1 Question:1

Make a scatterplot of the data, Title the plot and label the axes appropriately

Index<-c(16.7,17.1,18.2,18.1,17.2,18.2,16.0,17.2,18.0,17.2,16.9,17.1,18.2,17.3,17.5,16.6)
Days<-c(91,105,106,108,88,91,58,82,81,65,61,48,61,43,33,36)
plot(Index,Days,title(main = 'Scatterplot'),xlab = 'Index', ylab = 'Days')

--> As you can see the Scatterplot with two Quantative variable. The Variable on X-axis is Index and the other variable on Y-axis is Days

2 Question 2:

What are the Least Squares estimates of the parameters of a simple linear regression?

model <- lm(Days~Index)
summary(model)
## 
## Call:
## lm(formula = Days ~ Index)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -41.70 -21.54   2.12  18.56  36.42 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)
## (Intercept) -192.984    163.503  -1.180    0.258
## Index         15.296      9.421   1.624    0.127
## 
## Residual standard error: 23.79 on 14 degrees of freedom
## Multiple R-squared:  0.1585, Adjusted R-squared:  0.09835 
## F-statistic: 2.636 on 1 and 14 DF,  p-value: 0.1267
coefficients(model)
## (Intercept)       Index 
##  -192.98383    15.29637
--> As You can see estimates.$$\beta_{0} = -192.98383$$ $$\beta_{1}=15.29637$$

3 Question 3:

Add the least squares line to the scatter plot

plot(Index,Days,title(main = 'Scatterplot'),xlab = 'Index', ylab = 'Days')

plot(Days~Index)+abline(model)

## integer(0)
--> As you can see the above is the scattered data points and line fitted to the data

4 Question 4:

Check for model adequacy using diagnostic plots. Show the plots and comment on the assumptions of Constant Variance and Normality.

plot(model,2)

plot(model,1)

i=seq(1,16,1)
plot(i,model$residuals,ylab = 'Residuals',xlab = 'i')

-->
As you can see, here from the graph we can tell that it is normal distribution because the graph is steadily increasing.

Graph2) we can see that there is a lot of difference in residual and fitted graph cause there is a sudden steep.

Graph 3) we can say that the graph is linear.

5 Question 5:

Does the regression appear to be significant? Why or why not?

summary(model)
## 
## Call:
## lm(formula = Days ~ Index)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -41.70 -21.54   2.12  18.56  36.42 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)
## (Intercept) -192.984    163.503  -1.180    0.258
## Index         15.296      9.421   1.624    0.127
## 
## Residual standard error: 23.79 on 14 degrees of freedom
## Multiple R-squared:  0.1585, Adjusted R-squared:  0.09835 
## F-statistic: 2.636 on 1 and 14 DF,  p-value: 0.1267
--> Because P - value is less than 0.05, x is not significant. Moreover there are no ** asterisks in P value