You are on the market for a new car. You want to check whether there is a significant difference between the fuel economy of mid-size domestic cars and mid-size import cars. You sample 17 domestic car makes and find an average fuel economy of 34.904 MPG with a standard deviation of 4.6729 MPG. For imports, you sample 15 cars and find an average MPG of 28.563 MPG with a standard deviation of 8.4988. Construct a 90% confidence interval for the difference between the true average fuel economies in question. Assume the difference will represent (domestic - import). You can also assume that the standard deviations are statistically the same between the two populations.

Step-by-step explanation:

Hello!

The objective is to test if there is a difference between the fuel economy of mid-size domestic cars and mid-size import cars.

For this there are two samples taken:

X₁: Fuel economy of a domestic car.

Sample 1

n₁= 17 domestic cars

X[bar]₁= 34.904 MPG

S₁= 4.6729 MPG

X₂: Fuel economy of an import car.

Sample 2

n₂= 15 import cars

X[bar]₂= 28.563 MPG

S₂= 8.4988 MPG

To estimate the difference between the average economic fuel of domestic cars and import cars, assuming both variables have a normal distribution and both population variances are unknown but equal, the statistic to use is a t-test for two independent samples with pooled sample variance:

(X[bar]₁-X[bar]₂)±

Sa= 6.73

(34.904-28.563)±

6.341±1.697*2.38

[2.30;10.38]

With a confidence level of 90%, you'd expect that the difference between the average economic fuel of domestic cars and import cars will be contained in the interval [2.30;10.38].

I hope it helps!

answer: add the !

step-by-step explanation: then we will answer

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497