Consider regressing the number of pages in a book versus the number of words in

Question

Consider regressing the number of pages in a book versus the number of words in a book. The data would have _______________ correlation.

Question 16 options:

positive

negative

no

undetermined

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Question 17 (1 point)

A correlation coefficient, r, of 0.65 is stronger than a correlation coefficient of -0.65.

Question 17 options:

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Question 18 (1 point)

Choose the value of the Pearson's Correlation Coefficient (r) that best describes the two plots.

Question 18 options:

I: 0.179, II: 0.101

I: 0.821, II: 0.899

I: 0.101, II: 0.821

I: -0.821, II: -0.101

I: 0.821, II: 0.101

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Question 19 (1 point)

Suppose we would like to nd a relationship between the number of credits a student is enrolled in (X) and how many hours they work a week (Y ). Suppose the average number of credits a student is enrolled in is 13.2 with a standard deviation of 2.5 and the average number of hours a student works a week is 7.8 with a standard deviation of 4.2. The correlation between these two variables is 0.42. Calculate the regression equation for the data.

Question 19 options:

Y=-1.57-0.71X

Y=17.17+0.71X

Y=17.17-0.71X

Y=15.15-0.25X

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Question 20 (1 point)

Suppose we would like to nd a relationship between the number of credits a student is enrolled in (X) and how many hours they work a week (Y ). Suppose the average number of credits a student is enrolled in is 13.2 with a standard deviation of 2.5 and the average number of hours a student works a week is 7.8 with a standard deviation of 4.2. The correlation between these two variables is 0.42. Predict the number of hours a student works if they are enrolled in 15 credits.

Question 20 options:

10.8

9.1

6.5

5.6

positive

negative

no

undetermined

Explanation / Answer

16) it should be posiitve as more number of pages ; more number of words or vice cersa

17)False

18)I: 0.821, II: 0.101

19)

here slope =r*Sy/Sx =-0.42*4.2/2.5 =-0.71

and intercept =ybar -b1*Xbar =7.8-(-0.71)*13.2 =17.17

Y=17.17-0.71X

20)_

predicted value =17.17-0.71*15=6.5

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