linear regression

logistic_guy

Senior Member
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A study was conducted at Virginia Tech to determine if certain static arm-strength measures have an influence on the “dynamic lift” characteristics of an individual. Twenty-five individuals were subjected to strength tests and then were asked to perform a weightlifting test in which weight was dynamically lifted overhead. The data are given here.

Individual​
Arm Strength, \(\displaystyle x\)​
Dynamic Lift, \(\displaystyle y\)​
1​
17.3​
71.7​
2​
19.3​
48.3​
3​
19.5​
88.3​
4​
19.7​
75.0​
5​
22.9​
91.7​
6​
23.1​
100.0​
7​
26.4​
73.3​
8​
26.8​
65.0​
9​
27.6​
75.0​
10​
28.1​
88.3​
11​
28.2​
68.3​
12​
28.7​
96.7​
13​
29.0​
76.7​
14​
29.6​
78.3​
15​
29.9​
60.0​
16​
29.9​
71.7​
17​
30.3​
85.0​
18​
31.3​
85.0​
19​
36.0​
88.3​
20​
39.5​
100.0​
21​
40.4​
100.0​
22​
44.3​
100.0​
23​
44.6​
91.7​
24​
50.4​
100.0​
25​
55.9​
71.7​

\(\displaystyle \bold{(a)}\) Estimate \(\displaystyle \beta_0\) and \(\displaystyle \beta_1\) for the linear regression curve \(\displaystyle \mu_Y|_x = \beta_0 + \beta_1x\).
\(\displaystyle \bold{(b)}\) Find a point estimate of \(\displaystyle \mu_Y|_{30}\).
\(\displaystyle \bold{(c)}\) Plot the residuals versus \(\displaystyle x'\)s (arm strength). Comment.
 
A study was conducted at Virginia Tech to determine if certain static arm-strength measures have an influence on the “dynamic lift” characteristics of an individual. Twenty-five individuals were subjected to strength tests and then were asked to perform a weightlifting test in which weight was dynamically lifted overhead. The data are given here.

Individual​
Arm Strength, \(\displaystyle x\)​
Dynamic Lift, \(\displaystyle y\)​
1​
17.3​
71.7​
2​
19.3​
48.3​
3​
19.5​
88.3​
4​
19.7​
75.0​
5​
22.9​
91.7​
6​
23.1​
100.0​
7​
26.4​
73.3​
8​
26.8​
65.0​
9​
27.6​
75.0​
10​
28.1​
88.3​
11​
28.2​
68.3​
12​
28.7​
96.7​
13​
29.0​
76.7​
14​
29.6​
78.3​
15​
29.9​
60.0​
16​
29.9​
71.7​
17​
30.3​
85.0​
18​
31.3​
85.0​
19​
36.0​
88.3​
20​
39.5​
100.0​
21​
40.4​
100.0​
22​
44.3​
100.0​
23​
44.6​
91.7​
24​
50.4​
100.0​
25​
55.9​
71.7​

\(\displaystyle \bold{(a)}\) Estimate \(\displaystyle \beta_0\) and \(\displaystyle \beta_1\) for the linear regression curve \(\displaystyle \mu_Y|_x = \beta_0 + \beta_1x\).
\(\displaystyle \bold{(b)}\) Find a point estimate of \(\displaystyle \mu_Y|_{30}\).
\(\displaystyle \bold{(c)}\) Plot the residuals versus \(\displaystyle x'\)s (arm strength). Comment.
First - Please define
Linear regression
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem
 
[imath]\beta_0 = \frac{\sum \left(x_i - \overline x\right) \left(y_i - \overline y\right)}{\sum \left(x_i - \overline x \right)^2}[/imath]

[imath]\overline y = \beta_0 \overline x + c[/imath]
 
[imath]\beta_0 = \frac{\sum \left(x_i - \overline x\right) \left(y_i - \overline y\right)}{\sum \left(x_i - \overline x \right)^2}[/imath]

[imath]\overline y = \beta_0 \overline x + c[/imath]
It seems that you are an expert in data analysis. Teach me!

😭😭

\(\displaystyle \large \bold{(a)}\)


\(\displaystyle \large \mu_Y|_x = \beta_0 + \beta_1x\)


\(\displaystyle \large \beta_0 = \frac{\sum y\sum x^2 - \sum x\sum xy}{n\sum x^2 - \left(\sum x \right)^2}\)


\(\displaystyle \large \beta_1 = \frac{n\sum xy - \sum x\sum y}{n\sum x^2 - \left(\sum x\right)^2}\)

Individual​
Arm Strength, \(\displaystyle x\)​
Dynamic Lift, \(\displaystyle y\)​
\(\displaystyle xy\)​
\(\displaystyle x^2\)​
1​
17.3​
71.7​
1240.41​
299.29​
2​
19.3​
48.3​
932.19​
372.49​
3​
19.5​
88.3​
1721.85​
380.25​
4​
19.7​
75.0​
1477.50​
388.09​
5​
22.9​
91.7​
2099.93​
524.41​
6​
23.1​
100.0​
2310.00​
533.61​
7​
26.4​
73.3​
1935.12​
696.96​
8​
26.8​
65.0​
1742.00​
718.24​
9​
27.6​
75.0​
2070.00​
761.76​
10​
28.1​
88.3​
2481.23​
789.61​
11​
28.2​
68.3​
1926.06​
795.24​
12​
28.7​
96.7​
2775.29​
823.69​
13​
29.0​
76.7​
2224.30​
841.00​
14​
29.6​
78.3​
2317.68​
876.16​
15​
29.9​
60.0​
1794.00​
894.01​
16​
29.9​
71.7​
2143.83​
894.01​
17​
30.3​
85.0​
2575.50​
918.09​
18​
31.3​
85.0​
2660.50​
979.69​
19​
36.0​
88.3​
3178.80​
1296.00​
20​
39.5​
100.0​
3950.00​
1560.25​
21​
40.4​
100.0​
4040.00​
1632.16​
22​
44.3​
100.0​
4430.00​
1962.49​
23​
44.6​
91.7​
4089.82​
1989.16​
24​
50.4​
100.0​
5040.00​
2540.16​
25​
55.9​
71.7​
4006.03​
3125.81​
\(\displaystyle \textcolor{black}{\bold{Sum}}\)​
\(\displaystyle \textcolor{red}{\bold{778.7}}\)​
\(\displaystyle \textcolor{blue}{\bold{2050.0}}\)​
\(\displaystyle \textcolor{green}{\bold{65164.04}}\)​
\(\displaystyle \textcolor{purple}{\bold{26591.63}}\)​
 
We start by finding \(\displaystyle \beta_0\).

\(\displaystyle \large \beta_0 = \frac{\sum y\sum x^2 - \sum x\sum xy}{n\sum x^2 - \left(\sum x \right)^2}\)

Plug in numbers.

\(\displaystyle \large \beta_0 = \frac{(2050)(26591.63) - (778.7)(65164.04)}{25(26591.63) - (778.7)^2} = 64.5292\)
 
Then, we continue to find \(\displaystyle \beta_1\) as well.

\(\displaystyle \large \beta_1 = \frac{n\sum xy - \sum x\sum y}{n\sum x^2 - \left(\sum x\right)^2}\)

Plug in numbers.

\(\displaystyle \large \beta_1 = \frac{25(65164.04) - (778.7)(2050)}{25(26591.63) - \left(778.7\right)^2} = 0.5609\)
 
\(\displaystyle \bold{(a)}\) Estimate \(\displaystyle \beta_0\) and \(\displaystyle \beta_1\) for the linear regression curve \(\displaystyle \mu_Y|_x = \beta_0 + \beta_1x\).
\(\displaystyle \mu_Y|_x = \textcolor{red}{64.5292} + \textcolor{blue}{0.5609}x\)
 
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