Question

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?

Answer #1

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 R^{2} = r^{2}

or

or

=> r = 0.2989

Hence, the value of pearson correlation is 0.2989 for the model.

3.) Now, you are going to run the multivariable linear
regression model you just created.
For credit: Provide your model command and summary
command below along with all the output for your model
summary.
Model1 <- lm(LifeExpect2017~HouseholdIncome + Diabetic +
FoodInsecure + Uninsured + DrugOverdoseMortalityRate )
> summary(Model1)
Call:
lm(formula = LifeExpect2017 ~ HouseholdIncome + Diabetic + FoodInsecure +
Uninsured + DrugOverdoseMortalityRate)
Residuals:
Min 1Q Median 3Q Max
-5.4550 -0.8559 0.0309 0.8038 7.1801
Coefficients:
Estimate Std. Error t value Pr(>|t|)...

3.) Now, you are going to run the multivariable linear
regression model you just created.
For credit: Provide your model command and summary
command below along with all the output for your model
summary.
Model1 <- lm(LifeExpect2017~HouseholdIncome + Diabetic + FoodInsecure + Uninsured + DrugOverdoseMortalityRate )
> summary(Model1)
Call:
lm(formula = LifeExpect2017 ~ HouseholdIncome + Diabetic + FoodInsecure +
Uninsured + DrugOverdoseMortalityRate)
Residuals:
Min 1Q Median 3Q Max
-5.4550 -0.8559 0.0309 0.8038 7.1801
Coefficients:
Estimate Std. Error t value Pr(>|t|)...

Can you give me a simple interpretation of this output?
Call:
lm(formula = NOCRF ~ Mktrf + HML + SMB + SMB2)
Residuals:
Min
1Q Median
3Q Max
-10.1560 -0.6880 -0.0254 0.6660 21.9700
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.01163 0.02800
-0.415 0.678
Mktrf 1.25614
0.02389 53.540 <2e-16 ***
HML
2.01719 0.04238 47.602 <2e-16 ***
SMB -0.05150
0.04769 -1.080
0.280
SMB2 0.03180
0.03545 0.897 0.372
---
Signif. codes: 0 ‘***’ 0.001...

Consider the following regression run in R, which uses engine
size in liters, horsepower, weight, and domestic vs foreign
manufacturer to predict mileage:
------------------------------------------------------------------------------------------------------
> summary(lm(highwaympg~displacement+hp+weight+domestic))
Call:
lm(formula = highwaympg ~ displacement + hp + weight +
domestic)
Residuals:
Min 1Q Median 3Q
Max
-6.9530 -1.6997 -0.1708 1.6452 11.4028
Coefficients:
Estimate
Std. Error t value Pr(>|t|)
(Intercept) 53.849794 2.090657 25.757 < 2e-16
***
displacement 1.460873 0.748837
1.951 0.0543 .
hp -0.009802
0.011356 -0.863 0.3904
weight -0.008700
0.001094 -7.951 6.23e-12 ***
domestic -0.939918
0.762175 -1.233 0.2208
---...

> muncy = lm(hit_distance~launch_speed, data=muncy)
> summary(muncy)
Call:
lm(formula = hit_distance ~ launch_speed, data = muncy)
Residuals:
Min 1Q
Median 3Q
Max
-258.24 -105.23 23.29 116.06 174.73
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -240.8429 36.6769 -6.567 1.46e-10
***
launch_speed 4.8800
0.4022 12.134 < 2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' '
1
Residual standard error: 122.4 on 438 degrees of freedom
Multiple R-squared: 0.2516, Adjusted R-squared: 0.2499
F-statistic:...

4.-Interpret the following regression model
Call:
lm(formula = log(Sale.Price) ~ Lot.Size + Square.Feet + Num.Baths +
dis_coast + API.2011 + dis_fwy + dis_down + Pool, data = Training)
Residuals:
Min 1Q Median 3Q Max
-2.17695 -0.23519 -0.00112 0.26471 1.02810
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 9.630e+00 2.017e-01 47.756 < 2e-16 ***
Lot.Size -2.107e-06 3.161e-07 -6.666 4.78e-11 ***
Square.Feet 2.026e-04 3.021e-05 6.705 3.71e-11 ***
Num.Baths 6.406e-02 2.629e-02 2.437 0.015031 *
dis_coast -1.827e-05 6.881e-06 -2.655 0.008077 **
API.2011 3.459e-03 2.356e-04...

Marketing date on sales is presented for youtube. data are the
advertising budget in thousands of dollars along with the sales.
The experiment has been repeated 200 times with different budgets
and the observed sales have been recorded. The simple linear
regression model was fitted:
##
## Call:
## lm(formula = sales ~ youtube, data = marketing)
##
## Residuals:
## Min 1Q Median 3Q Max
## -10.06 -2.35 -0.23 2.48 8.65
##
## Coefficients:
## Estimate Std. Error t...

Residuals:
Min 1Q
Median 3Q
Max
-6249.5 -382.9 -139.3 25.6 31164.7
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.311e+02 2.219e+02 0.591
0.5550
debt 1.283e-01
3.288e-01 0.390
0.6966
sales 2.942e-01
1.366e-01 2.154 0.0321 *
income 1.546e+01
2.697e+00 5.730 2.42e-08 ***
assets -2.390e-05 4.839e-03
-0.005 0.9961
seo
2.973e+02 2.627e+02 1.132
0.2587
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’
1
Residual standard error: 2019 on 303 degrees of freedom
Multiple R-squared: 0.258, Adjusted...

Using the following data taking out of R (summary):
Call:
lm(formula = dys_detect ~ fin_loss, data = Lab5, na.action =
na.exclude)
Residuals:
Min 1Q Median 3Q Max
-582.66 -274.75 13.53 273.92 589.06
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 385.362 77.360 4.981 8.72e-07 ***
fin_loss 3.248 1.523 2.133 0.0334 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 316.6 on 498 degrees of freedom
Multiple R-squared: 0.009052, Adjusted R-squared:...

R Linear Model Summary. Based on the R output below, answer the
following:
(a) What can infer about β0 and/or β1 ?
(b) What is the interpretation of R2
. (Non-Adjusted) ? In particular, what does it say about how
“x explains y”
(c) Perform the test (α = 0.05): H0 : ρ = 0.5; Ha : ρ > 0.5
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.32632 0.24979 1.306 0.194
x 0.09521 0.01022 9.313 2.93e-15 ***
---
Signif....

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