Oxygen 47.01 23.91 43 61 170 190 8.1 1438 54.37 24.99 45 46 157 169 6.22 618 59.
ID: 2909671 • Letter: O
Question
Oxygen
47.01
23.91
43
61
170
190
8.1
1438
54.37
24.99
45
46
157
169
6.22
618
59.73
25.11662
44
39
165
173
5.626667
277
50.31
26.25858
39
56
179
181
6.326667
559
45.17
24.5392
48
59
177
177
8.32
4000
45.93
24.99592
41
71
177
181
7.993333
2961
49.14
25.10494
44
65
163
171
7.346667
1551
40.25
23.73396
45
64
176
177
9.1
8955
60.42
24.90943
39
49
171
187
6.246667
516
51.52
24.73187
45
46
169
169
7.153333
1278
37.93
23.93121
46
57
187
193
9.506667
13449
44.78
23.97361
46
52
177
177
7.766667
2360
48.13
26.22539
48
48
163
165
7.506667
1820
52.85
25.25057
55
51
167
171
6.893333
985
49.95
26.80163
50
45
181
186
8.1
551
41.04
26.29004
52
58
169
173
7.76
2344
47.27
24.21605
52
49
163
169
7.086667
1195
47.04
26.74587
49
49
163
165
6.933333
1025
51.36
24.24693
50
68
169
169
7.026667
1126
40.07
25.69204
58
59
175
177
8.726667
6165
46.08
24.28803
55
63
157
166
7.64
2079
46.35
24.46296
53
49
165
167
7.3
710
55.13
26.29201
51
49
147
156
6.466667
643
46.09
24.51372
52
49
173
173
7.813333
2473
39.88
25.96763
55
45
169
173
9.193333
9831
45.85
25.42663
52
60
187
189
7.42
1669
50.95
20.49481
58
50
149
156
6.686667
801
48.74
24.67682
50
57
187
189
6.373333
586
48.12
23.38058
49
53
171
177
8.2
2751
48.17
24.44704
53
54
171
173
9.3
1881
After looking at the graphs and the p-value from the Pearson correlation, the one best predictor for oxygen consumptionis the run time for one mile.
Build a simple linear regression model using the best predictor you found in question 1 and write the regression equation for estimating average oxygen consumption in the following space.
Rcmdr> RegModel.9 <- lm(Oxygen~RunTimeOneMile, data=Dataset)
Rcmdr> summary(RegModel.9)
Call:
lm(formula = Oxygen ~ RunTimeOneMile, data = Dataset)
Residuals:
Min 1Q Median 3Q Max
-5.9824 -2.5403 -0.3897 1.7629 7.8372
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 80.7309 4.4969 17.952 < 2e-16 ***
RunTimeOneMile -4.3439 0.5916 -7.343 0.0000000538 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 3.181 on 28 degrees of freedom
Multiple R-squared: 0.6582,Adjusted R-squared: 0.646
F-statistic: 53.92 on 1 and 28 DF, p-value: 0.00000005385
The Intercept (p-value = 2e-16)and the beta coefficient (p-value = 0.0000000538) are both significantly different from zero. The following equation is used to find the equation of the regression line, the variables ? and ? are highlighted above.
Equation of the Regression Line:
g= a+ b×x
g= 80.7309 + (-4.3439) ×x
Use the model you built in 2. to estimate theaverageoxygen consumption for all people whose values for all variables are in the following data table, with a 95% confidence interval. (Since only one significant predictor is to be used in this exercise, you would just use one of the following data values in the prediction.)
Age
BMI
Oxygen
RunTime
RestPulse
RunPulse
MaxPulse
Ranking
58
23
32.948
11
61
135
153
1200
Use the model you built in 2. to estimate the oxygen consumption for a personwhose values for all variables are in the data table above, with a 95% confidence interval.
Verify the assumptions for this regression model and comment on your findings.
[Copy and paste your R output here!]
Can the model you built earlier be used to estimate the oxygen consumption for subjects whose RunTime is 18, RestPulse is 80, RunPulse is 100, MaxPulse is 100, or Ranking is 15,000? Please answer yes or no and explain your answer. If your answer is “yes”, what would be your estimated average oxygen consumption for this type of subjects?
Oxygen
BMI Age Rest Pulse Run Pulse Max Pulse Run Time One Mile Rankings47.01
23.91
43
61
170
190
8.1
1438
54.37
24.99
45
46
157
169
6.22
618
59.73
25.11662
44
39
165
173
5.626667
277
50.31
26.25858
39
56
179
181
6.326667
559
45.17
24.5392
48
59
177
177
8.32
4000
45.93
24.99592
41
71
177
181
7.993333
2961
49.14
25.10494
44
65
163
171
7.346667
1551
40.25
23.73396
45
64
176
177
9.1
8955
60.42
24.90943
39
49
171
187
6.246667
516
51.52
24.73187
45
46
169
169
7.153333
1278
37.93
23.93121
46
57
187
193
9.506667
13449
44.78
23.97361
46
52
177
177
7.766667
2360
48.13
26.22539
48
48
163
165
7.506667
1820
52.85
25.25057
55
51
167
171
6.893333
985
49.95
26.80163
50
45
181
186
8.1
551
41.04
26.29004
52
58
169
173
7.76
2344
47.27
24.21605
52
49
163
169
7.086667
1195
47.04
26.74587
49
49
163
165
6.933333
1025
51.36
24.24693
50
68
169
169
7.026667
1126
40.07
25.69204
58
59
175
177
8.726667
6165
46.08
24.28803
55
63
157
166
7.64
2079
46.35
24.46296
53
49
165
167
7.3
710
55.13
26.29201
51
49
147
156
6.466667
643
46.09
24.51372
52
49
173
173
7.813333
2473
39.88
25.96763
55
45
169
173
9.193333
9831
45.85
25.42663
52
60
187
189
7.42
1669
50.95
20.49481
58
50
149
156
6.686667
801
48.74
24.67682
50
57
187
189
6.373333
586
48.12
23.38058
49
53
171
177
8.2
2751
48.17
24.44704
53
54
171
173
9.3
1881
Explanation / Answer
The Regression Model that has been drawn you in first part using one variable is,
RegModel = lm(Oxygen~RunTimeOneMile, data=Dataset)
Summary Model:
Coefficients: Estimate Std. Error t value Pr(>|t|)
(Intercept) 80.7309 4.4969 17.952 < 2e-16 ***
RunTimeOneMile -4.3439 0.5916 -7.343 0.0000000538 ***
The p-value is less than 0.05 so variable is RunTimeOneMile significant.
The Fitted Model is:
Oxygen = 80.7309 - 4.3439 * RunTimeOneMile
The Adjusted R-squared: 0.646 so model is good
For estimating the value of given table we want to build the Regression model using all variable:
The output of model is,
lm(formula = Oxygen ~ ., data = data)
therefore the fitted model is,
Oxygen = 97.3060 -0.33616 * BMI - 0.22732 * Age -0.15973 * RestPulse -0.18965 * RunPulse + 0.15690 * MaxPulse -1.97343 * RunTimeOneMile -0.0006023 * Rankings
Summary of model:
Coefficients: Estimate Std. Error t value Pr(>|t|)
(Intercept) 97.3060623 18.5833390 5.236 0.0000298 ***
BMI -0.3361646 0.4408102 -0.763 0.4538
Age -0.2273289 0.1262583 -1.801 0.0855 .
RestPulse -0.1597365 0.0758004 -2.107 0.0467 *
RunPulse -0.1896556 0.1179991 -1.607 0.1223
MaxPulse 0.1569042 0.1275520 1.230 0.2316
RunTimeOneMile -1.9734351 0.9016637 -2.189 0.0395 *
Rankings -0.0006023 0.0002642 -2.280 0.0327 *
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 2.767 on 22 degrees of freedom
Multiple R-squared: 0.7967,
Adjusted R-squared: 0.732
F-statistic: 12.31 on 7 and 22 DF, p-value: 0.000002553
The above results are drawn at 0.05 significance that is at 95% confidence level
The value of Adjusted R^2 is 0.732 and this is greater than the model 1 so model two is good than the model one.
The predicted value of Oxygen at this given table:
Age
BMI
Oxygen
RunTime
RestPulse
RunPulse
MaxPulse
Ranking
58
23
32.948
11
61
135
153
1200
Put the given value in the above model:
Oxygen = 97.3060 -0.33616 * BMI - 0.22732 * Age -0.15973 * RestPulse -0.18965 * RunPulse + 0.15690 * MaxPulse -1.97343 * RunTimeOneMile -0.0006023 * Rankings
> Oxygen=97.3060-0.33616*23-0.22732*58-0.15973*61-0.18965*135+0.15690*153-0.000602*1200
> Oxygen
[1] 64.32678
So predicted value of oxygen consumption is 64.3267
The predicted value of consumed oxygen at the given value,
RunTime is 18, RestPulse is 80, RunPulse is 100, MaxPulse is 100, or Ranking is 15,000
Oxygen = 97.3060 -0.33616 * BMI - 0.22732 * Age -0.15973 * RestPulse -0.18965 * RunPulse + 0.15690 * MaxPulse -1.97343 * RunTimeOneMile -0.0006023 * Rankings
= -109.5736
i.e. not possible to consume oxygen at this value
>>>>>>>>>>>>> Best Luck >>>>>>>>>>>>>
Coefficients Value Intercept 97.306 BMI -0.3361646 Age -0.2273289 RestPulse -0.1597365 RunPulse -0.1896556 MaxPulse 0.1569042 RunTimeOneMile -1.9734351 Rankings -0.0006023Related Questions
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