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3 years ago

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The regression equation y = 1.3x + 6.8 approximates the number of minutes it takes an employee to drive to work, y, given the nu

mber of miles the employeehas to drive, x. Which statement is true? For every extra mile an employee drives, the driving time increases by 6.8 minutes. For every extra mile an employee drives, the driving time increases by 1.3 minutes. For every extra minute an employee drives, the distance increases by 6.8 minutes. For every extra minute an employee drives, the distance increases by 1.3 minutes. For every extra mile an employee drives, the driving time increases by 1.3 minutes

1 answer:

Rama09 [41]3 years ago

5 0

Answer:

for every extra mile an employee drives, the driving time increases by 1.3 minutes

Step-by-step explanation:

Step 1: The given regression equation y = 1.3x + 6.8, where "x" is the number of miles and "y" is the time taken.

Step 2: Here the constant 6.8 remains the same. For each extra mile, the time increases by 1.3

For example, let's take x = 2 miles, the time taken y = 1.3(2) + 6.8

= 2.6 + 6.8

y = 9.4

Let's take x = 3, the time taken y = 1.3(3) + 6.8

= 3.9 + 6.8

y =10.7

You can see the difference in the timing by 1.3.

Therefore, for each extra mile, the time is increases by 1.3 minutes.

Thank you.

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$H_o :$ There is no significant relationship between y and $x_1, x_2, x_3, x_4$

$H_a :$ There is a significant relationship between y and $x_1, x_2, x_3, x_4$

First, we calculate the t-score using:

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The p value when $t = 244.44$ and $df = 25$ is:

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Solving (b): $SSE(x_1 ,x_2 ,x_3 ,x_4)$

To calculate SSE, we use:

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$SSR = 1760$ ----------- $(x_1 ,x_2 ,x_3 ,x_4)$

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To calculate SSE, we use:

$SSE = SST - SSR$

Given that:

$SSR = 1705$ ----------- $(x_2 ,x_3)$

$SST = 1805$

So:

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The null and alternate hypothesis are:

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$t = \frac{100-45}{2} \div \frac{45}{30 - 4 - 1}$

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The p value when $t = 15.28$ and $df = 25$ is:

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