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 $13 million. Is the result close to the actual number of viewers, 5.7 million? Use a significance level of 0.05. 98 14 Salary (millions of $) Viewers (millions) 2 11 11 7 14 1 D 9.1 4.6 10.8 10.7 6.3 2.9 4.4 3.2 Click the icon to view the critical values of the Pearson correlation coefficient r. Critical Values of the Pearson Correlation Coefficient r What is the regression equation? Critical Values of the Pearson Correlation Coefficient r n α = 0.05 x = 0.01 NOTE: To test Ho: p=0 ŷ=+x (Round to three decimal places as needed.) 4 0.950 0.990 against H₁: p#0, reject Ho 5 0.878 0.959 What is the best predicted number of viewers for a television star with a salary of $13 million? if the absolute value of ris 6 0.811 0.917 greater than the critical 7 0.754 millio 0.875 value in the table. The best predicted number of viewers for a television star with a salary of $13 million is (Round to one decimal place as needed.) 8 0.707 0.834 9 0.666 0.798 Is the result close to the actual number of viewers, 5.7 million? 10 0.632 0.765 11 0.602 0.735 12 0.576 0.708 OA. The result is not very close to the actual number of viewers of 5.7 million. OB. The result is very close to the actual number of viewers of 5.7 million. 13 0.553 0.684 14 0.532 0.661 15 0.514 0.641 O C. The result is exactly the same as the actual number of viewers of 5.7 million. O D. The result does not make sense given the context of the data. 16 0.497 0.623 17 0.482 0.606 18 0.468 0.590 19 0.456 0.575 20 0.444 0.561 25 0.396 0.505 30 0.361 0.463 35 0.335 0.430 40 0.312 0.402 45 0.294 0.378 50 0.279 0.361 60 0.254 0.330 70 0.236 0.305 80 0.220 0.286 0.207 0.269 90 100 n 0.196 0.256 α = 0.05 α = 0.01
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 $13 million. Is the result close to the actual number of viewers, 5.7 million? Use a significance level of 0.05. 98 14 Salary (millions of $) Viewers (millions) 2 11 11 7 14 1 D 9.1 4.6 10.8 10.7 6.3 2.9 4.4 3.2 Click the icon to view the critical values of the Pearson correlation coefficient r. Critical Values of the Pearson Correlation Coefficient r What is the regression equation? Critical Values of the Pearson Correlation Coefficient r n α = 0.05 x = 0.01 NOTE: To test Ho: p=0 ŷ=+x (Round to three decimal places as needed.) 4 0.950 0.990 against H₁: p#0, reject Ho 5 0.878 0.959 What is the best predicted number of viewers for a television star with a salary of $13 million? if the absolute value of ris 6 0.811 0.917 greater than the critical 7 0.754 millio 0.875 value in the table. The best predicted number of viewers for a television star with a salary of $13 million is (Round to one decimal place as needed.) 8 0.707 0.834 9 0.666 0.798 Is the result close to the actual number of viewers, 5.7 million? 10 0.632 0.765 11 0.602 0.735 12 0.576 0.708 OA. The result is not very close to the actual number of viewers of 5.7 million. OB. The result is very close to the actual number of viewers of 5.7 million. 13 0.553 0.684 14 0.532 0.661 15 0.514 0.641 O C. The result is exactly the same as the actual number of viewers of 5.7 million. O D. The result does not make sense given the context of the data. 16 0.497 0.623 17 0.482 0.606 18 0.468 0.590 19 0.456 0.575 20 0.444 0.561 25 0.396 0.505 30 0.361 0.463 35 0.335 0.430 40 0.312 0.402 45 0.294 0.378 50 0.279 0.361 60 0.254 0.330 70 0.236 0.305 80 0.220 0.286 0.207 0.269 90 100 n 0.196 0.256 α = 0.05 α = 0.01
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Question
Transcribed Image Text: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 $13 million. Is the result close to the actual number of viewers, 5.7 million? Use a significance level of 0.05.
98
Salary (millions of $)
Viewers (millions)
14 2
3.2 9.1 4.6
11 11 7 14 1 D
10.8 10.7 6.3 2.9 4.4
Click the icon to view the critical values of the Pearson correlation coefficient r.
Critical Values of the Pearson Correlation Coefficient r
What is the regression equation?
Critical Values of the Pearson Correlation Coefficient r
n
α = 0.05
α = 0.01
NOTE: To test H₂: p=0
=
+ x (Round to three decimal places as needed.)
4
0.950
0.990
against H₁: p‡0, reject Ho
5
0.878
0.959
if the absolute value of r is
6
0.811
0.917
greater than the critical
What is the best predicted number of viewers for a television star with a salary of $13 million?
The best predicted number of viewers for a television star with a salary of $13 million is millio
(Round to one decimal place as needed.)
7
0.754
0.875
value in the table.
8
0.707
0.834
9
0.666
0.798
Is the result close to the actual number of viewers, 5.7 million?
10
0.632
0.765
11
0.602
0.735
12
0.576
0.708
A. The result is not very close to the actual number of viewers of 5.7 million.
B. The result is very close to the actual number of viewers of 5.7 million.
13
0.553
0.684
14
0.532
0.661
15
0.514
0.641
C. The result is exactly the same as the actual number of viewers of 5.7 million.
D. The result does not make sense given the context of the data.
16
0.497
0.623
17
0.482
0.606
18
0.468
0.590
19
0.456
0.575
20
0.444
0.561
25
0.396
0.505
30
0.361
0.463
35
0.335
0.430
40
0.312
0.402
45
0.294
0.378
50
0.279
0.361
60
0.254
0.330
70
0.236
0.305
80
0.220
0.286
0.207
0.269
90
100
n
0.196
0.256
α = 0.05
α = 0.01
00
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