Question

The data show the number of viewers for television stars with certain salaries. Find the regression​ equation, letting salary be the independent​ (x) variable. Find the best predicted number of viewers for a television star with a salary of ​$10 million. Is the result close to the actual number of​ viewers, 10.1 ​million? Use a significance level of 0.05.

The data show the number of viewers for television stars with certain salaries. Find the regression equation, letting salary be the independent (x) variable. Find the best predicted number of viewers for a television star with a salary of $10 million the actual number of viewerS, 10.1 million? Use a significance level of 0.05 Is the result close Salary (millions of $) Viewers (millions) 1: 106 13 12 6 6 6.6 10.4 7 . 4.2 .5 Click the icon to view the critical values of the Pearson correlation coefficient r. What is the regression equation? y x (Round to three decimal places as needed.) What is the best predicted number of viewers for a television star with a salary of $10 million? number a television star with a salary of $10 million ismillion. ( predi .. Oaded) one decimal Is the result close the actual number of viewers, 10.1 million? O A. The result is exactly the same as the actual number of viewers of 10.1 million. O B. The result is very close to the actual number of viewers of 10,1 million. viewers of 10,1 million. C. The result is not very close to the actual number O D. The result does not make sense given the context of the data

Answer 1

SOLUTION:

From given data,

The data show the number of viewers for television stars with certain salaries. Find the regression​ equation, letting salary be the independent​ (x) variable. Find the best predicted number of viewers for a television star with a salary of ​$10 million. Is the result close to the actual number of​ viewers, 10.1​million? Use a significance level of 0.05.

observation	Salary ()	Viewer ()	(-$\bar{x}$)2	(-$\bar{y}$)2	(-$\bar{x}$) * (-$\bar{y}$)
1	106	6.6	7267.5625	0.16	85.25*(-0.4) = -34.1
2	13	4.2	60.0625	7.84	-7.75*(-2.8) = 21.7
3	12	7.6	76.5625	0.36	-8.75*(0.6) = -5.25
4	7	5.4	189.0625	2.56	-13.75*(-1.6) = 22
5	11	10.4	95.0625	11.56	-9.75*(3.4) = -33.15
6	5	7.4	248.0625	0.16	-15.75*(0.4) = -6.3
7	6	7.9	217.5625	0.81	-14.75*(0.9) = -13.275
8	6	6.5	217.5625	0.25	-14.75*(-0.5)=7.375
Sum	$\Sigma$ = 166	$\Sigma$ = 56	$\Sigma$(-$\bar{x}$)2 = 8371.5	$\Sigma$(-$\bar{y}$)2 =23.7	$\Sigma$(-$\bar{x}$) * (-$\bar{y}$) = -41

$\bar{x}$ = $\Sigma$ /n = 166/8 = 20.75

$\bar{y}$ = $\Sigma$ /n = 56/8 = 7

From above

S_xx = $\Sigma$(-$\bar{x}$)² = 8371.5

S_xy = $\Sigma$(-$\bar{x}$) * (-$\bar{y}$) = -41

Slope = = S_xy / S_xx =  -41 / 8371.5 = -0.004

31

Intercept = $\beta _{0}$ = $\bar{y}$ -
31
($\bar{x}$)

$\beta _{0}$ = 7 - (-0.00489)(20.75 ) = 7.101

What is the regression equation?

$\hat{y}$ = $\beta _{0}$ +
31

$\hat{y}$ = 7.101 -0.004

What is the best predicted number of viewers for a television star with a salary of $10 million?

Where,

= 10 then

$\hat{y}$ = 7.101 -0.004 (10)

$\hat{y}$ = 7.061

The result close to the actual number of viewers, 10.1 million?

Where,

= 10.1 then

$\hat{y}$ = 7.101 -0.004 (10.1)

$\hat{y}$ = 7.0606

The result is very close to the actual number of viewers of 10.1 million.

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