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lbvjy [14]
3 years ago
14

A statistician wants to determine if there is a difference in the fuel efficiency of cars between model years. To do this, he se

lects random makes and models of cars and compares the fuel efficiency in miles per gallon of the current model year and the previous model year. Suppose that data were collected for a random sample of 9 cars, where each difference is calculated by subtracting the fuel efficiency in miles per gallon of the previous model year from the fuel efficiency in miles per gallon of the current model year. Assume that the fuel efficiencies are normally distributed. The statistician uses the alternative hypothesis Ha:μd≠0. Using a test statistic of t≈6.163, which has 8 degrees of freedom, determine the range that contains the p-value.

Mathematics
1 answer:
Elodia [21]3 years ago
4 0

Answer:

P-value = 0.00028

Step-by-step explanation:

The statician performs an hypothesis test for the mean difference.

As the alternative hypothesis is<em> Ha: μd ≠ 0</em>, we know that it is a two-tailed test.

The test statistic is t=6.163.

For a two-tailed test and 8 degrees of freedom, this corresponds to a P-value of:

\text{P-value}=2\cdot P(t>6.163)=0.00028

This P-value is smaller enough to reject the null hypothesis at almost any significance level.

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4.67

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0.25 + 0.50 + 0.75 + 1.25 + 1.875 = 4.675

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It cost a farmer 43.60 pre acre to harvest corn. How much does it cost to harvest 1325.5 acres
KATRIN_1 [288]

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$57,791.8

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Each acre cost $43.60.

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3 years ago
Find f(-3) if f(x) = x^2.
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9

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f(-3)=-3×-3

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Prove that if x is an positive real number such that x + x^-1 is an integer, then x^3 + x^-3 is an integer as well.
Shkiper50 [21]

Answer:

By closure property of multiplication and addition of integers,

If x + \dfrac{1}{x} is an integer

∴ \left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right ) is an integer

From which we have;

x^3 + \dfrac{1}{x^3} is an integer

Step-by-step explanation:

The given expression for the positive integer is x + x⁻¹

The given expression can be written as follows;

x + \dfrac{1}{x}

By finding the given expression raised to the power 3, sing Wolfram Alpha online, we we have;

\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot x + \dfrac{3}{x}

By simplification of the cube of the given integer expressions, we have;

\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )

Therefore, we have;

\left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= x^3 + \dfrac{1}{x^3}

By rearranging, we get;

x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )

Given that  x + \dfrac{1}{x} is an integer, from the closure property, the product of two integers is always an integer, we have;

\left ( x + \dfrac{1}{x} \right) ^3 is an integer and 3\cdot \left (x + \dfrac{1}{x} \right ) is also an integer

Similarly the sum of two integers is always an integer, we have;

\left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right  ) is an integer

\therefore x^3 + \dfrac{1}{x^3} =   \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= \left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right  ) is an integer

From which we have;

x^3 + \dfrac{1}{x^3} is an integer.

4 0
3 years ago
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