Question

PLEASE HELP ASAP

WILL GIVE A THUMBS UP

A website that rents movies online recorded the age and the number of movies rented during the past month for some of their customers. The data are shown below and the best fitting regression line has bo = 18 and b1 = -0.247. a) Use the estimated regression equation to predict Rentals for all six values of Age. b) Find the residuals ej. c) Calculate the residual standard deviation, se 35 43 49 64 42 30 Age Rentals 8 9 7 8 10 64 42 30 a) Input the predicted values for Rentals below. Age 35 43 49 Rentals (Round to two decimal places as needed.)

Answer 1

Given,

x = age

y = Rentals

The regression line is,

$\hat y=18-0.247x$

a)

We predict the values by using regreesion equation,

Age(x)	Rental(y)	Predicted values($\hat y$)
35	8	9.355
43	9	7,379
49	7	5.897
64	1	2.192
42	8	7.626
30	10	10.59

b)Residual

(fi-) = ?

Age(x)	Rental(y)	Predicted values($\hat y$)	Residuals$(y-\hat y)$
35	8	9.355	-1.355
43	9	7,379	1.621
49	7	5.897	1.103
64	1	2.192	-1.192
42	8	7.626	0.374
30	10	10.59	-0.59

c)Residual standard deviation

  $Residual\ standard\ Deviation=\sqrt{\sum \frac{(y-\hat y)^2}{df}}$

where, df = n - number of parameters.

Age(x)	Rental(y)	Predicted values($\hat y$)	Residuals$(y-\hat y)$	$(y-\hat y)^2$
35	8	9.355	-1.355	1.8360
43	9	7,379	1.621	2.6276
49	7	5.897	1.103	1.2166
64	1	2.192	-1.192	1.4208
42	8	7.626	0.374	0.1398
30	10	10.59	-0.59	0.3481

$\sum (y-\hat y)^2=7.5891$ and

$Residual\ standard\ deviation=\sqrt{\frac{7.5891}{4}}$

$Residual\ standard\ deviation=\sqrt{1.8973}$