Statistical Inference Course Project

Synopsis

These sheet has one goal :

Basic Inferential Data Analysis

Analysis of the ToothGrowth data in the R datasets package.

Initialization code

knitr::opts_chunk$set(echo = TRUE)
library(ggplot2)
library(tidyr)
library(rstatix)

## 
## Attachement du package : 'rstatix'

## L'objet suivant est masqué depuis 'package:stats':
## 
##     filter

library(ggpubr)
# Load data
data("ToothGrowth")
#Sample of data grouped by the variable dose
ToothGrowth %>% sample_n_by(dose, size = 3)

## # A tibble: 9 × 3
##     len supp   dose
##   <dbl> <fct> <dbl>
## 1  16.5 OJ      0.5
## 2  10   OJ      0.5
## 3  10   VC      0.5
## 4  15.5 VC      1  
## 5  14.5 VC      1  
## 6  26.4 OJ      1  
## 7  33.9 VC      2  
## 8  26.7 VC      2  
## 9  29.5 VC      2

# Quick summary of the data
summary(ToothGrowth)

##       len        supp         dose      
##  Min.   : 4.20   OJ:30   Min.   :0.500  
##  1st Qu.:13.07   VC:30   1st Qu.:0.500  
##  Median :19.25           Median :1.000  
##  Mean   :18.81           Mean   :1.167  
##  3rd Qu.:25.27           3rd Qu.:2.000  
##  Max.   :33.90           Max.   :2.000

# Exploration of ToothGrowth$dose and ToothGrowth$len, since we know exactly what ToothGrowth$supp contains
unique(ToothGrowth$dose)

## [1] 0.5 1.0 2.0

unique(ToothGrowth$len)

##  [1]  4.2 11.5  7.3  5.8  6.4 10.0 11.2  5.2  7.0 16.5 15.2 17.3 22.5 13.6 14.5
## [16] 18.8 15.5 23.6 18.5 33.9 25.5 26.4 32.5 26.7 21.5 23.3 29.5 17.6  9.7  8.2
## [31]  9.4 19.7 20.0 25.2 25.8 21.2 27.3 22.4 24.5 24.8 30.9 29.4 23.0

Exploratory data analysis

Below, the code for the base of each graph.

my_plot_fun <- function(my_data,my_x) {
  my_data <- as.data.frame(my_data)
  my_data %>%
  ggplot(aes(x=my_x, y=len)) +
theme(plot.title = element_text(color="#ADA717", size=18, face="bold.italic",hjust=0.5),
 axis.title.x = element_text(color="#993333", size=18, face="bold"),
 axis.text.x = element_text(color="#993333", size=14,vjust = 0),
 axis.title.y = element_text(color="darkgreen", size=18, face="bold"),
 axis.text.y = element_text(face="bold", color="darkgreen", size=14),
 legend.text = element_text(size=12),
 legend.title = element_text(size=16))}

Box-plots from data, using `dose` and `len`, grouped by `supp`.

# Conversion of ToothGrowth$dose into a factor
ToothGrowth$dose<-as.factor(ToothGrowth$dose)

mpt <-my_plot_fun(ToothGrowth,my_x = ToothGrowth$dose) + 
  geom_boxplot(aes(color = supp,group=dose), width = 0.6) +
  geom_dotplot(aes(fill = as.factor(dose), color = supp,group=dose), binaxis='y', stackdir='center', 
               dotsize = 0.8,position = position_dodge(0.8),binwidth=1)+
  scale_fill_manual( values = c("#00AFBB", "#E3B166","#A1A861"))+
  scale_color_manual(values = c("red", "black")) + facet_grid(~ supp)+
  labs(x = "Doses", y ="Toothe Length", fill = "Doses", color="Supplement delivery",
    title = "Tooth length \n by dose amount",
  caption = "Plot tooth length ('len') by the dose amount ('dose'), \n grouped by supplement delivery method ('supp')") +
  theme(plot.caption = element_text(color = "darkblue", face = "italic", size = 12))
  
  mpt

We see this :

for the same supplement delivery method, the more the dose increases the more the tooth length increases.

Box-blots from data, using `supp` and `len`, grouped by `dose`.

# Conversion of ToothGrowth$dose into a factor
ToothGrowth$dose<-as.factor(ToothGrowth$dose)

mpt2 <-my_plot_fun(ToothGrowth,my_x = ToothGrowth$supp) + 
  geom_boxplot(aes(color = dose,group=supp), width = 0.6) +
  geom_dotplot(aes(fill = as.factor(supp), color = dose,group=supp), binaxis='y', stackdir='center', 
               dotsize = 0.8,position = position_dodge(0.8),binwidth=1)+
  scale_fill_manual( values = c("#00AFBB", "#E3B166"))+
  scale_color_manual(values = c("red", "black","darkgreen")) + facet_grid(~ dose)+
  labs(x = "Supplement delivery method", y ="Toothe Length", fill = "Supplement delivery", color="Doses",
    title = "Tooth length \n by supplement delivery method",
  caption = "Plot tooth length ('len') by the supplement delivery method  ('supp'), \n grouped by dose amount ('dose')") +
  theme(plot.caption = element_text(color = "darkblue", face = "italic", size = 12))
  
  mpt2

We see this:

the more the dose increases, for the same intake of supplement, the more the tooth length increases.

Hypotheses tests

First question : does the supplement method increase tooth size ?

There is two groups in supp, two samples :

tooth size with supplement delivery OJ ;

tooth size with supplement delivery VC.

Let see some statistics for each one of this groups

ToothGrowth %>% group_by(supp) %>%
  get_summary_stats(len,type = "mean_sd")

## # A tibble: 2 × 5
##   supp  variable     n  mean    sd
##   <fct> <chr>    <dbl> <dbl> <dbl>
## 1 OJ    len         30  20.7  6.61
## 2 VC    len         30  17.0  8.27

ToothGrowth %>% levene_test(len ~ supp)

## # A tibble: 1 × 4
##     df1   df2 statistic     p
##   <int> <int>     <dbl> <dbl>
## 1     1    58      1.21 0.275

The p-value of the LEVENE test is not significant, which suggests that there is not a significant difference between the variances of the two groups. Indeed the p-value of LEVENE is .275 > .05. Therefore, we will use STUDENT’s T-test, which assumes the equality of the two variances.
Extreme Data Determination

ToothGrowth %>% group_by(supp) %>%
  identify_outliers(len)

## [1] supp       len        dose       is.outlier is.extreme
## <0 lignes> (ou 'row.names' de longueur nulle)

The conclusion of this test is there is no extreme value.
Check the normality assumption by group

# Test d de SHAPIRO-WILK par groupes
ToothGrowth %>% group_by(supp) %>%
  shapiro_test(len)

## # A tibble: 2 × 4
##   supp  variable statistic      p
##   <fct> <chr>        <dbl>  <dbl>
## 1 OJ    len          0.918 0.0236
## 2 VC    len          0.966 0.428

# QQ-plot by group
ToothGrowth %>% ggqqplot(x = "len",facet.by = "supp")

3. The conclusion of these two tests is the two groups are normally distributed. However, the group VC has better normality than OJ.

Calculations : we assume independent groups

We want to know if the length of the teeth does not depend on the supplement delivery.

\[H_{0} : There ~is ~no ~difference ~between ~the ~two ~groups\]

\[H_{A} : There ~is ~a ~difference ~between ~the ~two ~groups\]

T-test :

my_stats <- ToothGrowth %>% group_by(dose) %>%
  t_test(formula = len ~ supp,var.equal = TRUE, detailed = TRUE, paired = FALSE)
my_stats

## # A tibble: 3 × 16
##   dose  estimate estimate1 estimate2 .y.   group1 group2    n1    n2 statistic
## * <fct>    <dbl>     <dbl>     <dbl> <chr> <chr>  <chr>  <int> <int>     <dbl>
## 1 0.5     5.25        13.2      7.98 len   OJ     VC        10    10    3.17  
## 2 1       5.93        22.7     16.8  len   OJ     VC        10    10    4.03  
## 3 2      -0.0800      26.1     26.1  len   OJ     VC        10    10   -0.0461
## # … with 6 more variables: p <dbl>, df <dbl>, conf.low <dbl>, conf.high <dbl>,
## #   method <chr>, alternative <chr>

dose .5 : T-value of 3.1697, we get a p-value of 0.0053 ;
dose 1.0 : T-value of 4.0328, we get a p-value of 7.81^{-4};
dose 2.0 : T-value of -0.0461, we get a p-value of 0.964 ;

Therefor, the p-value is strongly significant (\(p~<~.01\)) for dose .5 and dose 1.0. And is not significant for the dose 2.0 (\(p~>~.10\)).

COHEN’s d :

subset(ToothGrowth, ToothGrowth$dose %in% c(.5)) %>% cohens_d(formula = len ~ supp, var.equal = TRUE, paired = FALSE)

## # A tibble: 1 × 7
##   .y.   group1 group2 effsize    n1    n2 magnitude
## * <chr> <chr>  <chr>    <dbl> <int> <int> <ord>    
## 1 len   OJ     VC        1.42    10    10 large

subset(ToothGrowth, ToothGrowth$dose %in% c(1)) %>% cohens_d(formula = len ~ supp, var.equal = TRUE, paired = FALSE)

## # A tibble: 1 × 7
##   .y.   group1 group2 effsize    n1    n2 magnitude
## * <chr> <chr>  <chr>    <dbl> <int> <int> <ord>    
## 1 len   OJ     VC        1.80    10    10 large

subset(ToothGrowth, ToothGrowth$dose %in% c(2)) %>% cohens_d(formula = len ~ supp, var.equal = TRUE, paired = FALSE)

## # A tibble: 1 × 7
##   .y.   group1 group2 effsize    n1    n2 magnitude 
## * <chr> <chr>  <chr>    <dbl> <int> <int> <ord>     
## 1 len   OJ     VC     -0.0206    10    10 negligible

The “d” statistic redefines the difference of means as the number of standard deviations between these means.
Here :
- d is 1.4175476 for dose .5, which is saying there is a strong effect ;
- d is 1.8035094 for dose 1.0, which is saying there is a strong effect ;
- d is -0.0206327 for dose 2.0, which is saying there is a negligible effect.

First conclusion

The mean of the group OJ and is 20.663 with a standard deviation of 6.606. Whereas, the mean of the group VC is 16.963 with a standard deviation of 8.266. The STUDENT’s T-test show there is :

for dose .5 : strong evidence against \(H_{0}\) given that \(p\) value is 0.0053 (less than \(.01\)) with \(t\) value 3.17 and a degree of freedom \(df~=~\) 18. The difference is statistically strongly significant (strong evidence) ;
for dose 1.0 : strong evidence against \(H_{0}\) given that \(p\) value is 8^{-4} (less than \(.01\)) with \(t\) value 4.033 and a degree of freedom \(df~=~\) 18. The difference is statistically strongly significant (strong evidence) ;
for dose 2.0 : no evidence against \(H_{0}\) given that \(p\) value is 0.964 (more than \(.10\)) with \(t\) value -0.046 and a degree of freedom \(df~=~\) 18. The difference is statistically not significant (no evidence).

Finally, we consider \(H_{0}\) true for `dose 2.0` and false for `dose .5` and `1.0`. In addition to what was said above, we can notice that : - for `dose .5` : the value 0 is excluded of the confidence interval of 95% which is [1.7702617,8.7297383] ;

for dose .5 : the value 0 is excluded of the confidence interval of 95% which is [2.8406919,9.0193081] ;
for dose .5 : the value 0 is included in the confidence interval of 95% which is [-3.7229985,3.5629985].
But it can perhaps be a \(Type ~II\) error for dose 2.0 and \(Type ~I\) error for dose .5 or 1.0.

my_stats <- my_stats %>% add_xy_position(x = "supp")
my_bxp <- ggboxplot(ToothGrowth, x = "supp", y = "len", add = "jitter", facet.by = "dose",
  fill = "supp", palette = c("#00AFBB", "#E7B800")) +
  theme(legend.position="bottom") +
  stat_pvalue_manual(my_stats, tip.length = 0, label = "T-test p = {p}") +
  scale_y_continuous(expand = expansion(mult = c(0.1, 0.15))) +
  labs(x = "Supplement delivery", y ="Tooth length", fill = "Supplement delivery",
    title = "Boxplot of the tooth length by group (supp) \n",
  caption = "The plot resume some important statistics values and show \n
   the distribution of the observations") +
  theme(plot.caption = element_text(color = "darkblue", face = "italic", size = 12))
my_bxp

Second question : does the dose increase tooth size ?

There is three groups in dose, three samples :

tooth length with dose .5 ;
tooth length with dose 1 ;
tooth length with dose 2.

Important note : here, it’s possible to compare the three groups (one for each dose) at the same time. But, maybe it’s easier to investigate two by two.This is the choice made below.

Method of conducting the investigation

We will compare one group to another two by two. In each group, the research hypothesis will be :

\[H_{0} : There ~is ~no ~difference ~between ~the ~two ~groups\]

\[H_{A} : There ~is ~a ~difference ~between ~the ~two ~groups\]

Let see some statistics for each one of this three groups

ToothGrowth %>% group_by(dose) %>%
  get_summary_stats(len,type = "mean_sd")

## # A tibble: 3 × 5
##   dose  variable     n  mean    sd
##   <fct> <chr>    <dbl> <dbl> <dbl>
## 1 0.5   len         20  10.6  4.5 
## 2 1     len         20  19.7  4.42
## 3 2     len         20  26.1  3.77

ToothGrowth_sub1 <- subset(ToothGrowth, ToothGrowth$dose %in% c(.5,1))
ToothGrowth_sub2 <- subset(ToothGrowth, ToothGrowth$dose %in% c(1.0,2.0))
ToothGrowth_sub2$dose<-as.factor(ToothGrowth_sub2$dose)
ToothGrowth_sub3 <- subset(ToothGrowth, ToothGrowth$dose %in% c(.5,2.0))
ToothGrowth_sub3$dose<-as.factor(ToothGrowth_sub3$dose)
# dose .5 and 1.0
ToothGrowth_sub1 %>% levene_test(len ~ dose)

## # A tibble: 1 × 4
##     df1   df2 statistic     p
##   <int> <int>     <dbl> <dbl>
## 1     1    38     0.287 0.595

# dose 1.0 and 2.0
ToothGrowth_sub2 %>% levene_test(len ~ dose)

## # A tibble: 1 × 4
##     df1   df2 statistic     p
##   <int> <int>     <dbl> <dbl>
## 1     1    38      1.61 0.212

# dose .5 and 2.0
ToothGrowth_sub3 %>% levene_test(len ~ dose)

## # A tibble: 1 × 4
##     df1   df2 statistic     p
##   <int> <int>     <dbl> <dbl>
## 1     1    38     0.300 0.587

The three p-value of the LEVENE test are not significant, which suggests that there is not a significant difference between the variances of the two groups. Indeed the p-values of LEVENE are 0.595, 0.212 and 0.587. Therefore, we will use STUDENT’s T-test, which assumes the equality of the two variances.

Extreme Data Determination

ToothGrowth %>% group_by(dose) %>%
  identify_outliers(len)

## # A tibble: 1 × 5
##   dose    len supp  is.outlier is.extreme
##   <fct> <dbl> <fct> <lgl>      <lgl>     
## 1 0.5    21.5 OJ    TRUE       FALSE

The conclusion of this test is there is only one extreme value, with the group dose = .5.

Check the normality assumption by group

# Test d de SHAPIRO-WILK par groupes
ToothGrowth %>% group_by(dose) %>%
  shapiro_test(len)

## # A tibble: 3 × 4
##   dose  variable statistic     p
##   <fct> <chr>        <dbl> <dbl>
## 1 0.5   len          0.941 0.247
## 2 1     len          0.931 0.164
## 3 2     len          0.978 0.902

# QQ-plot by group
ToothGrowth %>% ggqqplot(x = "len",facet.by = "dose")

The conclusion of these two tests is the two groups are normally distributed. However, the group VC has better normality than OJ.

Calculations : we assume independent groups

T-test :

# dose .5 and 1.0
ToothGrowth_sub1 %>% t_test(formula = len ~ dose,var.equal = TRUE)

## # A tibble: 1 × 8
##   .y.   group1 group2    n1    n2 statistic    df           p
## * <chr> <chr>  <chr>  <int> <int>     <dbl> <dbl>       <dbl>
## 1 len   0.5    1         20    20     -6.48    38 0.000000127

# dose 1.0 and 2.0
ToothGrowth_sub2$dose<-as.numeric(ToothGrowth_sub2$dose)
ToothGrowth_sub2 %>% t_test(formula = len ~ dose,var.equal = TRUE)

## # A tibble: 1 × 8
##   .y.   group1 group2    n1    n2 statistic    df         p
## * <chr> <chr>  <chr>  <int> <int>     <dbl> <dbl>     <dbl>
## 1 len   2      3         20    20     -4.90    38 0.0000181

# dose .5 and 2.0
ToothGrowth_sub3$dose<-as.numeric(ToothGrowth_sub3$dose)
ToothGrowth_sub3 %>% t_test(formula = len ~ dose,var.equal = TRUE)

## # A tibble: 1 × 8
##   .y.   group1 group2    n1    n2 statistic    df        p
## * <chr> <chr>  <chr>  <int> <int>     <dbl> <dbl>    <dbl>
## 1 len   1      3         20    20     -11.8    38 2.84e-14

group dose .5 and 1.0 : with a T-value -6.4766, we get a p-value of 1.27^{-7}. Therefor, the p-value is significant (\(p~<~.01\)) ;
group dose 1.0 and 2.0 : with a T-value -4.9005, we get a p-value of 1.81^{-5}. Therefor, the p-value is significant (\(p~<~.01\)) ;
group dose .5 and 2.0 : with a T-value -11.799, we get a p-value of 2.84^{-14}. Therefor, the p-value is significant (\(p~<~.01\)).

COHEN’s d :

ToothGrowth %>% cohens_d(formula = len ~ dose, var.equal = TRUE)

## # A tibble: 3 × 7
##   .y.   group1 group2 effsize    n1    n2 magnitude
## * <chr> <chr>  <chr>    <dbl> <int> <int> <ord>    
## 1 len   0.5    1        -2.05    20    20 large    
## 2 len   0.5    2        -3.73    20    20 large    
## 3 len   1      2        -1.55    20    20 large

The “d” statistic redefines the difference of means as the number of standard deviations between these means. Here, d values are :
- -2.0480958, which is saying, there is a strong effect ;
- -3.7311859, which is saying, there is a strong effect ;
- -1.5496692, which is saying, there is a strong effect.

Second conclusion

The mean of the group dose = .5 is 10.605 with a standard deviation of 4.5. Whereas, the mean of the group dose = 1.0 is 19.735 with a standard deviation of 4.415. Finally, the mean of the group dose = 2.0 is 26.1 with a standard deviation of 3.774.

The STUDENT’s T-test shows :

dose .5 and 1.0 : there is strong evidence against \(H_{0}\) given that the \(p\) 1.27^{-7} value is with \(t\) value -6.477 and a degree of freedom \(df~=~\) 38. The difference is statistically strongly significant (strong evidence).
dose 1.0 and 2.0 : there is strong evidence against \(H_{0}\) given that the \(p\) 1.81^{-5} value is with \(t\) value -4.9 and a degree of freedom \(df~=~\) 38. The difference is statistically strongly significant (strong evidence).
dose 1.0 and 2.0 : there is strong evidence against \(H_{0}\) given that the \(p\) 2.84^{-14} value is with \(t\) value -11.799 and a degree of freedom \(df~=~\) 38. The difference is statistically strongly significant (strong evidence).

Finally, we consider \(H_{0}\) false in each test. In addition to what was said above, we can notice that the value 0 is not included in the confidence intervals of 95% which are respectively [-11.9837477,-6.2762523], [-8.9943868,-3.7356132] and [-18.153519,-12.836481].

But it can perhaps be a Type I error, with a very low probability.

all_stats <- ToothGrowth %>% t_test(formula = len ~ dose,var.equal = TRUE, detailed = TRUE,paired = FALSE)
all_stats <- all_stats %>% add_xy_position(x = "dose")
all_stats <- all_stats %>% mutate(y.position = c(29, 35, 39))
my_bxp2 <- ggboxplot(ToothGrowth, x = "dose", y = "len", add = "jitter",
  fill = "dose", palette = c("#00AFBB", "#E7B800","#CF2406")) +
  theme(legend.position="bottom") +
  stat_pvalue_manual(all_stats, tip.length = 0, label = "T-test, p = {p}") +
  labs(x = "Dose", y ="Tooth length", fill = "Dose",
    title = "Boxplot of the tooth length by group (dose) \n",
  caption = "The plot resume some important statistics values and show \n
   the distribution of the observations") +
  theme(plot.caption = element_text(color = "darkblue", face = "italic", size = 12))
my_bxp2

Statistical Inference Course Project - Part II

Idriss .S

25/02/2022

Synopsis

These sheet has one goal :

Basic Inferential Data Analysis

Initialization code

Exploratory data analysis

Below, the code for the base of each graph.

Box-plots from data, using `dose` and `len`, grouped by `supp`.

We see this :

for the same supplement delivery method, the more the dose increases the more the tooth length increases.

Box-blots from data, using `supp` and `len`, grouped by `dose`.

We see this:

the more the dose increases, for the same intake of supplement, the more the tooth length increases.

Hypotheses tests

First question : does the supplement method increase tooth size ?

Let see some statistics for each one of this groups

Extreme Data Determination

Check the normality assumption by group

Calculations : we assume independent groups

\[H_{0} : There ~is ~no ~difference ~between ~the ~two ~groups\]

\[H_{A} : There ~is ~a ~difference ~between ~the ~two ~groups\]

First conclusion

Finally, we consider \(H_{0}\) true for `dose 2.0` and false for `dose .5` and `1.0`. In addition to what was said above, we can notice that : - for `dose .5` : the value 0 is excluded of the confidence interval of 95% which is [1.7702617,8.7297383] ;

But it can perhaps be a \(Type ~II\) error for `dose 2.0` and \(Type ~I\) error for `dose .5` or `1.0`.

Second question : does the dose increase tooth size ?

Method of conducting the investigation

\[H_{0} : There ~is ~no ~difference ~between ~the ~two ~groups\]

\[H_{A} : There ~is ~a ~difference ~between ~the ~two ~groups\]

Let see some statistics for each one of this three groups

Extreme Data Determination

Check the normality assumption by group

Calculations : we assume independent groups

Second conclusion

Finally, we consider \(H_{0}\) false in each test. In addition to what was said above, we can notice that the value 0 is not included in the confidence intervals of 95% which are respectively [-11.9837477,-6.2762523], [-8.9943868,-3.7356132] and [-18.153519,-12.836481].

But it can perhaps be a Type I error, with a very low probability.

Statistical Inference Course Project - Part II

Idriss .S

25/02/2022

Synopsis

These sheet has one goal :

Basic Inferential Data Analysis

Initialization code

Exploratory data analysis

Below, the code for the base of each graph.

Box-plots from data, using dose and len, grouped by supp.

We see this :

for the same supplement delivery method, the more the dose increases the more the tooth length increases.

Box-blots from data, using supp and len, grouped by dose.

We see this:

the more the dose increases, for the same intake of supplement, the more the tooth length increases.

Hypotheses tests

First question : does the supplement method increase tooth size ?

Let see some statistics for each one of this groups

Extreme Data Determination

Check the normality assumption by group

Calculations : we assume independent groups

\[H_{0} : There ~is ~no ~difference ~between ~the ~two ~groups\]

\[H_{A} : There ~is ~a ~difference ~between ~the ~two ~groups\]

First conclusion

Finally, we consider \(H_{0}\) true for dose 2.0 and false for dose .5 and 1.0. In addition to what was said above, we can notice that : - for dose .5 : the value 0 is excluded of the confidence interval of 95% which is [1.7702617,8.7297383] ;

But it can perhaps be a \(Type ~II\) error for dose 2.0 and \(Type ~I\) error for dose .5 or 1.0.

Second question : does the dose increase tooth size ?

Method of conducting the investigation

\[H_{0} : There ~is ~no ~difference ~between ~the ~two ~groups\]

\[H_{A} : There ~is ~a ~difference ~between ~the ~two ~groups\]

Let see some statistics for each one of this three groups

Extreme Data Determination

Check the normality assumption by group

Calculations : we assume independent groups

Second conclusion

Finally, we consider \(H_{0}\) false in each test. In addition to what was said above, we can notice that the value 0 is not included in the confidence intervals of 95% which are respectively [-11.9837477,-6.2762523], [-8.9943868,-3.7356132] and [-18.153519,-12.836481].

But it can perhaps be a Type I error, with a very low probability.

Box-plots from data, using `dose` and `len`, grouped by `supp`.

Box-blots from data, using `supp` and `len`, grouped by `dose`.

Finally, we consider \(H_{0}\) true for `dose 2.0` and false for `dose .5` and `1.0`. In addition to what was said above, we can notice that : - for `dose .5` : the value 0 is excluded of the confidence interval of 95% which is [1.7702617,8.7297383] ;

But it can perhaps be a \(Type ~II\) error for `dose 2.0` and \(Type ~I\) error for `dose .5` or `1.0`.