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3 years ago

8

Suppose an automobile manufacturer designed a radically new lightweight engine and wants to recommend the grade of gasoline that

will have the best fuel economy. The four grades are regular, economy, premium, and super premium. The test car made three trial runs on the test track using each of the four grades. The miles per gallon were recorded for each grade. At the 0.05 level, what is the critical value of F used to test the hypothesis that the miles per gallon for each fuel are the same?

1 answer:

Deffense [45]3 years ago

8 0

Answer:

The F-statistic used to test the hypothesis that the miles per gallon for each fuel are the same is 4.07.

Step-by-step explanation:

There are four treatments in the data given, i.e. k = 4.

Total number of observations, n = 12.

Note: degrees of freedom is denoted as df.

For treatment, the degrees of freedom = k-1 = 4-1 =3 df.

The total degrees of freedom = n-1 = 12-1 = 11 df.

The error in degrees of freedom = df (total) - df(treatment)

The error in degrees of freedom = 11 - 3 = 8 df

At α = 0.05 level,from the F table, the F-statistic with (3 , 8)df is 4.07.

Therefore, the F-statistic used to test the hypothesis that the miles per gallon for each fuel are the same is 4.07.

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alexandr1967 [171]

Answer:

$Y' = (-2, 4\frac{2}{3})$

Step-by-step explanation:

Given

$Y = (-3,7)$

$Scale\ Factor = (\frac{2}{3}x,\frac{2}{3}y)$

Required

Determine Y'

Y' can be solved by multiplying the scale factor by Y

i.e.

$Y' = Scale\ Factor * Y$

For, the x coordinates.

$Y' = \frac{2}{3}x$

Where

$x = -3$

$Y' = \frac{2}{3} * -3$

$Y' = \frac{-6}{3}$

$Y = -2$

For the y coordinates:

$Y' = \frac{2}{3}y$

Where

$y = 7$

$Y' = \frac{2}{3} * 7$

$Y' = \frac{14}{3}$

$Y' = 4\frac{2}{3}$

Hence:

$Y' = (-2, 4\frac{2}{3})$

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3 years ago

High concentrations of carbon monoxide CO can cause coma and possible death. The time required for a person to reach a COHb leve

ale4655 [162]

Answer:

A) 10.38hours

B) 617.95 is the concerntration of CO

Step-by-step explanation:

A) when x=600

T= 0.0002x^2 - 0.316x + 127.9

T= 0.0002(600)^2 -0.316(600) + 127.9

T= 10.38secs

B) to find the concerntration, use Almighty formular

X=-b+-sqrt(b^2-4ac)/2a

0.0002x^2-0.316x-127.9=9

0.0002x^2 - 0.316x + 118.9=0

X= 0.316 +- sqrt(0.316^2) -4(0.0002×118.9)/(2×0.0002)

X= 0.316+(0.68817/0.0004)=962.05 or

X=-0.316 -0.68817/0.0004 =617.95

Therefore X is within the domain of 617.95 is the concerntration

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3 years ago

Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.

fredd [130]

$\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

$\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

so the ODE is indeed exact and there is a solution of the form $F(x,y)=C$ . We have

$\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)$

$\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)$

$f'(y)=0\implies f(y)=C$

$\implies F(x,y)=x^2y^3+\ln(x+y^2)=C$

With $y(1)=2$ , we have

$8+\ln9=C$

so

$\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}$

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