install.packages(“dplyr”) install.packages(“tidyr”) library(dplyr) library(tidyr)

--> Here is the given Data of Bacteria and Exposure. 
Bacteria<-c(175,108,95,82,71,50,49,31,28,17,16,11)
Exposure<-c(1,2,3,4,5,6,7,8,9,10,11,12)
dat<-data.frame(Bacteria,Exposure)
print(dat,row.names=FALSE)
##  Bacteria Exposure
##       175        1
##       108        2
##        95        3
##        82        4
##        71        5
##        50        6
##        49        7
##        31        8
##        28        9
##        17       10
##        16       11
##        11       12

1 Question 1:

Fit a simple linear regression model to the data. What Is the value of R^2?

model=lm(Bacteria~Exposure)
summary(model)
## 
## Call:
## lm(formula = Bacteria ~ Exposure)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -17.323  -9.890  -7.323   2.463  45.282 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   142.20      11.26  12.627 1.81e-07 ***
## Exposure      -12.48       1.53  -8.155 9.94e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 18.3 on 10 degrees of freedom
## Multiple R-squared:  0.8693, Adjusted R-squared:  0.8562 
## F-statistic: 66.51 on 1 and 10 DF,  p-value: 9.944e-06
summary(model)$r.square
## [1] 0.8693011
--> Here you can see, we used lm function to create regression model
As you can see the value of R Square is 0.86930 .

2 Question 2:

Check for model adequacy

plot(model)

--> Here you can see, the adequacy of the model.
As we can see, there is no random pattern in the residuals plots above against the fitted plots. We draw the conclusion that the regression model we developed in response to question no. 1 does not accurately reflect the variance in our data. More data points are required to provide the most accurate result. As can be seen, our graph is not linear since the data is not normally distributed.

3 Question 3:

Use Box-Cox to perform a power transformation, transform the data as appropriate

library(MASS)
b<-boxcox(model)

lambda<-b$x
likelihood<-b$y
z<-lambda[which.max(likelihood)]
print(z)
## [1] 0.1010101
--> Here you can see, With the help of Box-cox Transformation we eliminate the nonlinearity.
We got the value of power transform is 0.1, which is very near to zero. We can conclude that the residuals for the regression model are as normal as possible.

4 Question 4:

Fit a simple linear regression model to the transformed data. What Is the value of R^2?

y<-Bacteria^z
newmodel<-lm(y~Exposure)
summary(newmodel)
## 
## Call:
## lm(formula = y ~ Exposure)
## 
## Residuals:
##       Min        1Q    Median        3Q       Max 
## -0.026103 -0.008334 -0.003293  0.012661  0.025559 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  1.700726   0.010226  166.31  < 2e-16 ***
## Exposure    -0.034957   0.001389  -25.16 2.25e-10 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.01662 on 10 degrees of freedom
## Multiple R-squared:  0.9844, Adjusted R-squared:  0.9829 
## F-statistic: 632.9 on 1 and 10 DF,  p-value: 2.255e-10
--> Here you can see, we created newmodel.
After running the summary() command we are able to see the value of R Square that is 0.9844

5 Question 5:

Check for model adequacy using the transformed data (comment)

plot(newmodel)

--> Here you can see, the adequacy of the model.
As we can see, there is no pattern in the residuals plots above against the fitted plots. We draw the conclusion that variance is constant. may be the result is not precised as we discussed in above answer, More data points are required to provide the most accurate result. As can be seen, this transformed model (newmodel) provides better result than our previous model becuase the normality graph also shows the data points nearly in a straight line.

6 Question 6:

Estimate the number of bacteria at 10 minutes of exposure, how does this compare with the observed value?

pred<-fitted.values(newmodel,Exposure)
pred[10]^(1/z)
##       10 
## 19.67812
--> As you can see the number of Bacteria at 10 min of Exposure is 19.67812

7 Question 7:

Provide a 95% prediction interval on the number of bacteria at 10 minutes of exposure.

predict(newmodel,data.frame(exposure=10),interval="prediction")
##         fit      lwr      upr
## 1  1.665769 1.623640 1.707898
## 2  1.630812 1.589836 1.671788
## 3  1.595855 1.555826 1.635884
## 4  1.560898 1.521594 1.600202
## 5  1.525941 1.487127 1.564754
## 6  1.490984 1.452418 1.529549
## 7  1.456027 1.417461 1.494592
## 8  1.421070 1.382256 1.459883
## 9  1.386113 1.346809 1.425417
## 10 1.351156 1.311127 1.391185
## 11 1.316199 1.275223 1.357174
## 12 1.281242 1.239113 1.323371
pred_interval<-predict(newmodel,data.frame(exposure=10),interval="prediction")
pred_interval^(1/z)
##          fit        lwr       upr
## 1  156.30969 121.296382 200.15799
## 2  126.70629  98.489163 161.99169
## 3  102.24322  79.508493 130.66207
## 4   82.11213  63.790666 105.03414
## 5   65.61802  50.844880  84.14271
## 6   52.16533  40.244614  67.17052
## 7   41.24552  31.619920  53.42879
## 8   32.42594  24.650598  42.33982
## 9   25.33996  19.060242  33.42142
## 10  19.67812  14.611076  26.27301
## 11  15.18038  11.099490  20.56332
## 12  11.62914   8.352124  16.01944
--> As you can see the prediction interval on the number of bacteria at 10 minutes of exposure is: Upper= 26.27301 and Lower= 14.611076 and fit= 19.67812