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harina [27]
3 years ago
11

To avoid the problem of having access to tables of the F distribution with values for the lower tail when a one-tailed test is r

equired, let the _____ variance be the numerator of the test statistic. a. sample variance from the population with the larger hypothesized b. larger sample c. sample variance from the population with the smaller hypothesized d. smaller sample
Mathematics
1 answer:
omeli [17]3 years ago
6 0

Answer:

The answer is "Option b".

Step-by-step explanation:

The significant variance shows that the numbers in the set are far from the mean and far from each other as well. Alternatively, a little variation implies the reverse. The variance value of 0, on either hand, shows that all values inside a set of numbers are the same. If you need yet another test, use a greater sample variance as the numerator of the test statistic to avoid having to reference tables of the F distribution that primary device from of the lower tail.

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Mr Smith's art class took a bus trip to an art museum. The bus averaged 65 miles per hour on the highway and 25 miles per hour i
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Let x be the distance traveled on the highway and y the distance traveled in the city, so:
\left \{ {{x+y=375} \atop { \frac{1}{65}x+ \frac{1}{25}y =7}} \right.
 
Now, the system of equations in matrix form will be:
\left[\begin{array}{ccc}1&1&\\ \frac{1}{65} & \frac{1}{25} &\end{array}\right]   \left[\begin{array}{ccc}x&\\y&\end{array}\right] =  \left[\begin{array}{ccc}375&\\7&\end{array}\right]

Next, we are going to find the determinant:
D=  \left[\begin{array}{ccc}1&1\\ \frac{1}{65} & \frac{1}{25} \end{array}\right] =(1)( \frac{1}{25}) - (1)( \frac{1}{65} )= \frac{8}{325}
Next, we are going to find the determinant of x:
D_{x} =  \left[\begin{array}{ccc}375&1\\7& \frac{1}{25} \end{array}\right] = (375)( \frac{1}{25} )-(1)(7)=8

Now, we can find x:
x=  \frac{ D_{x} }{D} = \frac{8}{ \frac{8}{325} } =325mi

Now that we know the value of x, we can find y:
y=375-325=50mi

Remember that time equals distance over velocity; therefore, the time on the highway will be:
t_{h} = \frac{325}{65} =5hours
An the time on the city will be:
t_{c} = \frac{50}{25} =2hours

We can conclude that the bus was five hours on the highway and two hours in the city. 

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3 years ago
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