Question

Here are the results of a regression of Car Deaths in the UK by month from...

Here are the results of a regression of Car Deaths in the UK by month from Jan 1969 to Dec 1984 on a dummy variable: 0 = no seatbelt law, and 1 = seat belt law (the law was instituted in February 1983)

Coefficients:

Estimate Std. Error t value Pr(> | t |)
(Intercept) 125.870 1.849 68.082 < 2e-16 ***
Seatbelts -25.609 5.342 -4.794 3.29e-06 ***

R-squared = 0.11

Justify your answer with numbers.

#1. How would you justify your answer with numbers?

#2. Is there a need to add more variables to the model?

#3. Did the seat belt law make a difference?

Homework Answers

Answer #1

Que.1

Least square regression equation is,

car deaths = 125.87 - 25.609 * Seat belts

Que.2

Since our fitted model explain only 11% (R2) variation in the car deaths, we need to add more variable to the model in order to explain more variation in the car deaths in the UK.

Que.3

For seat belt variable,

test statistic , t = -4.794 and

p-value = 3.29e-06 which is less than 0.05, hence we reject null hypothesis () and conclude that seat belt variable has significant effect on car deaths. It means that seat belt law make a difference.

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
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...
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|)...
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|)...