The following partial MINITAB regression output for the Fresh detergent data rel
ID: 3233555 • Letter: T
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
(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 answers to 3 decimal places.)
(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 answers to 3 decimal places.)
dv =
(d) For this case: n = 30, b0 = 8.388664, b1 = -.091904, and s = .587952. 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.)
Explanation / Answer
Solution:
(a)
From the given table the point estimate is 8.3795
And a 95% confidence interval for the mean demand for fresh in all sales period when the price difference is 0.10 is [8.088, 8.671]
(b)
From the given table the point estimate is 8.3795
And a 95% prediction interval for the actual demand for fresh in all sales period when the price difference is 0.10 is [7.140, 9.619]
(c)
Calculate the distance value when x = 0.10. we have given that s = 0.587952 and sy = 0.1425 from the output
Distance value = (sy/s)2 = (0.1425/0.587952)2 = 0.0587
(d)
Calculate the 99% confidence interval and prediction interval x = 0.10 for the mean and lobor costs when:
We have given that n = 30, b0 = 8.388664, b1 = -0.091904, and s = .587952
Confidence Interval is:
= b0 + b1x
= 8.388664 + (-0.091904) × 0.10 = 8.3794
= ± t/2 s(Distance value)
= 8.3794 ± 2.763(0.587952) 0.0587
= 8.3794 ± 0.39
= [8.7694, 7.9894]
This comparison (within rounding) to the computer generated output. For the prediction interval with the same quantities, we get:
± t/2 s(1 + Distance value)
8.3794 ± 2.763(0.587952) 1.0587
= 8.3794 ±1.67
= [10.0494, 6.7094]
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