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

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Chapter1: Starting With Matlab
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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: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
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
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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