Management of a soft-drink bottling company wished to develop a method for alloc

Question

Management of a soft-drink bottling company wished to develop a method for allocating delivery costs to customers. Although one aspect of cost clearly relates to travel time within a particular route, another type of cost reflects the time required to unload the cases of soft drink at the delivery point. A sample of 20 customers was selected from routes within a territory and the delivery time and the number of cases delivered were measured with the following results: CUSTOMER NUMBER OF CASES DELIVERY TIME (MINUTES) CUSTOMER NUMBER OF CASES DELIVERY TIME (MINUTES) 1 52 32.1 11 161 43.0 2 64 34.8 12 184 49.4 3 73 36.2 13 202 57.2 4 85 37.8 14 218 56.8 5 95 37.8 15 243 60.6 6 103 39.7 16 254 61.2 7 116 38.5 17 267 58.2 8 121 41.9 18 275 63.1 9 143 44.2 19 287 65.6 10 157 47.1 20 298 67.3 Assuming that we wanted to develop a model to predict delivery time based on the number of cases delivered: (a) Use the least-squares method to compute the regression coefficients b0 and bv. (b) Interpet the meaning of b0 and b1 in this problem. (c) Predict the delivery time for 150 cases of soft drink. (d) Should you use the model to predict the delivery time for a customer who is receiving 500 cases of soft drink? Why or why not? (e) Determaine the coefficient of determination, r^2 and explain its meaning in this problem. (f) Preform 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?.

Explanation / Answer

Regression Regression analysis Example : 1. A company sets different prices for a particular stereo system in eight different regions of the country. The accompanying table shows the numbers of units sold and the corresponding prices (in hundreds of dollars). SALES 420 380 350 400 440 380 450 420 PRICE 5.5 6.0 6.5 6.0 5.0 6.5 4.5 5.0 Using Microsoft Excel, the following output is obtained: SUMMARY OUTPUT Regression Statistics Multiple R 0.937137027 R Square 0.878225806 Adjusted R Square 0.857930108 Standard Error 12.74227575 Observations 8 ANOVA df SS MS F Significance F Regression 1 7025.806452 7025.806452 43.27152318 0.000592135 Residual 6 974.1935484 162.3655914 Total 7 8000 Coefficients Standard Error t Stat P-value Intercept 644.516129 36.68873299 17.56714055 2.18343E-06 PRICE -42.58064516 6.473082556 -6.578109392 0.000592135 (a) Plot these data, and estimate the linear regression of sales on price. Here is an Excel-generated scatter plot. You could unscientifically estimate a regression line with a ruler and a pencil, drawing the line so that it “fits” the pattern of dots. The estimated regression line, from the Excel output, is: Sales (Units) = 644.52 – 42.58(Price in $100) (b) What effect would you expect a $100 increase in price to have on sales? A $100 increase in the price will be expected to cause a 42.58 unit drop in sales. Example : 2. On Friday, November 13, 1989, prices on the New York Stock Exchange fell steeply; the Standard and Poors 500-share index was down 6.1% on that day. The accompanying table shows the percentage losses (y) of the twenty-five largest mutual funds on November 13, 1989. Also shown are the percentage gains (x), assuming reinvested dividends and capital gains, for these same funds for 1989, through November 12. y x y x y x 4.7 38.0 6.4 39.5 4.2 24.7 4.7 24.5 3.3 23.3 3.3 18.7 4.0 21.5 3.6 28.0 4.1 36.8 4.7 30.8 4.7 30.8 6.0 31.2 3.0 20.3 4.4 32.9 5.8 50.9 4.4 24.0 5.4 30.3 4.9 30.7 5.0 29.6 3.0 19.9 3.8 20.3 3.3 19.4 4.9 24.6 3.8 25.6 5.2 32.3 (a) Estimate the linear regression of November 13 losses on pre-November 13, 1989, gains. Here is the Excel output: SUMMARY OUTPUT Regression Statistics Multiple R 0.733725713 R Square 0.538353422 Adjusted R Square 0.518281832 Standard Error 0.642482917 Observations 25 ANOVA df SS MS F Significance F Regression 1 11.07156114 11.07156114 26.82166251 2.99579E-05 Residual 23 9.494038861 0.412784298 Total 24 20.5656 Coefficients Standard Error t Stat P-value Intercept 1.885344634 0.506748146 3.72047663 0.001123232 Gains 0.089565882 0.017294171 5.178963459 2.99579E-05 The estimated regression line is: Losses = 1.885 + 0.0896(Gains) (b) Interpret the slope of the sample regression line. Large mutual funds lost about 1.885% on November 13 (the intercept), plus an additional loss of about 0.09% for every 1% in value gained in 1989 before November 13 (the slope). In other words, the amount of value a large mutual fund lost on November 13 depended on how much value had been gained before November 13. Example : 3. For a period of 11 years, the figures in the accompanying table were found for annual change in unemployment rate and annual change in mean employee absence rate due to own illness. Year Change In Unemployment Rate Change In Mean Employee Absence Rate Due To Own Illness 1 -.2 +.2 2 -.1 +.2 3 +1.4 +.2 4 +1.0 -.4 5 -.3 -.1 6 -.7 +.2 7 +.7 -.1 8 +2.9 -.8 9 -.8 +.2 10 -.7 +.2 11 -1.0 +.2 Excel Regression output: SUMMARY OUTPUT Regression Statistics Multiple R 0.805179413 R Square 0.648313886 Adjusted R Square 0.609237652 Standard Error 0.207325489 Observations 11 ANOVA df SS MS F Significance F Regression 1 0.713145275 0.713145275 16.5910019 0.002786228 Residual 9 0.386854725 0.042983858 Total 10 1.1 Coefficients Standard Error t Stat P-value Intercept 0.044851904 0.063473424 0.706624935 0.497684443 Unemployment -0.22425952 0.055057259 -4.073205359 0.002786228 (a) Estimate the linear regression of change in mean employee absence rate due to own illness on change in unemployment rate. Change in Absence Rate = 0.0449 – 0.2243(Change in Unemployment Rate) (b) Interpret the estimated slope of the regression line. A one-percent increase in the unemployment rate is associated with a 0.2243% decrease in the absence rate. (Note that the intercept is not statistically significant.) Example : 4. Refer to the data of Exercise 2. Test against a two-sided alternative the null hypothesis that mutual fund losses on Friday, November 13, 1989, did not depend linearly on previous gains in 1989. We perform this test by looking at the p-value associated with the regression coefficient for the variable “Gains” (see regression output; this p-value is 2.996E-05, or 0.00002996). The null hypothesis that the “Gains” coefficient is zero can be rejected with a very small probability of Type I error (alpha ? 0.00003). Example : 5. An attempt was made to evaluate the forward rate as a predictor of the spot rate in the Canadian treasury bill market. For a sample of seventy-nine quarterly observations, the estimated linear regression: y = .00027 + .7916x was obtained, where y = Actual change in the spot rate x = Change in the forward rate The coefficient of determination was 0.097, and the estimated standard error of the estimator of the slope of the population regression line was 0.2759. (a) Interpret the slope of the estimated regression line. For every 1% change in the forward rate, the spot rate actually changes by about 0.7916%. (b) Interpret the coefficient of determination. About 9.7% of the variation in the spot rate is explained by variation in the forward rate. (c) Test the null hypothesis that the slope of the population regression line is 0 against the alternative that the true slope is positive, and interpret your result. t Our t-table doesn’t go far enough to give us probabilities with 77 degrees of freedom. However, we know that the z distribution will provide a good approximation. Note that we have a one-tailed alternative hypothesis. We reject H0 at any ? > 0.0021 (d) Test against a two-sided alternative the null hypothesis that the slope of the population regression line is 1, and interpret your result. t Our p-value is 0.4472 because this is a two-sided alternative hypothesis. We cannot reject H0 at any ? < 0.4472. Example : 6. For a sample of 306 students in a basic business communications course, the sample regression line y = 58.813 + 0.2875x was obtained, where y = Final student score at the end of the course x = Score on a diagnostic writing skills test given at the beginning of the course The coefficient of determination was 0.1158, and the estimated standard error of the estimated slope of the population regression line was 0.04566. (a) Interpret the slope of the sample regression line. The final score tends to be about 0.2875 higher for every unit of increase in the diagnostic test score. (b) Interpret the coefficient of determination. About 11.58% of the variability in final test scores is explained by variation in diagnostic test scores. (c) The information given allows the null hypothesis that the slope of the population regression line is 0 to be tested against the alternative that it is positive. Carry out this test and state your conclusion. t ridiculously small

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