Sujit , Nitish And Sai
Sujit Thakur
2/8/2022
Getting data
days <- c(91,105,106,108,88,91,58,82,81,65,61,48,61,43,33,36)
ind <- c(16.7,17.1,18.2,18.1,17.2,18.2,16,17.2,18,17.2,16.9,17.1,18.2,17.3,17.5,16.6)Q1) Scatter plot of data
plot(ind,days, main = "Scatter plot of Days to index ")
Q2) Least square estimate of the parameter
model <- lm(days~ind)
model##
## Call:
## lm(formula = days ~ ind)
##
## Coefficients:
## (Intercept) ind
## -193.0 15.3Estimate of our fitted regression line came out to be as intercept is “-193” and slope is “15.3”
Q3) Add Least square line to scatter plot
plot(ind,days, main = "Scatter plot of Days to index ")
abline(model)
## Q4) Model Adequacy
plot(model)
As we can see that in residual vs fitted plot we are able to see a pattern of decresing funnel shape , hence we can say our assumption of constant variance doesnt holds true here .
We can see from the normal plot that th data points fairly fall in a straight line , hence we can say that they are fairly normally distributed
We can see from the Sqrt(standardized residual) vs fitted values , we can see that there are none of the points which seems to far away and they also lie within the value of 1.5 on y axis . And hence we claim that there are no outliers which are significantlty affecting our regression model
Q5) Does Regression appers to be significant ?
using hypothesis test
\(Null Hypothesis :Ho : B_1 = 0\)
\(ALternative Hypothesis : Ha : B_1 \neq 0\)
summary(model)##
## Call:
## lm(formula = days ~ ind)
##
## 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
## ind 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