Question

13.74 Please answer only question H. Also, please include how Sigma can be founded. Thank you for your time. (show all work)

13.74 Management of a soft-drink bottling company has the business objective of developing a method for allocating delivery costs to customers. Although one cost clearly relates to travel time within a particular route, another variable cost reflects the time required to unload the cases of soft drink at the delivery point. To begin, management decided to develop a regression model to predict delivery time based on the number of cases delivered. A sample of 20 deliveries within a territory was selected. The deliv- ery times and the number of cases delivered were organized in the 3 following table and stored in Delivery 184 73 Delivery Delivery Number Time Number Time Customer of Cases minutes) Customer of Cases (minutes) 32.1 161 43.0 64 34.8 49.4 36.2 57.2 37.8 56.8 37.8 60.6 39.7 61.2 38.5 121 41.9 63.1 44.2 65.6 157 47.1 67.3 58,2 143 287 298 a. Use the least-squares method to compute the regression coeffi- cients by and by b. Interpret the meaning of b, and b, in this problem. c. Predict the mean delivery time for 150 cases of soft drink. d. Should you use the model to predict the delivery time for a cus- tomer who is receiving 500 cases of soft drink? Why or why not? e. Determine the coefficient of determination, and explain its meaning in this problem. f. Perform a residual analysis. Is there any evidence of a pattern in the residuals? Explain. g. At the 0.05 level of significance, is there evidence of a linear relationship between delivery time and the number of cases delivered? h. Construct a 95% confidence interval estimate of the mean delivery time for 150 cases of soft drink and a 95% prediction interval of the delivery time for a single delivery of 150 cases of soft drink.

Answer 1

Ʃx =	3398
Ʃy =	972.5
Ʃxy =	182677.8
Ʃx² =	701940
Ʃy² =	49802.31
Sample size, n =	20
x̅ = Ʃx/n = 3398/20 =	169.9
y̅ = Ʃy/n = 972.5/20 =	48.625
SSxx = Ʃx² - (Ʃx)²/n = 701940 - (3398)²/20 =	124619.8
SSyy = Ʃy² - (Ʃy)²/n = 49802.31 - (972.5)²/20 =	2514.4975
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 182677.8 - (3398)(972.5)/20 =	17450.05

Slope, b = SSxy/SSxx = 17450.05/124619.8 =    0.140026304

y-intercept, a = y̅ -b* x̅ = 48.625 - (0.14003)*169.9 =    24.83453095

Regression equation :   

ŷ = 24.8345 + (0.14) x  

Sum of Square error, SSE = SSyy -SSxy²/SSxx = 2514.4975 - (17450.05)²/124619.8 = 71.03149

Standard error, σ = √(SSE/(n-2)) = √(71.03149/(20-2)) =    1.98650

Predicted value of y at x =    150

ŷ = 24.8345 + (0.14) * 150 = 45.8385  

Significance level, α =    0.05

Critical value, t_c = T.INV.2T(0.05, 18) = 2.1009  

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95% confidence interval for the mean:

1 (150-169.912 } (45.8385 +2.1009 + 1.98651页: 124619.8 , (44.8761,46,8009)

95% prediction interval for the mean:

11 (150-169.92 ( 45.8385 +2.1000 + 1.9865 11 + + - 124619.8 (41.5555, 50.1215)

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