Question

The following partial MINITAB regression output for the Fresh detergent data relates to predicting demand for future sales periods in which the price difference will be. 10 Predicted Values for New Observations New ObsFit SE Fit 1 8 .0806 0.0648 2 8 .4804 0.0586 95% CI (7.9479, 8.2133) (8.3604, 8.6004) 95% PI (7.4187, 8.7425) (7.8209, 9.1398) Click here for the Excel Data File (a) Report a point estimate of and a 95 percent confidence interval for the mean demand for Fresh in all sales periods when the price difference is . 10. (Round your Cl answer to 3 decimal places and other answer to 4 decimal places.) Point estimate = Confidence interval = [ (b) Report a point prediction of and a 95 percent prediction interval for the actual demand for Fresh in an individual sales period when the price difference is . 10. (Round your Pl answer to 3 decimal places and other answer to 4 decimal places.) Point estimate = Confidence interval = [ (c) Remembering that s = .316561 and that the distance value equals (s/s), use so from the computer output to hand calculate the distance value when x = .10. (Round your answer to 4 decimal places.) dv = (d) For this case: n = 30, bo = 7.81409, bn = 2.6652, and s = .316561. Using this information, and your result from part (c), find 99 percent confidence and prediction intervals for mean and individual demands when x = 10. (Round your answers to 4 decimal places.) 99% C.I. 99% P.I.:

Answer 1

a) Answer: --- Date: -3/12/2019 From the output, the 95 percent confidence interval for the mean demand for Fresh in all sales periods when the price difference is 10 is [7.948, 8.213]. And, the point estimate is 8.0806 From the output, the 95 percent prediction interval for the mean demand for Fresh in all sales periods when the price difference is .10 is (7.419, 8.743). And, the point estimate is 8.0806 Standard deviation s =0.316561 From the output, when x =10 the value of s, is 0.0648 Now compute the distance value when x =10 as follows: 0.0648 0316561 = (0.2047 = 0.041902 = 0.0419 Therefore, dvz 0.0419 First compute the predicted value when x=0.10. y = b +bx = 7.81409+2.6652(0.10) = 8.08061 Degrees of freedom df = n-2 = 30-2 = 28 At 99% confidence (a=0.01) with df = 28. the tabulated value is 12/24 =0.01221 = 2.763 Then, the 99% confidence interval for mean demand when price difference is 0.10 is, Û21.245V = 8.08061: (2.763)(0.316561) 70.041902 = 8.08061: (2.763)(0.0648) = 8.08061: 0.179042 = [7.9016, 8.2597] The 99% prediction interval for actual demand when price difference is 0.10 ýt 2 1 + = 8.08061+(2.7630.316561) 1+0,041902 = 8.08061: (2.763)(0.323125) = 8.08061 0.892795 = [7.1878, 8.9734] Thus, 99% CI is [7.9016, 8.2597] & 99%PI is [7.1878, 8.9734] From the output, the 95 percent confidence interval for the mean demand for Fresh in all sales periods when the price difference is 0.25 is [8.360, 8.600). And, the point estimate is 8.4804 From the output, the 95 percent prediction interval for the mean demand for Fresh in all sales periods when the price difference is 0.25 is [7.821, 9.140). And the point estimate is 8.4804 Standard deviation 5 =0.316561 From the output, when x =0.25 the value of s, is 0.0586 Now compute the distance value when x-10 as follows: 0.05861 (0.316561 = (0.185114) = 0.034267 <0.0343 Therefore, dvz 0.0343 First compute the predicted value when x = 0.25. û = bo + bx = 7.81409+2.6652(025) = 8.48039 Degrees of freedom df = -2 = 30 -2 = 28 At 99% confidence (a=0.01) with df = 28. the tabulated value is 1.24 -0.01 228 -2.763 Then, the 99% confidence interval for mean demand when price difference is 0.25 is, ýtta 2.5 vay = 8.48039 (2.763) (0.316561) 0.034267 = 8.48039 (2.763)(0.0586) = 8.48039 +0.161912 = [8.3185, 8.6423] The 99% prediction interval for actual demand when price difference is 0.25 is, ýt 2 sv1 + 4 = 8.48039 (2.763) (0.316561) V1 +0.034267 = 8.48039 (2.763) (0.321939) = 8.48039 0.889518 = 17.5909, 9.36991 Thus, 99% CI is [8.3185, 8.6423] & 99%PI is (7.5909.9.3699]