library(readr)
d12_16<-read_csv("~/Downloads/exp12-16.csv")
## Rows: 16 Columns: 2
## ── Column specification ────────────────────────────────────────────────────────
## Delimiter: ","
## dbl (2): x, y
## 
## 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.
x<-d12_16$x
y<-d12_16$y
plot(x, y)+abline(lm(y ~ x, data = d12_16), col = "blue")

## integer(0)

###It does seem like the data has a linear relationship, thus it does support the use of simplear linear regression model.

LinearMod<-lm(y~x, data=d12_16)

Point Estimate of Slope and Intercept: -1163.5 and 245.2

-1163.5+245.2(50)=11,096.5

summary(LinearMod)
## 
## Call:
## lm(formula = y ~ x, data = d12_16)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -224.80 -153.36   -6.17  120.30  388.90 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -1163.45     783.14  -1.486 0.175682    
## x             245.15      45.91   5.340 0.000694 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 205.3 on 8 degrees of freedom
##   (6 observations deleted due to missingness)
## Multiple R-squared:  0.7809, Adjusted R-squared:  0.7535 
## F-statistic: 28.52 on 1 and 8 DF,  p-value: 0.0006941

SD for Intercept: 783.14, SD for x: 45.91

Proportion is 0.7535, thus 75.35% of the proportion variation between runoff and rainfall can be explained through this model.

12.19

d12_19<-read_csv("~/Downloads/exp12-19.csv")
## Rows: 16 Columns: 2
## ── Column specification ────────────────────────────────────────────────────────
## Delimiter: ","
## dbl (2): x, y
## 
## 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.
x<-d12_19$x
y<-d12_19$y

LM<-lm(y~x, data=d12_19)
summary(LM)
## 
## Call:
## lm(formula = y ~ x, data = d12_19)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -2.0507 -0.9754 -0.4972  0.8232  3.5319 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -12.7369     7.8359  -1.625    0.126    
## x             1.1942     0.1135  10.525 4.94e-08 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.61 on 14 degrees of freedom
## Multiple R-squared:  0.8878, Adjusted R-squared:  0.8798 
## F-statistic: 110.8 on 1 and 14 DF,  p-value: 4.94e-08

y=-12.7369+1.1942x

y=-12.7369+1.1942(250)= 285.8131

50*1.1942=59.71

Estimated regression line cannot be used to predict the emission rate for a liberation rate of 500 because the value does not lie within the range of x values causing extrapolation.

12.35 ### B_1=(2759.6-((222.1)(193))/17)/(3056.69-((222.1)^2/17))=1.536 B_0=((193/17)-(1.536(222.1/17))) SSE=2975-(-8.7145)193-1.536(2759.6)=418.1529 sd=5.2799 SEslope=0.4241 CI=(0.6327, 2.4393 ###H0:B_1=0 Ha:B_1≠0 t=((1.536)/0.4241)=3.6218 ### It would not be sensible to use linear model for predicting percentage of nausea when the does equals 5 as this dose is out of range (extrapolation) ###B_1=(2744.6-((216.1)(190.5))/16)/(3020.69-((216.1)^2/16))=1.683 B_0=((216.1/16)-(1.638(190.5/16))) SSE=2968.75-(-10.83)190.5-1.683(2744.6)=412.7 sd=5.4294 SEslope=0.5376 CI=(0.5298, 2.8362)