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 ​$33 million. Is the result close to the actual number of​ viewers, 3.03.0 ​million? Use a significance level of 0.05.

Answer 1

Answer:

y = 5.064 - 0.010x

Here, in this question x = 33

y = 4.734 million views

In this question data is missing and I filled it out and calculated the questions accordingly. Please refer to the attachment for the data calculated.

Step-by-step explanation:

First of all, this question is incomplete. It lacks the data needed to calculate the required things.

So, I  figured out the question and I will try my best to solve the question at hand.

I have calculated the some data sheet in excel. Please refer to the attachment for that data.

let a be the salaries

b be the viewers

$S_{aa}$ = ∑$a^{2}$  - $\frac{ Summtion a^{2} }{n}$

$S_{aa}$ = 7522.8571

$S_{bb}$ = ∑$b^{2}$  - $\frac{ Summtion b^{2} }{n}$

$S_{bb}$ = 47.1943

$S_{ab}$ = ∑$ab^{}$  - $\frac{ Summtion a x Summation b^{} }{n}$

$S_{ab}$ = -76.8286

Now the slope of the required regression equation is:

Slope =  $S_{ab}$/$S_{aa}$ = -76.8286/7522.8571 = -0.010

And the intercept of the required regression equation:

intercept = $\frac{1}{n}$ x (∑b - slope x ∑a) = 5.064

So, the regression equation will be:

y = intercept + slope x

y = 5.064 - 0.010x

Here, in this question x = 33

y = 5.064 - 0.010 (33)

y = 4.734 million views