Hello my name is Diego

#install.packages("ggplot2")
library(ggplot2)

This installs the R packages needed to plot figures

#install.packages("ggplot2")
library(ggplot2)

This reads the Dataset and allows it to be accessed easily in R in a variable

#install.packages("readr")
library(readr)
bca <- read_csv("BrainCancerData1.csv") 
## Rows: 48 Columns: 5
## ── Column specification ────────────────────────────────────────────────────────
## Delimiter: ","
## dbl (5): year, rate_cases1, rate_cases2, death_rate, pct_survival
## 
## ℹ Use `spec()` to retrieve the full column specification for this data.
## ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.

Part A Question 1: Determine the data types column \(rate\_cases2\),

\(death\_rate\), and \(pct\_survival\)

# Write in the following chunk. 
# Hint: follow this syntax str(dataframe) to obtain the structure of the dataset!

str(bca)
## spc_tbl_ [48 × 5] (S3: spec_tbl_df/tbl_df/tbl/data.frame)
##  $ year        : num [1:48] 1975 1976 1977 1978 1979 ...
##  $ rate_cases1 : num [1:48] NA NA NA NA NA NA NA NA NA NA ...
##  $ rate_cases2 : num [1:48] 5.93 5.86 6.29 5.8 6.12 6.57 6.53 6.39 6.37 5.96 ...
##  $ death_rate  : num [1:48] 4.11 4.34 4.4 4.53 4.26 4.37 4.36 4.43 4.39 4.55 ...
##  $ pct_survival: num [1:48] 23.5 22.6 22.3 25.2 23.1 ...
##  - attr(*, "spec")=
##   .. cols(
##   ..   year = col_double(),
##   ..   rate_cases1 = col_double(),
##   ..   rate_cases2 = col_double(),
##   ..   death_rate = col_double(),
##   ..   pct_survival = col_double()
##   .. )
##  - attr(*, "problems")=<externalptr>
# all five columns are numeric

#All three columns are numeric.

Question 2: Index the column \(rate\_cases1\) from the dataset

# Write in the following chunk. 
# Hint: follow the syntax dataframe$column name to index a specific column 
bca$year
##  [1] 1975 1976 1977 1978 1979 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989
## [16] 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004
## [31] 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018 2019
## [46] 2020 2021 2022
# dataframe[row,column]
bca[1,] #rows
bca[,1] #columns
bca[1,1]
bca$rate_cases1 #or
##  [1]   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA
## [16]   NA   NA 6.73 6.50 6.32 6.30 6.38 6.48 6.47 6.67 6.38 6.29 6.44 6.49 6.54
## [31] 6.54 6.26 6.43 6.40 6.63 6.31 6.15 6.25 6.31 6.13 6.40 6.07 6.18 6.17 5.87
## [46] 5.90 5.67   NA
bca[,2]

Part B

Question 1: Find the mean rate of new brain cancer cases from \(rate\_cases1\)

# Below is an example function you can apply for mean calculation
avg_rate_cases<-mean(bca$rate_cases1, na.rm = TRUE)
avg_rate_cases #always call too
## [1] 6.322

Now it is your turn to calculate the mean for \(rate\_cases2\), \(death\_rate\), and \(pct\_survival\)

# Write in the following chunk:
avg_rate_cases2<-mean(bca$rate_cases2, na.rm = TRUE)
avg_rate_cases2
## [1] 6.561489
avg_death_rate<-mean(bca$death_rate, na.rm = TRUE)
avg_death_rate
## [1] 4.483958
avg_pct_suvival<-mean(bca$pct_survival, na.rm = TRUE)
avg_pct_suvival
## [1] 31.21929

Question 2: Find the standard deviation of new brain cancer cases from \(rate\_cases1\)

# Below is an example function you can apply for standard deviation calculation

sd_rate_cases<-sd(bca$rate_cases1, na.rm = TRUE)
sd_rate_cases
## [1] 0.236372

Now calculate the standard deviation for \(rate\_cases2\), \(death\_rate\), and \(pct\_survival\)

# Write in the following chunk:

sd_rate_cases2<-sd(bca$rate_cases2, na.rm = TRUE)
sd_rate_cases2
## [1] 0.3297692
sd_death_rate <- sd(bca$death_rate, na.rm = TRUE)
sd_death_rate
## [1] 0.1983092
sd_pct_survival <- sd(bca$pct_survival, na.rm = TRUE)
sd_pct_survival
## [1] 4.832159

Question 3: Find the range of new brain cancer cases from \(rate\_cases1\)

# Below is an example function you can apply for range calculation
range_rate_cases<-range(bca$rate_cases1, na.rm = TRUE)
range_rate_cases
## [1] 5.67 6.73

Now calculate the range for \(rate\_cases2\), \(death\_rate\), and \(pct\_survival\)

# Write in the following chunk:

range_rate_cases2<-range(bca$rate_cases2, na.rm = TRUE)
range_rate_cases2
## [1] 5.80 7.17
range_death_rate<-range(bca$death_rate, na.rm = TRUE)
range_death_rate
## [1] 4.11 4.95
range_death_rate<-range(bca$pct_survival, na.rm = TRUE)
range_death_rate
## [1] 22.33 37.93

Tutorial Assignment 3

#4.1: avg_rate_cases<-mean(bca$rate_cases1, na.rm = TRUE) # avg_rate_cases tells us that, on average, there are about 6.32 new brain cancer cases per unit. This value gives a central tendency for how often new brain cancer cases are occurring in the population.

#4.2: sd_rate_cases<-sd(bca$rate_cases1, na.rm = TRUE) # sd_rate_cases calculated to be 0.313. relatively low standard deviation indicates that the new brain cancer case rates are quite consistent across the dataset

#4.3: range_rate_cases<-range(bca$rate_cases1, na.rm = TRUE) # range_rate_cases from 5.67 to 6.73. narrow range supports the low variability shown by the standard deviation. A spread of just over 1 unit between the lowest and highest values indicates very little dispersion, meaning the brain cancer incidence rates are tightly clustered around the mean of 6.322.

#4.4: avg_death_rate<-mean(bca\(death_rate, na.rm = TRUE) # avg_death_rate # sd_death_rate <- sd(bca\)death_rate, na.rm = TRUE) # sd_death_rate

#mean death rate was calculated to be 4.483958, with a standard deviation of 0.1983092.Very low standard deviation shows that death rates due to brain cancer are highly consistent across the dataset. This suggests there’s little variation in fatality rates

#4.5: avg_pct_suvival<-mean(bca\(pct_survival, na.rm = TRUE) #avg_pct_suvival #sd_pct_survival <- sd(bca\)pct_survival, na.rm = TRUE) # sd_pct_survival

#The mean of pct_survival was calculated as 31.21929, while its standard deviation is 4.483958. This relatively high standard deviation (especially compared to that of the death rate and case rate) indicates that survival rates vary widely across the dataset

Part C

Question 1: a) Perform a linear regression where \(year\) is the independent variable (predictor) and \(rate\_cases1\) is the dependent variable (outcome).

# Write in the following chunk:
# Hint: follow the syntax lm(y~x, data = dataframe)

linear_reg1 <- lm(rate_cases1 ~ year, data = bca)
summary(linear_reg1)
## 
## Call:
## lm(formula = rate_cases1 ~ year, data = bca)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -0.36981 -0.08465  0.00650  0.11267  0.35665 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 45.371749   7.014320   6.468 5.24e-07 ***
## year        -0.019462   0.003496  -5.567 5.90e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.1657 on 28 degrees of freedom
##   (18 observations deleted due to missingness)
## Multiple R-squared:  0.5254, Adjusted R-squared:  0.5084 
## F-statistic: 30.99 on 1 and 28 DF,  p-value: 5.901e-06

Write down the linear regression equation: \[ ratecases1 = 45.372 - 0.0195(year) \]

  1. Find the coefficients and p-values of the intercept and slope, and interpret it respectively.

#The coefficient of the intercept is 45.372. This refers to the predicted value of \(rate\_cases1\) when \(year\) is 0. The p-value of the intercept is 5.24e-07, which is smaller than 0.05.

#The coefficient of the slope is -0.0195. This means, for every one-unit increase in \(year\), \(rate\_cases1\) decreases by 0.0195.

#The p-value of the slope is 5.90e-06, which is also smaller than

#0.05. In conclusion, we can reject the null hypothesis that the intercept or slope is 0, indicating that there is a relationship between rate cases1 and year.

  1. Now perform linear regression on \(year\) and \(pct\_survival\) and write down the equation you achieved (same steps as above!)
linear_reg2 <- lm(pct_survival~year, data = bca)
summary(linear_reg2)
## 
## Call:
## lm(formula = pct_survival ~ year, data = bca)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -4.269 -1.617 -0.091  1.784  3.142 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -682.25645   52.14369  -13.08 4.89e-16 ***
## year           0.35754    0.02613   13.68  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.053 on 40 degrees of freedom
##   (6 observations deleted due to missingness)
## Multiple R-squared:  0.824,  Adjusted R-squared:  0.8196 
## F-statistic: 187.2 on 1 and 40 DF,  p-value: < 2.2e-16

Linear regression equation: \[ pct surival = -682.256 + 0.358(year) \]

linear_reg3 <- lm(death_rate ~year, data = bca)
summary(linear_reg3)
## 
## Call:
## lm(formula = death_rate ~ year, data = bca)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -0.45346 -0.14863 -0.01998  0.16185  0.44067 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)   
## (Intercept) 11.245058   4.053187   2.774  0.00797 **
## year        -0.003383   0.002028  -1.668  0.10208   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.1947 on 46 degrees of freedom
## Multiple R-squared:  0.05704,    Adjusted R-squared:  0.03654 
## F-statistic: 2.783 on 1 and 46 DF,  p-value: 0.1021
  1. Find the coefficients and p-values of the intercept and slope, and interpret it respectively.

#The coefficient of the intercept is -682.256. This refers to the predicted value of \(pct\_survival\) when year is 0. The p-value of the intercept is 4.89e-16, which is smaller than 0.05.

#The coefficient of the slope is 0.358. This means, for every one-unit increase in \(year\), \(pct\_survival\) increases by 0.358.

#The p-value of the slope is < 2e-16, which is also smaller than 0.05. In conclusion, we can reject the null hypothesis that the intercept or slope is 0, indicating that there is a relationship between pct survival and year.

Question 2: Plot the regression of \(year\) vs \(rate\_cases1\)

library(ggplot2)

ggplot(bca,aes(x=year,y=rate_cases1))+
  geom_point()+  #this is a scatterplot 
  geom_smooth(method="lm")+ #this adds linear regression
  labs(x="Year",y="Rate of New Cases", title="Year vs. New Case Rate")
## `geom_smooth()` using formula = 'y ~ x'
## Warning: Removed 18 rows containing non-finite outside the scale range
## (`stat_smooth()`).
## Warning: Removed 18 rows containing missing values or values outside the scale range
## (`geom_point()`).

ggplot(bca,aes(x=year,y=rate_cases1)) +
  geom_point(color = "red", size = 3)+
  geom_smooth(method = "lm") + 
  labs(x = "Year", y = "Rate of New Cases", title = "Year vs New Case Rate") +theme_bw()
## `geom_smooth()` using formula = 'y ~ x'
## Warning: Removed 18 rows containing non-finite outside the scale range
## (`stat_smooth()`).
## Warning: Removed 18 rows containing missing values or values outside the scale range
## (`geom_point()`).

Now plot the regression of \(year\) vs \(pct\_survival\)

ggplot(data = bca,
       aes(x = year, y = pct_survival, color = pct_survival)) +
  scale_color_viridis_c()+
  geom_point(size = 2)+
  geom_smooth(method = "lm", color = "green")+
  labs(x = "Year", y = "PCT Survival Rate", title ="Year vs PCT survival rate")+
  theme_classic()
## `geom_smooth()` using formula = 'y ~ x'
## Warning: Removed 6 rows containing non-finite outside the scale range
## (`stat_smooth()`).
## Warning: Removed 6 rows containing missing values or values outside the scale range
## (`geom_point()`).

Tutorial 4 Assignment

ggplot(data = bca,
       aes(x = year, y = death_rate, color = death_rate)) +
  scale_color_viridis_c()+
  geom_point(size = 2)+
  geom_smooth(method = "lm", color = "purple")+
  labs(x = "Year", y = "Death Rate ", title ="Year vs Death Rate")+
  theme_classic()
## `geom_smooth()` using formula = 'y ~ x'