8.) Now, do a simple linear regression model for LifeExpect2017 vs. AverageDailyPM2.5. For credit, provide the summary output for this simple linear regression model.
> Model2 <- lm(LifeExpect2017~ AverageDailyPM2.5)
> summary(Model2)
Call:
lm(formula = LifeExpect2017 ~ AverageDailyPM2.5)
Residuals:
Min 1Q Median 3Q Max
-17.1094 -1.7516 0.0592 1.7208 18.4604
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 81.6278 0.2479 329.23 <2e-16 ***
AverageDailyPM2.5 -0.4615 0.0267 -17.29 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 2.837 on 3042 degrees of freedom
(98 observations deleted due to missingness)
Multiple R-squared: 0.08944, Adjusted R-squared: 0.08914
F-statistic: 298.8 on 1 and 3042 DF, p-value: < 2.2e-16
9.) For credit: For the PM2.5 model, what is the R-square value for the model for #8 and how is this value related to the correlation test Pearson r value?
Solution:
We are given the summary output of a simple linear regression model for LifeExpect2017 vs. AverageDailyPM2.5 and we have to determine the R square value and how does it related to pearson correlation r value.
From the summary output, we see that the R square value is 0.08944, this means that 8.94% of total vaariation of response variable is esxplained by the independent variable.
The R square value is also define as the square of correlation coeffiecient i.e
R Square = (correlation coeficient )2
or R2 = r2
or
or
=> r = 0.2989
Hence, the value of pearson correlation is 0.2989 for the model.
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