My proposal is to find 30 different pair of shoes and count the number of eyelets on the shoe and then measure the length of the shoe lace. I think it will be a positive association and semi strong because I think the more eyelets on the shoe the longer the shoe lace.



Shirley Hayward
11/20/17
Biometrika Project


Link to my spreadsheet: https://docs.google.com/a/oxbowhs.org/spreadsheets/d/1ErEpyBAsCnfpDzqAQFK8MCOar9ajV2gGCUyPCc8DxE4/edit?usp=sharing


Simple linear regression results:
Dependent Variable: Length_of_shoe_lace
Independent Variable: No_of_eyelets
Length_of_shoe_lace = 18.91935 + 6.0575663 No_of_eyelets
Sample size: 30
R (correlation coefficient) = 0.98126245
R-sq = 0.96287599
Estimate of error standard deviation: 2.4310422


Parameter estimates:
Parameter
Estimate
Std. Err.
Alternative
DF
T-Stat
P-value
Intercept
18.91935
1.4269671
≠ 0
28
13.258435
<0.0001
Slope
6.0575663
0.22478195
≠ 0
28
26.948632
<0.0001


Analysis of variance table for regression model:
Source
DF
SS
MS
F-stat
P-value
Model
1
4291.9876
4291.9876
726.22879
<0.0001
Error
28
165.47905
5.9099662


Total
29
4457.4667



R and R^2 are very strong suggesting that this data is linear.
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external image BKj8Ehc_JDUEN8gHWbpbAC9ckH6eOtT8-bqaEtU8rmy1zAG6MIv_glXL4fwbjHEyHS_DZI_ncoEXZRDCVoGKv0NGsbYccR5t5NpTWw4jZ-EQkTip-m2V2QqZ4p_gymeETxB5GSKc


I’m seeing that there is a pattern in both the scatter plot and the residual scatter plot. I am going to try and straighten the data by graphing the other points.


external image -YMjBrtVtaP7tyEmMoCCM6pSRrmduxLxU3QvFDyI46HpMPoppJgE6AEugkjqBKW50LZsjNSK_-LHBXLeLJ-2x7KcrlieHGIAd_ACUxI-uf92WaXUAGOwCtgiuQ84GGNYz8DZpSLw
The first graph is the number of eyelets versus the reciprocal of length. The middle graph is the number of eyelets versus The reciprocal sqrt(length). The last one is the number of eyelets versus the Log(length). Any of these can straighten the data however I feel that the middle one is the best because the residuals are very small on the y-axis.


Simple linear regression results:
Dependent Variable: TheReciprocalSqrt(Legnth
Independent Variable: No_of_eyelets
TheReciprocalSqrt(Legnth = 0.18449247 - 0.0078804985 No_of_eyelets
Sample size: 30
R (correlation coefficient) = -0.95933519
R-sq = 0.920324
Estimate of error standard deviation: 0.0047391355


Parameter estimates:
Parameter
Estimate
Std. Err.
Alternative
DF
T-Stat
P-value
Intercept
0.18449247
0.002781766
≠ 0
28
66.322067
<0.0001
Slope
-0.0078804985
0.00043819566
≠ 0
28
-17.983972
<0.0001


This links back to my mini proposal by proving it correct. As the number of eyelets in shoes increased as did the length of the shoelace in inches. I thought that it would be a positive linear, strong relationship, which was proven with all the analysis I did.






Here is a pictures
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