Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a marble with a diameter of 1.3 cm. How does the result compare to the actual circumference of 4 Use a significance level of 0.05. Baseball Basketball Golf Soccer Tennis Ping-Pong|Volleyball 4.4 22.1 13.8 69.4 21.7 Critical Values of the Pearson Correlation Coefficient r Diameter Circumference 22.9 23.6 74.1 7.3 6.9 4.1 21.2 12.9 66.6 Click the icon to view the critical values of the Pearson correlation coefficient r. a= 0.05 10.950 0.878 |0.811 10.754 a= 0.01 |0.990 0.959 0.917 10.875 ]NOTE: To test H,:p = 0 Jagainst H,: p # 0, reject H, if the absolute value of r is greater than the critical Ivalue in the table n The regression equation is y =0+Dx. (Round to five decimal places as needed.)
Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a marble with a diameter of 1.3 cm. How does the result compare to the actual circumference of 4 Use a significance level of 0.05. Baseball Basketball Golf Soccer Tennis Ping-Pong|Volleyball 4.4 22.1 13.8 69.4 21.7 Critical Values of the Pearson Correlation Coefficient r Diameter Circumference 22.9 23.6 74.1 7.3 6.9 4.1 21.2 12.9 66.6 Click the icon to view the critical values of the Pearson correlation coefficient r. a= 0.05 10.950 0.878 |0.811 10.754 a= 0.01 |0.990 0.959 0.917 10.875 ]NOTE: To test H,:p = 0 Jagainst H,: p # 0, reject H, if the absolute value of r is greater than the critical Ivalue in the table n The regression equation is y =0+Dx. (Round to five decimal places as needed.)
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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please see attached - thank you in advance
Transcribed Image Text:0.482
0.468
0.456
0.444
0.396
0.361
0.335
0.312
17
18
0.606
0.590
0.575
0.561
0.505
0.463
0.430
19
20
25
30
35
40
45
50
60
70
80
90
0.402
0.294
0.279
0.254
0.236
0.220
0.207
0.196
0.378
0.361
0.330
0.305
0.286
0.269
0.256
100
In
a= 0.05
a= 0.01
Transcribed Image Text:Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a marble with a diameter of 1.3 cm. How does the result compare to the actual circumference of 4.1 o
Use a significance level of 0.05.
Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball
22.1 6.9
21.7
Critical Values of the Pearson Correlation Coefficient r
21.2
66.6
Diameter
7.3
23.6
4.1
Circumference 22.9
74.1
13.8 69.4
12.9
A Click the icon to view the critical values of the Pearson correlation coefficient r.
a= 0.05
0.950
0.878
0.811
|0.754
10.707
0.666
0.632
0.602
NOTE: To test H: p = 0
against H,: p # 0, reject H,
if the absolute value of r is
Igreater than the critical
|value in the table.
In
a= 0.01
0.990
0.959
0.917
0.875
0.834
0.798
0.765
0.735
The regression equation is y =D+Dx.
(Round to five decimal places as needed.)
4
15
16
8
19
10
11
12
13
14
0.576
0.553
0.532
0.514
0.497
0.482
Io 468
0.708
0.706
0.684
15
16
17
0.661
0.641
|0.623
Enter your answer in the edit fields and then click Check Answer.
parts
remaining
0.606
In 590
18
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