Chapter 10 Correlation and Regression Section 10-3 Correlation.
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Transcript of Chapter 10 Correlation and Regression Section 10-3 Correlation.
![Page 1: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/1.jpg)
Chapter 10Correlation and Regression
Section 10-3Correlation
![Page 2: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/2.jpg)
Section 10-3
Exercise #13
Chapter 10Correlation and Regression
![Page 3: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/3.jpg)
a. Draw the scatter plot for the variables.
b. Compute the value of the correlation coefficient.
c. State the hypotheses.
d. Test the significance of the correlation coefficient at = 0.05, using Table I.
e. Give a brief explanation of the type of relationship.
For the following exercise, complete these steps.
![Page 4: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/4.jpg)
A researcher wishes to determine if a person’s age is related to the number of hours he or she exercises per week. The data for the sample are shown below.
Age x 18 26 32 38 52 59
Hours y 10 5 2 3 1.5 1
a. Draw the scatter plot for the variables.
2
4
6
810
0 10 20 3040 50 60Age
Hou
rs
70
![Page 5: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/5.jpg)
b. Compute the value of the correlation coefficient.Age x 18 26 32 38 52 59
Hours y 10 5 2 3 1.5 1
y = 22.5
y2 = 141.25
n = 6
x = 225
x 2 = 9653
xy = 625
![Page 6: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/6.jpg)
y = 22.5 y2 = 141.2 n = 6
x = 225 x 2 = 9653 xy = 625
– =
2 22 2 –
n xy x yr
n x x n y y
![Page 7: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/7.jpg)
y = 22.5 y2 = 141.2 n = 6
x = 225 x 2 = 9653 xy = 625
– 6 625 225 22.5=
2 2 – –6 69653 225 141.25 22.5
r
r = – 0.832
![Page 8: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/8.jpg)
0 1: = 0 and : 0H Hc. State the hypotheses.
Age x 18 26 32 38 52 59
Hours y 10 5 2 3 1.5 1
![Page 9: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/9.jpg)
0 1: = 0 and : 0H H
d. Test the significance of the correlation coefficient at = 0.05, using Table I.
Age x 18 26 32 38 52 59
Hours y 10 5 2 3 1.5 1
n = 6 d.f . = 4 r = – 0.832
C.V . = ± 0.811
Decision: Reject H0 .
![Page 10: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/10.jpg)
0 1: = 0 and : 0H H
e. Give a brief explanation of the type of relationship.
Age x 18 26 32 38 52 59
Hours y 10 5 2 3 1.5 1
n = 6 d.f . = 4 r = – 0.832
There is a significant linear relationship between a person’s age and the number of hours he or she exercises per week.
Decision: Reject H0 .
![Page 11: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/11.jpg)
Section 10-3
Exercise #15
Chapter 10Correlation and Regression
![Page 12: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/12.jpg)
For the following exercise, complete these steps.
a. Draw the scatter plot for the variables.
b. Compute the value of the correlation coefficient.
c. State the hypotheses.
d. Test the significance of the correlation coefficient at = 0.05, using Table I.
e. Give a brief explanation of the type of relationship.
![Page 13: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/13.jpg)
The director of an alumni association for a small college wants to determine whether there is any type of relationship between the amount of an alumnus’s contribution (in dollars) and the years the alumnus has been out of school. The data are shown here.
Years x 1 5 3 10 7 6
Contribution y 500 100 300 50 75 80
![Page 14: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/14.jpg)
a. Draw the scatter plot for the variables.Years x 1 5 3 10 7 6
Contribution y 500 100 300 50 75 80
100
200
300
400500
0 2 4 86 10 20Years
Con
trib
utio
n
30
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b. Compute the value of the correlation coefficient.Years x 1 5 3 10 7 6
Contribution y 500 100 300 50 75 80
y = 1105
y2 = 364,525
n = 6
x = 32
x 2 = 220
xy = 3405
![Page 16: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/16.jpg)
b. Compute the value of the correlation coefficient.
y = 1105 y2 = 364,52 n = 6 x = 32 x 2 = 220 xy = 3405
– =
2 22 2 –
n xy x yr
n x x n y y
![Page 17: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/17.jpg)
b. Compute the value of the correlation coefficient.
y = 1105 y2 = 364,52 n = 6 x = 32 x 2 = 220 xy = 3405
3405 – 32 11056=
2 2 6 220 – 32 364,525 – 11056
r
r = – 0.883
![Page 18: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/18.jpg)
c. State the hypotheses.
0 1: = 0 and : 0H H
Years x 1 5 3 10 7 6
Contribution y 500 100 300 50 75 80
![Page 19: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/19.jpg)
d. Test the significance of the correlation coefficient at = 0.05, using Table I.
Years x 1 5 3 10 7 6
Contribution y 500 100 300 50 75 80
0 1: = 0 and : 0H H
n = 6 d.f . = 4 r = – 0.883
Decision: Reject H0 .
C.V . = ± 0.811
![Page 20: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/20.jpg)
e. Give a brief explanation of the type of relationship.
0 1: = 0 and : 0H H
Years x 1 5 3 10 7 6
Contribution y 500 100 300 50 75 80
n = 6 d.f . = 4 r = – 0.883
There is a significant linear relationship between a person’s age and his or her contribution.
Decision: Reject H0 .
![Page 21: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/21.jpg)
Section 10-3
Exercise #17
Chapter 10Correlation and Regression
![Page 22: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/22.jpg)
For the following exercise, complete these steps.
a. Draw the scatter plot for the variables.
b. Compute the value of the correlation coefficient.
c. State the hypotheses.
d. Test the significance of the correlation coefficient at = 0.05, using Table I.
e. Give a brief explanation of the type of relationship.
![Page 23: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/23.jpg)
A criminology student wishes to see if there is a relationship between the number of larceny crimes and the number of vandalism crimes on college campuses in Southwestern Pennsylvania. The data are shown. Is there a relationship between the two types of crimes?
Number of larceny crimes, x
24 6 16 64 10 25 35
Number of vandalism crimes y
21 3 6 15 21 61 20
![Page 24: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/24.jpg)
a. Draw the scatter plot for the variables.Number of larceny
crimes, x24 6 16 64 10 25 35
Number of vandalism crimes y
21 3 6 15 21 61 20
20
4060
80
0 10 20 4030 50 60larceny crimes
vand
alis
m c
rimes
70 80
![Page 25: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/25.jpg)
b. Compute the value of the correlation coefficient.Number of larceny
crimes, x24 6 16 64 10 25 35
Number of vandalism crimes y
21 3 6 15 21 61 20
y = 147
y2 = 5273
n = 7
x = 180
x 2 = 6914
xy = 4013
![Page 26: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/26.jpg)
y = 147 y2 = 527 n = 7 x = 180 x 2 = 6914 xy = 4013
– =
2 22 2 –
n xy x yr
n x x n y y
![Page 27: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/27.jpg)
y = 147 y2 = 527 n = 7 x = 180 x 2 = 6914 xy = 4013
4013 – 180 1476=
2 2 6 6914 – 180 5273 – 1476
r
r = 0.104
![Page 28: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/28.jpg)
c. State the hypotheses.
0 1: = 0 and : 0H H
Number of larceny crimes, x
24 6 16 64 10 25 35
Number of vandalism crimes y
21 3 6 15 21 61 20
![Page 29: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/29.jpg)
n = 7 d.f .= 5 r = 0.104
d. Test the significance of the correlation coefficient at = 0.05, using Table I.
Number of larceny crimes, x
24 6 16 64 10 25 35
Number of vandalism crimes y
21 3 6 15 21 61 20
C.V. = ± 0.754
Decision: Do not reject H0 .
![Page 30: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/30.jpg)
e. Give a brief explanation of the type of relationship.
Number of larceny crimes, x
24 6 16 64 10 25 35
Number of vandalism crimes y
21 3 6 15 21 61 20
There is not a significant linear relationship between the number of larceny crimes and the number of vandalism crimes.
Decision: Do not reject H0 .
n = 7 d.f .= 5 r = 0.104
![Page 31: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/31.jpg)
Section 10-3
Exercise #23
Chapter 10Correlation and Regression
![Page 32: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/32.jpg)
a. Draw the scatter plot for the variables.
b. Compute the value of the correlation coefficient.
c. State the hypotheses.
d. Test the significance of the correlation coefficient at = 0.05, using Table I.
e. Give a brief explanation of the type of relationship.
For the following exercise, complete these steps.
![Page 33: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/33.jpg)
The average daily temperature (in degrees Fahrenheit) and the corresponding average monthly precipitation (in inches) for the month of June are shown here for seven randomly selected cities in the United States. Determine if there is a relationship between the two variables.
Average daily temperature, x
86 81 83 89 80 74 64
Average monthly precipitation, y
3.4 1.8 3.5 3.6 3.7 1.5 0.2
![Page 34: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/34.jpg)
a. Draw the scatter plot for the variables.Average daily temperature, x
86 81 83 89 80 74 64
Average monthly precipitation, y
3.4 1.8 3.5 3.6 3.7 1.5 0.2
1
2
3
4
0 60Temperature
Prec
ipita
tion
70 80 90 100
5
![Page 35: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/35.jpg)
b. Compute the value of the correlation coefficient.Average daily temperature, x
86 81 83 89 80 74 64
Average monthly precipitation, y
3.4 1.8 3.5 3.6 3.7 1.5 0.2
y = 17.7
y2 = 55.99
n = 7
x = 557
x 2 = 44,739
xy = 1468.9
![Page 36: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/36.jpg)
y = 17.7 y2 = 55.99 n = 7 x = 557 x 2 = 44,739 xy = 1468.9
– =
2 22 2 –
n xy x yr
n x x n y y
![Page 37: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/37.jpg)
y = 17.7 y2 = 55.99 n = 7 x = 557 x 2 = 44,739 xy = 1468.9
r = 7(1468.9) – (557)(17.7)
7(44,739) – (557)2
7(55.99) – (17.7)2
r = 0.883
![Page 38: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/38.jpg)
c. State the hypotheses.
Average daily temperature, x
86 81 83 89 80 74 64
Average monthly precipitation, y
3.4 1.8 3.5 3.6 3.7 1.5 0.2
0 1: = 0 and : 0H H
![Page 39: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/39.jpg)
d. Test the significance of the correlation coefficient at = 0.05, using Table I.Average daily temperature, x
86 81 83 89 80 74 64
Average monthly precipitation, y
3.4 1.8 3.5 3.6 3.7 1.5 0.2
0 1: = 0 and : 0H H
C.V. = ± 0.754
Decision: Reject H0 .
n =7 d.f . = 5 r = 0.883
![Page 40: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/40.jpg)
e. Give a brief explanation of the type of relationship.
Average daily temperature, x
86 81 83 89 80 74 64
Average monthly precipitation, y
3.4 1.8 3.5 3.6 3.7 1.5 0.2
0 1: = 0 and : 0H H
There is a significant linear relationship between temperature and precipitation.
Decision: Reject H0 .
n =7 d.f . = 5 r = 0.883
![Page 41: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/41.jpg)
Chapter 10Correlation and Regression
Section 10-4Regression
![Page 42: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/42.jpg)
Section 10-4
Exercise #13
Chapter 10Correlation and Regression
![Page 43: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/43.jpg)
Find the equation of the regression line and find the y value for the specified x value. Remember that no regression should be done when r is not significant.
Ages and Exercise
11.532510Hours y
595238322618Age x
![Page 44: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/44.jpg)
2
22
– =
–
y x x xya
n x x
Find y when x = 35 years.Ages and Exercise
11.532510Hours y
595238322618Age x
2
22.5 9653 – 225 625 =
6 9653 – 225a
![Page 45: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/45.jpg)
a = 10.499
2
22
– =
–
y x x xya
n x x
Find y when x = 35 years.Ages and Exercise
11.532510Hours y
595238322618Age x
![Page 46: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/46.jpg)
Find y when x = 35 years.Ages and Exercise
11.532510Hours y
595238322618Age x
22
– =
–
n xy x yb
n x x
2
6 625 – 225 22.5 =
6 9653 – 225b
![Page 47: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/47.jpg)
Find y when x = 35 years.Ages and Exercise
11.532510Hours y
595238322618Age x
22
– =
–
n xy x yb
n x x
b = – 0.18
![Page 48: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/48.jpg)
Find y when x = 35 years.Ages and Exercise
11.532510Hours y
595238322618Age x
a = 10.499 b = – 0.18
y = a + bx
y = 10.499 – 0.18x
y = 10.499 – 0.18(35)
y = 4.199 hours
![Page 49: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/49.jpg)
Section 10-4
Exercise #15
Chapter 10Correlation and Regression
![Page 50: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/50.jpg)
Find the equation of the regression line and find the y value for the specified x value. Remember that no regression should be done when r is not significant.Years and Contribution
807550300100500Contribution y, $
6710351Years x
![Page 51: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/51.jpg)
2
22
– =
–
y x x xya
n x x
2
1105 220 – 32 3405 =
6 220 – 32a
Find y when x = 4 years.Years and Contribution
807550300100500Contribution y, $
6710351Years x
![Page 52: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/52.jpg)
a =
243,100 – 108,9601320 – 1024
a = 134,140
296
2
1105 220 – 32 3405 =
6 220 – 32a
a = 453.176
![Page 53: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/53.jpg)
b = 6(3405) – (32)(1105)
6(220) – (32)2
22
– =
–
n xy x yb
n x x
Find y when x = 4 years.Years and Contribution
807550300100500Contribution y, $
6710351Years x
![Page 54: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/54.jpg)
b = 6(3405) – (32)(1105)
6(220) – (32)2
b = – 50.439
b =
20,430 – 35,360296
b =
–14,930296
![Page 55: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/55.jpg)
b = – 50.439
y = a + bx
y = 453.176 – 50.439x
y = $251.42
y = 453.176 – 50.439(4)
a = 453.176
Find y when x = 4 years.Years and Contribution
807550300100500Contribution y, $
6710351Years x
![Page 56: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/56.jpg)
Section 10-4
Exercise #23
Chapter 10Correlation and Regression
![Page 57: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/57.jpg)
Find the equation of the regression line and find the y value when x = 70 ºF. Remember that no regression should be done when r is not significant.
0.2
64
1.53.73.63.51.83.4Avg. mo. Precip. y
748089838186Avg. daily temp. x
Temperatures ( in. F ) and precipitation (in.)
x = 557
x2 = 44,739
y = 17.7xy = 1468.9
![Page 58: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/58.jpg)
x = 557
x2 = 44,739
y = 17.7xy = 1468.9
2
22
– =
–
y x x xya
n x x
a = (17.7)(44,739) – (557)(1468.9)
7(44,739) – (557)2
a = – 8.994
![Page 59: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/59.jpg)
22
– =
–
n xy x yb
n x x
x = 557
x2 = 44,739
y = 17.7xy = 1468.9
b = 7(1468.9) – (557)(17.7)
7(44,739) – (557)2
b = 0.1448
![Page 60: Chapter 10 Correlation and Regression Section 10-3 Correlation.](https://reader035.fdocuments.us/reader035/viewer/2022081415/56649d605503460f94a40c6b/html5/thumbnails/60.jpg)
y = a+ bx
y = – 8.994 + 0.1448x
y = 1.1 inches
y = – 8.994+ 0.1448(70)
a = – 8.994
b = 0.1448
x = 557
x2 = 44,739
y = 17.7xy = 1468.9
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Chapter 10Correlation and Regression
Section 10-5Coefficient of Determination and Standard Error of the Estimate
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Section 10-5
Exercise #9
Chapter 10Correlation and Regression
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Find the coefficients of determination and non-determination when and explain the meaning of:
r2 = 0.49
r = 0.70
49% of the variation of y is due to the variation of x.
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Find the coefficients of determination and non-determination when and explain the meaning of:
1– r 2 = 0.51
r = 0.70
51% of the variation of y is due to chance.
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Section 10-5
Exercise #15
Chapter 10Correlation and Regression
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Compute the standard error of the estimate.
x = 225
x2
= 9653
xy = 625
y = 22.5
y2 = 141.25
n = 6
a = 10.499 b = – 0.18
sest =
y 2 – a y – b xy
n – 2
s
est=
141.25 – 10.499(22.5) – (– 0.18)(625)
6 – 2
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sest =
141.25 – 10.499(22.5) – (– 0.18)(625)
6 2
sest = 4.380625
sest = 2.09
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Section 10-5
Exercise #19
Chapter 10Correlation and Regression
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Find the 90% prediction interval when x = 20 years.
x = 225
y = 22.5
x2
= 9653
y2 = 141.25
xy = 625
n = 6
a = 10.499
b = – 0.18
Age x 18 26 32 38 52 59
Hours y 10 5 2 3 1.5 1
y = 10.499 – 0.18x
= 10.499– 0.18(20) = 6.899
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x = 225
y = 22.5
x2
= 9653
y2 = 141.25
xy = 625
n = 6
a = 10.499
b = – 0.18
y = 6.899
2
2 2( )11 + +
( )– < 2
n x Xyy t s nest n x x
2
2 2– + +
( )1< + 1 2( )
n x Xy t sest n n x x
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1.60 < y < 12.20
6.899 – (2.132)(2.09) 1 + 16
+ 6(20 – 37.5)2
6(9653) 2252< y
< 6.899 + (2.132)(2.09) 1+ 16
+ 6(20 – 37.5)2
6(9653) – 2252
6.899– (2.132)(2.09)(1.19)< y
< 6.899 + (2.132)(2.09)(1.19)
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Section 10-5
Exercise #21
Chapter 10Correlation and Regression
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Find the 90% prediction interval when x = 4 years.
Years x 1 5 3 10 7 6
Contributions y, $ 500 100 300 50 75 80
x = 35
y = 1105
x2
= 220
y2 = 364,525
xy = 3405
n = 6
a = 453.176
b = – 50.439
y = 453.176 – 50.439x
= 251.42 = 453.176 – 50.439(4)
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2
2 2( )11 + + 2 ( )
–– <
–
n x Xyy t s nest n x x
2
2 2 + +
( – )1< + 1 2– ( )
n x Xy t sest n n x x
x = 35
y = 1105
x2
= 220
y2 = 364,525
xy = 3405
n = 6
a = 453.176
b = – 50.439
y = 251.42
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$30.46 < y < $472.38
251.42 – (2.132)(94.22) 1 + 16
+ 6(4 – 5.33)2
6(220) – 322< y
< 251.42+ (2.132)(94.22) 1 + 16
+ 6(4 – 5.33)2
6(220) – 322
251.42 – (2.132)(94.22)(1.1) < y
< 251.42+ (2.132)(94.22)(1.1)
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Chapter 10Correlation and Regression
Section 10-6Multiple Regression
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Section 10-6
Exercise #7
Chapter 10Correlation and Regression
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A manufacturer found that a significant relationship
exists among the number of hours an assembly line
employee works per shift x1, the total
number of items produced x2, and the
number of defective items produced y.
The multiple regression equation is
. Predict the
number of defective items
produced by an employee who has
worked 9 hours and produced
24 items.
y = 9.6 + 2.2x1 – 1.08x2
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y = 3.48 or 3 items
y = 9.6 + 2.2x1 – 1.08x2
= 9.6 + 2.2 9 – 1.08 24y
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Section 10-6
Exercise #9
Chapter 10Correlation and Regression
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An educator has found a significant relationship among a college graduate’s IQ x1, score on the verbal section
of the SAT x2, and income for the first year
following graduation from college y. Predict the income of a college graduate whose IQ is 120 and verbal SAT score is650. The regression equation is
y ' = 5000 + 97x1+ 35x2 .
y = 5000+ 97(120) – 35(650)
y = $39,390