise 8.4
```{r}
# Size of a test; 
dpois(0:2, lambda=6.5) %>% sum() # Probability of a type one error is the probability or incorrectly accepting H0, or the probability that X <=2 given the hypothesis is actually true. 
ppois(q=2, lambda=6.5) # 0.430
dt <- data.table(a=1:6, b=ppois(q=2, lambda=1:6))
dt # s=0.6766764, t = 0.1246520	
dt[b<0.5]
ggplot(data=dt, mapping=aes(x=a, y=b)) + geom_point() + geom_line() + theme_simple

# solve(ppois(q=2, lambda=x), 0.5) # ????

# Question 2
dbinom(x=0:2, size=12, prob=0.45) %>% sum() # 0.04214198
pbinom(q=2, size=12, prob = 0.3) # 0.2528153

# Question 3 
dbinom(x=8:10, size=10, prob=0.4) %>% sum() #.01229455
dt <- data.table(a=seq(0.1, 1, by=0.1),z= 1-pbinom(q=7, size=10, prob=seq(0.1, 1, by=0.1))) # 0.054687500, 0.6777995264
dt # Power increases as the value of p moves further from zero. 

# Question 4
# Test A, reject p<0.5 if number of heads is two or fewer
pbinom(2, size=10, prob=0.5) # 0.0546875
# Test B, coin is spun 5 times and if no heads reject, if there is a head try again and if no heads reject.
# Probability of rejecting each time given Null of fairness
p <- 0.5
(1-p)^5 + (1-(1-p)^5)*(1-p)^5 # 0.06152344

dt <- data.table(p=c(0.1, 0.2, 0.25, 0.3, 0.35, 0.4))
dt <- dt %>% mutate(A= pbinom(2, size=10, prob=p),
                    B=(1-p)^5 + (1-(1-p)^5)*(1-p)^5 )
# dt
# For the same number of spins the gambler will have better power using Binomial, so test A.
ggplot(data=dt) + 
    geom_line(data = dt, mapping = aes(x=p, y=A), colour=BITBlue) + 
    geom_line(data = dt, mapping= aes(x=p, y=B), colour=BITOrange) + 
    theme_simple + 
    geom_text(x=0.2, y=0.75, label="Test A", colour=BITBlue) +
    geom_text(x=0.2, y=0.5, label="Test B", colour=BITOrange)
```

```{r}
# Exercise 5
dt <- data.table(a=1:100, b=cumsum(dgeomw(trials=1:100, prob=0.15)))
dt[b>0.99] # a = 29 + 1 = 30; size = 0.9910226; 0.0089774
# (1-p)^29

# Exercise 6 - Shit question
dt <- data.table(a=0:10, b=cumsum(dbinom(x=0:10, size=10, prob=0.9)))
# Think only evidence if x=10? 
0.9^10 # 0.3486784
options(scipen = 999)
dt <- data.table(a=0:12, b=cumsum(dbinom(x=0:12, size=12, prob=0.9)))
dt
```
## Chapter 8 mixed exercises
```{r}
dt <- data.table(a=0:15, b=cumsum(dbinom(0:15, size=15, prob=0.35)))
dt[b>0.95] # Cu = 8+1
1-dt[a==8] # 0.04219384	

sum(dbinom(9:15, size=15, prob=0.5)) # 0.3036194

# Question 2
dt <- data.table(a=0:3, b=cumsum(dpois(x=0:3, lambda=3.5)))
dt[b<0.05] # Only if P[X=0]
# 0.03019738	
dpois(0, lambda=3) # 0.04978707; power

# Question 3
qnorm(p=0.025, mean=8, sd=sqrt(1/2)) # 6.614096
qnorm(p=0.975, mean=8, sd=sqrt(1/2)) # 9.385904
# 5 %
pnorm(q=qnorm(p=0.975, mean=8, sd=sqrt(1/2)), mean=7, sd=sqrt(1/2)) - pnorm(q=qnorm(p=0.025, mean=8, sd=sqrt(1/2)), mean=7, sd=sqrt(1/2)) # 0.7070111

# Power is 1-prob of type 2
# 0.2929889


# Question 4
(1-sum(dpois(0:17, lambda=10)))*(1-sum(dpois(0:18, lambda=10))) + sum(dpois(0:3, lambda=10))*sum(dpois(0:2, lambda=10)) # 0.0001312308

(1-sum(dpois(0:17, lambda=5)))*(1-sum(dpois(0:18, lambda=5))) + sum(dpois(0:3, lambda=5))*sum(dpois(0:2, lambda=5)) # 0.03303602

# Question 5
1-sum(dpois(0:7, lambda=4.5)) # 0.08658647
1-ppois(7, lambda=4.5)

dt <- data.table(a=1:10, b=1-ppois(7, lambda=1:10))
ggplot(data=dt, aes(x=a, y=b)) + 
    geom_point() + 
    geom_line() + 
    theme_simple

```

```{r}
# Question 6
dbinom(x=0:3, size=15, prob=0.45) %>% sum() # 0.04242127
dt <- data.table(a=1:5/10)
dt <- dt %>% mutate(b=pbinom(q=3, size=15, prob=a))
ggplot(data=dt, aes(x=a, y=b)) + geom_point() + geom_line() + theme_simple
```
```{r}
# Question 7
dt <- data.table(a=0:20, b=cumsum(dpois(0:20, lambda=2)))
dt[b>0.92] # Cu = 4+1
1-sum(dpois(0:4, lambda=4)) # 0.3711631

# Question 8
#HO: \lambda=2, H1: \lambda > 2
library(TREX)
dt <- data.table(a=0:15, 
                 b=cumsum(dpois(x=0:15, lambda=2)))
dt[b>0.92] # cu = 4+1

1-ppois(4, lambda=3) # 0.1847368; probability of producting 5 or more defects given lambda = 3

# Po(6)
dt <- data.table(a=0:50, 
                 b=cumsum(dpois(x=0:50, lambda=6)))
dt[b>0.9] # b=10+1, so upper limit of cu

1-ppois(10, lambda=9) # 0.2940117

# Question 9
ppois(2, lambda=6) # 0.0619688
dt <- data.table(a=c(1,1.5, 2, 4, 5,6,7))
dt <- dt %>% mutate(b=ppois(q=2, lambda=a))

ggplot(data=dt, mapping=aes(x=a, y=b)) + 
    geom_point() + 
    geom_line() + theme_simple

# Again, there's some solve command that will do this! Mathematica!
# lambda<1.5
```
```{r}
# Question 10
1-sum(dbinom(0:4, size=10, prob=0.1)) # 0.001634937
1-pbinom(4, size=10, prob=0.1)

dt <- data.table(a=seq(0.15, 0.4, by=0.05))
dt <- dt %>% mutate(b=1-pbinom(4, size=10, prob=a))
dt # matches

1-pbinom(2, size=5, prob=0.1) # 0.00856
dt <- dt %>% mutate(c=1-pbinom(2, size=5, prob=a))
dt

ggplot(data=dt) + 
    geom_point(data=dt, aes(x=a,y=b), colour=BITBlue) + 
    geom_line(data=dt, aes(x=a,y=b), colour=BITBlue) + 
    geom_point(data=dt, aes(x=a, y=c), colour=BITBlack) + 
    geom_line(data=dt, aes(x=a,y=c), colour=BITBlack) 

# Quicker, and has better power for lower values of p. 
```
```{r}
# Exercise 11
options(scipen = 99)
1-ppois(q=5, lambda=3) # 0.08391794
dt <- data.table(a=4:10/10) # multiply by 10!
dt <- dt %>% mutate(b=1-ppois(5, lambda=a*10))

# Probability of a type 1 error to be less than 5% for n=15
dt2 <- data.table(a=0:100, b=cumsum(dpois(x=0:100, lambda=4.5)))
dt2[b>0.95] # cu = 8+1
1-dt2[a==8] # size is 0.04025731

dt <- dt %>% mutate(c=1-ppois(8, lambda= 15 * a))
dt

ggplot(data=dt) + 
    geom_point(data=dt, aes(x=a,y=b), colour=BITBlue) + 
    geom_line(data=dt, aes(x=a,y=b), colour=BITBlue) + 
    geom_point(data=dt, aes(x=a, y=c), colour=BITBlack) + 
    geom_line(data=dt, aes(x=a,y=c), colour=BITBlack) 
# They cross around 0.63
```
```{r}
# Challenge
outer(1:6, 1:6)


# 1/36 chance given unbiased; testing against p>1/36; binomial with 12 attempts
1-sum(dbinom(size=12, prob=1/36, x=0:1)) # 0.042329
# 1 - (1-p)^11 * (1-11p)

sum(dbinom(4:6, size=6, prob=1/6)) + sum(dbinom(0:3, size=6, prob=1/6)) * sum(dbinom(4:6, size=6, prob=1/6)) # 0.01732825

```

# Example A2 past paper - pp.182

```{r}
dt <- data.table(a=1:6, O=c(52,31,12,7,1,1))
dt[,sum(O)] # 104
dt <- dt %>% mutate(E=c(dgeomw(1:5, prob=0.4), 
                        1-sum(dgeomw(1:5, prob=0.4)))*sum(O))

dt[,sum(((O-E)^2)/E)] # 14.87929

# Takeaways; the last Ei is 6 or above! 
# If E < 5 you have to merge it! The following code merges

merge <- c(a=5, dt %>%
                   slice(5:6) %>%
                   select(2:3)%>%
                   colSums()) %>%
    t()

dt2 <- rbind(dt %>% slice(1:4) %>% as.matrix(), merge) %>% 
    as.data.table()


qchisq(p=0.95, df=5) # 11.0705; apparently four DF? But unclear why?

# Calculate the test statistic

```
```{r}
# Question 4
pnorm(q=2.3, mean=2.1, sd=sqrt(2.1/200), lower.tail = F) # 0.02548097
sum(dpois(0:1, lambda=4.2)) # 0.077977


```
```{r}
# Question 5
dt <- matrix(data=c(1,3,2,
              -1,21,20,
              1,1,1), nrow=3, byrow=T)
solve(dt, c(1,4.9, 0.55))
library(matlib)
inv(dt) %*% c(1,4.9, 0.55)

```
```{r}
# Question 6
1-sum(dpois(0:17, lambda=15)) # 0.2511412
1- sum(dbinom(0:17, size=500, prob=0.03)) # 0.2485376
```
```{r}
# Question 7 
dt <- data.table(a=0:25, b=cumsum(dbinom(0:25, size=25, prob=1/6)))
dt[b<0.1] # X = 0 on lower tail, or 1 as a bit larger
dt[b>0.95] # a = 7

dt %>% filter(a==0) # 0.0104826	
dt %>% filter(a==1) # 0.06289558

dt[b<0.2]
1-dt %>% filter(a==7) # 0.04473193


# 0.05521453



# Again; 
(1-1/6)^12 # 0.1121567

options(scipen = 100)
dt <- data.table(a=0:100/100)
dt <- dt %>% mutate(philip=(1-a)^24 * (1+24*a),gemma=(1-a)^12)
dt[a==0.09] # The power of the first test

# Phillips power is higher when p=0.09 at 0.3286098 vs 0.3224755, and it has a lower actual significance level!

```
# June 2019 FS1 Past Paper
Let's fucking goooo! 
```{r}
1-sum(dbinom(x=0:2, size=40, prob=0.02)) # 0.04567023

dnbinomw(trial=40, size=3, prob=0.02)
0.02*choose(39, 2) * (0.02)^2 * (0.98)^37
3/0.02
```
```{r}
1-sum(dpois(0:3, lambda=20/3))
dbinom(x=3, size=4, prob=exp(-5/3))
dpois(1, lambda=5/3)*dpois(1, lambda=5)
```
```{r}
1-pnorm(q=3.25, mean=3, sd=sqrt(2.6/50)) # 0.1364688
1-pnorm(q=3.25, mean=3, sd=sqrt(2.6/80)) # 0.08275893
```
```{r}
dt <- data.table(a=0:50, b=cumsum(dpois(x=0:50, lambda=7.5)))
dt[b<0.98] # cu - 1 = 13; 
1-0.9784353490	# 0.02156465
dt[b<0.03]
p <- 0.0202567151 + 0.02156465

1-sum(dbinom(x=0:1, size=8, prob=p))
sum(dpois(x=3:13, lambda=6.3))
```
```{r}
(2*log(2)-1)/log(2)^2
```
```{r}
dgeomw(trials=4, prob=1/3)
sum(dgeomw(trials=1:5, prob=1/3))
1-(1-1/3)^5
```
```{r}
options(scipen = 99)
dt <- data.table(a=1:100, 
                 blue=dgeomw(trials = 1:100, prob=1/3), 
                 red=dgeomw(trial=1:100, prob=2/3))
dt %>% mutate(bluescore=a*blue, 
              redscore=a*red) 
```
```{r}
(exp(1)*1/3)/(1-exp(1)*(1-1/3))
```

```{r}
dt <- data.table(a=c(1,2,3,4,5),
                 O=c(5,21,23,13,18), 
                 E=c(5.53, 14.89, 24.26, 22.24, 13.08))
dt[,sum((O-E)^2/E)] # 8.312993

qchisq(p=0.95, df=3) # 7.814728

1/80 * (4+42+69+52+55+42)

3.3/6
```