Management of a soft-drink company wants to develop a method for allocating deli

Question

Management of a soft-drink company wants to develop a method for allocating delivery costs to customers. Although one cost clearly relates to travel time withing a particular route, another variable cost reflects the time required to unload the cases of soft drink at the delivery point. A sample of 20 deliveries within a territory was selected. The delivery times and the number of cases delivered were recorder and provided:

e) Determine the coefficient of determination, r2, and explain its imeaning in this problem.

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) Cpnstruct a 95% confidence interval estimate of the mean delivery time for 150 caes of soft drink and 95% confidence interval prediction interval of the delivery time for a single delivery of 150 cases of soft drink.

i) Determine residual standard deviation of the model. What does this number tell us?

j) Find the sample correlation coefficient between number of cases delivered and delivery time. Is there statistical evidence that a positive correlation between these variables exists? Test at 0.01 significance.

Customer Number of cases Delivery Time 1 52 32.1 2 64 34.8 3 73 36.2 4 85 37.8 5 95 37.8 6 103 39.7 7 116 38.5 8 121 41.9 9 143 44.2 10 157 47.1 11 161 43 12 184 49.4 13 202 57.2 14 218 56.8 15 243 60.6 16 254 61.2 17 267 58.2 18 275 63.1 19 287 65.6 20 298 67.3

Explanation / Answer

Here is the given regression summary output

(1) Here COefficient of Determination R2  = 0.971 and it tells that there is very strong realtionship between number of cases delivered and delivery time.

(2) From the given regression summary we can see that F value = 619.19 and P - value =2.15 * 10-15, which is under significance level alpha = 0.05 so, the regression is very very significant and that means the relationship is significant.

(3) For 95% confidence interval estimate for Mean delivery time cases for 150 cases is

Mean delivery time t150 = 24.84 + 0.14 * 150 = 45.84 min

95% confidence interval estimate = (41.88, 49.80 )

Lower interval = 22.62 + 0.1282 * 150 = 41.88

Upper interval = 27.04 + 0.1518 * 150 = 49.80

all above values can be taken from above.

95% confidence prediction interval = t150 +- 1.96 * (11.50/sqrt(20)) = 45.84 +- 1.96 * 2.5714 =  ( 40.80, 50.88)

where = 11.50 min , calculated by std deviatio of 20 cases and n = 20

= ( 40.80, 50.88)

(i) residual standard deviation =

residual std. deviation = 1.9335

Lower residual sstd. dev value tells us that data is statistically linear and fitted in linear line.

(f) sample correlation coefficient between number of cases delivered and delivery time = 6.94

Here t - stat = 24.88 and P - value = 2.2 * 10-15, which is under significance level 0.01 so we can say that the correlation is statisticall significant. And, yes there is positive correlation/.

Multiple R 0.985774 R Square 0.971751 Adjusted R Square 0.970182 Standard Error 1.986503 Observations 20 ANOVA df SS MS F Significance F Regression 1 2443.466 2443.466 619.1956 2.15E-15 Residual 18 71.03149 3.946194 Total 19 2514.498 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept 24.83453 1.054219 23.55729 5.61E-15 22.6197 27.04936 22.6197 27.04936 Number of cases 0.140026 0.005627 24.88364 2.15E-15 0.128204 0.151849 0.128204 0.151849

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