A national consumer agency selected independent random samples of 45 owners of newer cars (less than five years old) and 40 owners of older cars (more than five years old) to estimate the difference in mean dollar cost of yearly routine maintenance, such as oil changes, tire rotations, filters, and wiper blades. The agency found the mean dollar cost per year for newer cars was $195 with a standard deviation of $46. For older cars, the mean was $286 with a standard deviation of $58. Which of the following represents the 95 percent confidence interval to estimate the difference (newer minus older) in the mean dollar cost of routine maintenance between newer and older cars?
A. (195 - 286) + 1.992 underroot46/45 + 58/40
B. (286 - 195) + 1.992 underroot46^2/45+ 58^2/40
C. (195 - 286)+ 1.992 underrot 140² 158²/45+40
D. (286 - 195) + 1.992 46 1582 45 +40
E. (195-286) +1.992 462 + 58 45 40

Answer:

(195 - 286) ± 1.992 * sqrt((46^2/45) + (58^2/40))

Step-by-step explanation:

Given that:

NEWER CARS:

Sample size = n1 = 45

Standard deviation s1 = 46

Mean = m1 = 195

OLDER CARS:

Sample size = n2 = 40

Standard DEVIATION s2 = 58

Mean = m2 = 286

Confidence interval at 95% ; α = 1 - 0.95 = 0.05 ; 0.05 / 2 = 0.025

Confidence interval is calculated thus : (newer--older)

(m1 - m2) ± Tcritical * standard error

Mean difference = m1 - m2; (195 - 286)

Tcritical = Tn1+n2-2, α/2 = T(45+40)-2 = T83, 0.025 = 1.99 (T value calculator)

Standard error (E) = sqrt((s1²/n1) + (s2²/n2))

E = sqrt((46^2/45) + (58^2/40))

Hence, confidence interval:

(195 - 286) ± 1.992 * sqrt((46^2/45) + (58^2/40))

The 95% confidence interval for estimating the difference in the mean dollar cost of the routine maintenance between newer and older cars is given by:

[tex]CI=(\overline{x_1} - \overline{x_2}) \pm t_{critical} \sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}}[/tex]

The first sample consists of owners of newer cars

The second sample consists of owner of older cars.

95% confidence interval for difference between both samples' means.

Since the sample sizes are > 30, thus we can use the z table for finding the Confidence interval.

The CI is given as:

[tex]CI=(\overline{x_1} - \overline{x_2}) \pm t_{critical} \sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}}[/tex]

For 0.95 probability confidence, we have t at 40+45-3= t at 83 at 0.05/2 is 1.992 (from T tables)

Thus,

[tex]CI=(195-268) \pm 1.992 \sqrt{\dfrac{46^2}{45} + \dfrac{58^2}{40}}[/tex]

Learn more about confidence interval here:

https://brainly.com/question/4249560

The life span of a domestic cat is normally distributed with a mean of 15.7 years and a standard deviation of 1.6 years. a. Find the probability that a cat will live to be older than 14 years. b. Find the probability that a cat will live between 14 and 18 years. c. If a cat lives to be over 18 years, would that be unusual? Why or why not? d. How old would a cat have to be to be older than 90% of other cats?

A music store marks up the instruments it sells by 30%.
If the price tag on a trumpet says $104, how much did the store pay for it?

-3(1)+(-3)(-1) =
+()]
-3(1) + (-3)(-1)=-3[0]
-3(1)+(-3)(-1) =
_
+(-3)(-1) =
+

Pls help me with this ASAP

Which of the following is not a solution of the equation y=3x-4?
A. 2,2
B. 0,-4
C. 4,8
D. -1,-1

8. Given: AD perpendicullar DB, DB perpendicullar BC, AB = CD Prove: AB||DC​

At 5:00 p.m., Antonio turned on the oven. While the oven preheated, the temperature in the oven increased from 72°F to 400°F over a 10-minute period. The oven remained at 400°F for 45 minutes until Antonio turned it off. It took 60 minutes for the temperature in the oven to cool, returning to 72°F at 6:55 p.m. Which statement best explains whether or not the temperature in the oven is a function of the time?