In a statistics class, last spring, the students measured their height, their ar
ID: 3375072 • Letter: I
Question
In a statistics class, last spring, the students measured their height, their arm span (finger tip to fingertip), and the length of their forearm (elbow to finger tip). All distances were measured in inches We collected data to answer this question: Which is a better predictor of someone's height, their arm span or their forearm length? In other words, will someone's forearm length or their arm span more accurately predict their height? Listed below are the data that were collected TABLE OF DATA: MEASUREMENTS FROM STUDENTS Arm Forearm Student Height span ength 60.5 68 17.5 60 16.6 17 63.5 17 1 67 18 68 67 18.5 71.5 TO DO Use your skills from Chapter 8 to create the better linear regression line to predict a person's height. You'll need to do two linear regressions then determine and argue which equation is a "better" model than the otherExplanation / Answer
Result:
Regression Analysis
r²
0.777
n
8
r
0.881
k
1
Std. Error
1.857
Dep. Var.
Height
ANOVA table
Source
SS
df
MS
F
p-value
Regression
72.0326
1
72.0326
20.89
.0038
Residual
20.6861
6
3.4477
Total
92.7188
7
Regression output
confidence interval
variables
coefficients
std. error
t (df=6)
p-value
95% lower
95% upper
Intercept
-0.4538
14.3211
-0.032
.9757
-35.4963
34.5887
Arm span
1.0217
0.2235
4.571
.0038
0.4748
1.5687
The regression to predict height from Arm span,
Height = -0.4538+1.0217*Arm span
Regression Analysis
r²
0.877
n
8
r
0.936
k
1
Std. Error
1.380
Dep. Var.
Height
ANOVA table
Source
SS
df
MS
F
p-value
Regression
81.2944
1
81.2944
42.70
.0006
Residual
11.4243
6
1.9041
Total
92.7188
7
Regression output
confidence interval
variables
coefficients
std. error
t (df=6)
p-value
95% lower
95% upper
Intercept
-5.8948
10.8513
-0.543
.6065
-32.4469
20.6573
Forearm length
4.1332
0.6325
6.534
.0006
2.5854
5.6810
The regression to predict height from Forearm length,
Height = -5.8948 +4.1332 *Forearm length
The R square value, the coefficient of deamination with the model Arm span is 0.777. The percentage of variance explained is 77.7%.
The R square value, the coefficient of deamination with the model Forearm length
is 0.877. The percentage of variance explained is 87.7%.
The model with Forearm length is better model.
Regression Analysis
r²
0.777
n
8
r
0.881
k
1
Std. Error
1.857
Dep. Var.
Height
ANOVA table
Source
SS
df
MS
F
p-value
Regression
72.0326
1
72.0326
20.89
.0038
Residual
20.6861
6
3.4477
Total
92.7188
7
Regression output
confidence interval
variables
coefficients
std. error
t (df=6)
p-value
95% lower
95% upper
Intercept
-0.4538
14.3211
-0.032
.9757
-35.4963
34.5887
Arm span
1.0217
0.2235
4.571
.0038
0.4748
1.5687