Most individuals are aware of the fact that the average annual repair cost for an automobile depends on the age of the automobile. A researcher is interested in finding out whether the variance of the annual repair costs also increases with the age of the automobile. A sample of automobiles years old showed a sample standard deviation for annual repair costs of and a sample of automobiles years old showed a sample standard deviation for annual repair costs of . Let year old automobiles be represented by population . a. State the null and alternative versions of the research hypothesis that the variance in annual repair costs is larger for the older automobiles. b. Conduct the hypothesis test at a level of significance. Calculate the value of the test statistic (to 2 decimals).

Step-by-step explanation:

Hello!

The researcher's objective is to test if the variance of the annual repair costs increases with the age of the automobile, i.e. the older the car, the more the repairs costs. The parameters of the study are the population variances of the annual repair costs of 4 years old cars and 2 years old cars.

X₁: Costs of annual repair of a 4 years old car.

Assuming X₁~N(μ₁;δ₁²)

A sample of 26 automobiles 4 years old showed a sample standard deviation for annual repair costs of $170

n₁= 26 and S₁= $170

X₂: Costs of annual repair of a 2 years old car.

Assuming X₂~N(μ₂;σ₂²)

A sample of 25 automobiles 2 years old showed a sample standard deviation for the annual repair cost of $100.

n₂= 25 and S₂= $100

a. State the null and alternative versions of the research hypothesis that the variance in annual repair costs is larger for older automobiles.

H₀: δ₁² ≤ σ₂²

H₁: δ₁² > σ₂²

b. At a .01 level of significance, what is your conclusion? What is the p-value? Discuss the reasonableness of your findings.

This is a variance ratio test and you have to use a Snedecor's F-statistic:

This test is one-tailed to the right and so is the p-value, you have to calculate it under a F₂₅;₂₄

P(F₂₅;₂₄≥2.89)= 1 - P(F₂₅;₂₄<2.89)= 1 - 0.994= 0.006

Using the p-value approach the decision rule is:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

α: 0.01

The p-value is less than the level of significance, the decision is to reject the null hypothesis.

Then using a 1% level, you can conclude that the population variance of the cost of annual repairs for 4 years old cars is greater than the population variance of the cost of annual repairs for 2 years old cars.

I hope this helps!

i can't read the graph

step-by-step explanation:

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752