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.

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.

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.

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.

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:

Analysis of variance table for regression model:

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.

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:

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