Question

# 8.) Now, do a simple linear regression model for LifeExpect2017 vs. AverageDailyPM2.5. For credit, provide the...

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.

#### Earn Coins

Coins can be redeemed for fabulous gifts.