If f (x) = x2 – 2x, find:<br> f(6) = [?]

Most individuals are aware of the fact that the average annual repair cost for an automobile depends on the age of the automobil

o-na [289]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

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 26 automobiles 4 years old showed a sample standard deviation for annual repair costs of $120 and a sample of 23 automobiles 2 years old showed a sample standard deviation for annual repair costs of $100. Let 4 year old automobiles be represented by population 1.

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

At a 0.01 level of significance, what is your conclusion? What is the p-value?

Answer:

Null hypotheses = H₀ = σ₁² ≤ σ₂²

Alternative hypotheses = Ha = σ₁² > σ₂²

Test statistic = 1.44

p-value = 0.1954

0.1954 > 0.01

Since the p-value is greater than the given significance level therefore, we cannot reject the null hypothesis.

We can conclude that there is no sufficient evidence to support the claim that the variance in annual repair costs is larger for older automobiles.

Step-by-step explanation:

Let σ₁² denotes the variance of 4 years old automobiles

Let σ₂² denotes the variance of 2 years old automobiles

State the null and alternative hypotheses:

The null hypothesis assumes that the variance in annual repair costs is smaller for older automobiles.

Null hypotheses = H₀ = σ₁² ≤ σ₂²

The alternate hypothesis assumes that the variance in annual repair costs is larger for older automobiles.

Alternative hypotheses = Ha = σ₁² > σ₂²

Test statistic:

The test statistic is given by

Test statistic = σ₁²/σ₂²

Test statistic = 120²/100²

Test statistic = 1.44

p-value:

The degree of freedom corresponding to 4 years old automobiles is given by

df₁ = n - 1

df₁ = 26 - 1

df₁ = 25

The degree of freedom corresponding to 2 years old automobiles is given by

df₂ = n - 1

df₂ = 23 - 1

df₂ = 22

Using Excel to find out the p-value,

p-value = FDIST(F-value, df₁, df₂)

p-value = FDIST(1.44, 25, 22)

p-value = 0.1954

Conclusion:

When the p-value is less than the significance level then we reject the Null hypotheses

p-value < α (reject H₀)

But for the given case,

p-value > α

0.1954 > 0.01

Since the p-value is greater than the given significance level therefore, we cannot reject the null hypothesis.

We can conclude that there is no sufficient evidence to support the claim that the variance in annual repair costs is larger for older automobiles.

8 0

3 years ago

What is the solution to the equation 14 ^ (x + 3) = 21?

Rina8888 [55]

The answer to your question is b

8 0

3 years ago

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(-5z)³ as a positive exponent.

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To write this expression as a positive exponent we use this rule of exponents: x^color(red)(a) = 1/x^color(red)(-a) 5^-3 = 1/5^(- -3) .

3 0

3 years ago

Quadrilateral RSTQ is a parallelogram.

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<S ~= <Q AND TQ ~= SR

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3 years ago

2.The number of tickets sold to a play for each showing is 78, 84, 87, 80, 91, 95, and 80.

Luden [163]

Answer:

$Min =78$

The maximum is :

$Max = 95$

Now we can calculate the median since the sample size os n =7 the median would be the middle value at the position 4 from the dataset ordered and we got:

$Median =84$

Now we can find the quartile 1 we analyze the first 4 values 78, 80,80, 84 and the first quartile would be:

$Q_1= \frac{80+80}{2}= 80$

Now we can find the quartile 3 we analyze the last 4 values 84,87, 91, 95 and the third quartile would be:

$Q_3= \frac{87+91}{2}=89$

And finally the 5 number summary would be:

$Min = 78, Q_1 = 80, Median=84, Q_3 = 89, Max=96$

Step-by-step explanation:

We have the following data given:

78, 84, 87, 80, 91, 95, and 80.

The first step is order the dataset on increasing way and we got:

78, 80,80, 84,87, 91, 95

We can begin finding the minimum value and for this case is:

$Min =78$

The maximum is :

$Max = 95$

Now we can calculate the median since the sample size os n =7 the median would be the middle value at the position 4 from the dataset ordered and we got:

$Median =84$

Now we can find the quartile 1 we analyze the first 4 values 78, 80,80, 84 and the first quartile would be:

$Q_1= \frac{80+80}{2}= 80$

Now we can find the quartile 3 we analyze the last 4 values 84,87, 91, 95 and the third quartile would be:

$Q_3= \frac{87+91}{2}=89$

And finally the 5 number summary would be:

$Min = 78, Q_1 = 80, Median=84, Q_3 = 89, Max=96$

5 0

3 years ago

If f (x) = x2 – 2x, find: f(6) = [?]