Question

3.) Now, you are going to run the multivariable linear regression model you just created. For...

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|)    
(Intercept)                8.266e+01  4.016e-01 205.847  < 2e-16 ***
HouseholdIncome            5.483e-05  3.365e-06  16.291  < 2e-16 ***
Diabetic                  -4.088e-01  1.683e-02 -24.292  < 2e-16 ***
FoodInsecure              -1.542e-01  1.267e-02 -12.176  < 2e-16 ***
Uninsured                 -2.242e-02  7.041e-03  -3.184  0.00148 ** 
DrugOverdoseMortalityRate -5.240e-02  3.135e-03 -16.716  < 2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.374 on 1714 degrees of freedom
  (1422 observations deleted due to missingness)
Multiple R-squared:  0.7377,    Adjusted R-squared:  0.737 
F-statistic: 964.3 on 5 and 1714 DF,  p-value: < 2.2e-16

4.) Are all your model terms statistically significant from #3? For credit, yes or no, and if no, which ones are not significant and why?

5.) For credit: Overall, is your model significantly better than nothing? If yes, explain why using the p-value approach. If no, explain why using the p-value approach.

Homework Answers

Answer #1

4.) Are all your model terms statistically significant from #3? For credit, yes or no, and if no, which ones are not significant and why?

Ans: Yes. The all p-values are associated with the predictors are less than 0.05 i.e 5% level of significance. therefore all model term statistically significant.

5) Overall, is your model significantly better than nothing?

Ans: Yes. beacuse p value of F-statistics is p-value<2.2e-16, therefore overall model is significantly better than nothing.

 
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
3.) Now, you are going to run the multivariable linear regression model you just created. For...
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|)...
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 ‘...
4.-Interpret the following regression model Call: lm(formula = log(Sale.Price) ~ Lot.Size + Square.Feet + Num.Baths +...
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...
Residuals:     Min      1Q Median      3Q     Max -6249.5 -382.9 -139.3    25.6 31164.7 Coefficients:         &nbs
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...
Consider the following regression run in R, which uses engine size in liters, horsepower, weight, and...
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    ---...
Can you give me a simple interpretation of this output? Call: lm(formula = NOCRF ~ Mktrf...
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...
> muncy = lm(hit_distance~launch_speed, data=muncy) > summary(muncy) Call: lm(formula = hit_distance ~ launch_speed, data = muncy)...
> 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:...
If you run summary() command on a result of linear model fitting returned by lm(), you...
If you run summary() command on a result of linear model fitting returned by lm(), you will see a column Pr(>t). You could guess that it is a result of some t-test. How does it apply here? So far we have seen t-tests in the settings of testing sample locations. The ingredients were: 1) null hypothesis: in earlier cases we looked into the null that stated that two samples came from the distribution(s) with the same mean(s); 2) test statistic:...
The data set (Canvas: body.csv) contains records of CHEST_DIAM, , CHEST_DEPTH, ANKLE_DIAM,WAIST_GIRTH, WRIST_GIRTH, WRIST_DIAM (all in...
The data set (Canvas: body.csv) contains records of CHEST_DIAM, , CHEST_DEPTH, ANKLE_DIAM,WAIST_GIRTH, WRIST_GIRTH, WRIST_DIAM (all in cm.), AGE (years), WEIGHT (kg.), HEIGHT (cm.), andGENDER (1=male) for 108 individuals. We will be looking for the best set of variables to (parsimoniously?) modelWEIGHT. Even though 6 explanatory variables only gives 29=512 possibilities for “all possible” regressions, we’lltry to be more methodical about it. ##question2 library(MASS) ## ## Attaching package: 'MASS' ## The following object is masked from 'package:olsrr': ## ##     cement body =...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT