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In: Statistics and Probability

Dean made a statistical estimation of the cost-output relationship for a shoe store. The data for...

Dean made a statistical estimation of the cost-output relationship for a shoe store. The data for the firm is given in the following table. x 4.5 7 9 10 15 20 33 50 y 3 3.3 3.4 3.5 4.5 5.5 7.5 12 Here x is the output in thousands of pairs of shoes, and y is the cost in thousands of dollars. A. Determine the best-fitting line (using least squares). S1(x) = Incorrect: Your answer is incorrect. r2 = B. Determine the best-fitting quadratic (using the least squares) and the square of the correlation coefficient. S2(x) =

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Expert Solution

Dean made a statistical estimation of the cost-output relationship for a shoe store. The data for the firm is given in the following table.

x 4.5 7 9 10 15 20 33 50

y 3 3.3 3.4 3.5 4.5 5.5 7.5 12

Here x is the output in thousands of pairs of shoes, and y is the cost in thousands of dollars.

Excel Addon Megastat used.

Menu used: correlation/Regression ---- Regression Analysis.

  1. Determine the best-fitting line (using least squares).

y = 1.701+0.196*x

S1(x) = standard error = 0.4055

r2 = 0.9852

Regression Analysis

0.9852

n

8

r

0.9926

k

1

Std. Error of Estimate

0.4055

Dep. Var.

y

Regression output

confidence interval

variables

coefficients

std. error

   t (df=6)

p-value

95% lower

95% upper

Intercept

a =

1.701

x

b =

0.196

0.010

19.966

1.02E-06

0.172

0.220

ANOVA table

Source

SS

df

MS

F

p-value

Regression

65.552

1  

65.552

398.65

1.02E-06

Residual

0.987

6  

0.164

Total

66.539

7  

  1. Determine the best-fitting quadratic (using the least squares) and the square of the correlation coefficient.

Y=2.460+0.103*x+0.002*x2

S2(x) =0.2015

R square = 0.9969

Regression Analysis

0.9969

Adjusted R²

0.9957

n

8

R

0.9985

k

2

Std. Error of Estimate

0.2015

Dep. Var.

y

Regression output

confidence interval

variables

coefficients

std. error

   t (df=5)

p-value

95% lower

95% upper

Intercept

a =

2.460

x

b1 =

0.103

0.022

4.736

.0052

0.047

0.159

xx

b2 =

0.002

0.0003947

4.394

.0071

0.001

0.003

ANOVA table

Source

SS

df

MS

F

p-value

Regression

66.336

2  

33.168

817.03

5.14E-07

Residual

0.203

5  

0.041

Total

66.539

7  


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