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 Use a significance level of 0.05. Baseball Basketball Golf Soccer Tennis Ping-Pong|Volleyball 4.4 22.1 6.9 13.8 69.4 21.7 O Critical Values of the Pearson Correlation Coefficient r 4.1 12.9 Diameter 7.3 23.6 74.1 21.2 Circumference 22.9 66.6 ]NOTE: To test Ho:p = 0 Jagainst H,: p#0, reject H, ]if the absolute value of r is greater than the critical Jvalue in the table. Click the icon to view the critical values of the Pearson correlation coefficient r. a= 0.01 0.990 0.959 0.917 0.875 0.834 to 798 a= 0.05 0.950 0.878 0.811 0.754 0.707 0.666 0.632 0.602 0.576 0.553 0.532 0.514 to 497 n 4 The regression equation is y =+x. (Round to five decimal places as needed.) 6 17 8 |0.798 0.765 0.735 0.708 0.684 0.661 10.641 19 10 11 12 13 14 15 Enter your answer in the edit fields and then click Check Answer. 16 Io 623
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 Use a significance level of 0.05. Baseball Basketball Golf Soccer Tennis Ping-Pong|Volleyball 4.4 22.1 6.9 13.8 69.4 21.7 O Critical Values of the Pearson Correlation Coefficient r 4.1 12.9 Diameter 7.3 23.6 74.1 21.2 Circumference 22.9 66.6 ]NOTE: To test Ho:p = 0 Jagainst H,: p#0, reject H, ]if the absolute value of r is greater than the critical Jvalue in the table. Click the icon to view the critical values of the Pearson correlation coefficient r. a= 0.01 0.990 0.959 0.917 0.875 0.834 to 798 a= 0.05 0.950 0.878 0.811 0.754 0.707 0.666 0.632 0.602 0.576 0.553 0.532 0.514 to 497 n 4 The regression equation is y =+x. (Round to five decimal places as needed.) 6 17 8 |0.798 0.765 0.735 0.708 0.684 0.661 10.641 19 10 11 12 13 14 15 Enter your answer in the edit fields and then click Check Answer. 16 Io 623
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter4: Equations Of Linear Functions
Section4.5: Correlation And Causation
Problem 15PPS
Related questions
Question
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
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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
![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 c
Use a significance level of 0.05.
Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball
Critical Values of the Pearson Correlation Coefficient r
Diameter
7.3
23.6
4.4
22.1
6.9
4.1
21.2
Circumference 22.9
74.1
13.8 69.4
21.7
12.9
66.6
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
0.707
0.666
0.632
0.602
0.576
0.553
a= 0.01
0.990
0.959
0.917
0.875
0.834
0.798
0.765
10.735
]NOTE: To test H: p = 0
Jagainst H,: p # 0, reject H,
lif the absolute value of r is
greater than the critical
Ivalue in the table.
in
14
15
The regression equation is y =+x.
(Round to five decimal places as needed.)
16
7
8
9
10
1
0.735
0.708
0.684
0.661
0.641
0.623
0.606
11
12
13
14
15
16
17
0.532
0.514
0.497
0.482
Enter your answer in the edit fields and then click Check Answer.
2 parts
remaining
18
l0 468
Io 590
Answer](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F355ac174-1093-4f33-ab7a-528c02966840%2F4164e797-d22f-4f58-a65b-1ced11a6d4b7%2F02pxg9_processed.png&w=3840&q=75)
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 c
Use a significance level of 0.05.
Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball
Critical Values of the Pearson Correlation Coefficient r
Diameter
7.3
23.6
4.4
22.1
6.9
4.1
21.2
Circumference 22.9
74.1
13.8 69.4
21.7
12.9
66.6
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
0.707
0.666
0.632
0.602
0.576
0.553
a= 0.01
0.990
0.959
0.917
0.875
0.834
0.798
0.765
10.735
]NOTE: To test H: p = 0
Jagainst H,: p # 0, reject H,
lif the absolute value of r is
greater than the critical
Ivalue in the table.
in
14
15
The regression equation is y =+x.
(Round to five decimal places as needed.)
16
7
8
9
10
1
0.735
0.708
0.684
0.661
0.641
0.623
0.606
11
12
13
14
15
16
17
0.532
0.514
0.497
0.482
Enter your answer in the edit fields and then click Check Answer.
2 parts
remaining
18
l0 468
Io 590
Answer
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