Question

Question 5. Consider the demand for Fresh Detergent in a future sales period when Enterprise Industries' price for Fresh will be x₁ = 3.81, the average price of competitors’ similar detergents will be x₂ = 4.29, and Enterprise Industries' advertising expenditure for Fresh will be x₃ = 6.80. A 95 percent prediction interval for this demand is given on the following JMP output:

	Predicted Demand	Lower 95% Mean Demand	Upper 95% Mean Demand		StdErr Indiv Demand	Lower 95% Indiv Demand	Upper 95% Indiv Demand
31	8.5136623684	8.0014622793	9.0258624574		.7295926211	7.0139632575	10.0133614790

(a) Find and report the 95 percent prediction interval on the output. If Enterprise Industries plans to have in inventory the number of bottles implied by the upper limit of this interval, it can be very confident that it will have enough bottles to meet demand for Fresh in the future sales period. How many bottles is this? If we multiply the number of bottles implied by the lower limit of the prediction interval by the price of Fresh ($3.81), we can be very confident that the resulting dollar amount will be the minimum revenue from Fresh in the future sales period. What is this dollar amount? (Round 95% PI to 5 decimal places and dollar amount to 1 decimal place and Level of inventory needed to the nearest whole number.)

95% PI [   _______ , ______ ]

Level of inventory needed = ______ bottles

Lower dollar amount = _______

(b) Calculate a 99 percent prediction interval for the demand for Fresh in the future sales period. Hint: n = 30 and s = 0.686. Optional technical note needed. The distance value equals Leverage. (Round your answers to 5 decimal places.)

99% PI [ _______ , _______ ]

Answer 1

a)              
95% PI [   7.01396   ,   10.01336   ]
Level of inventory needed=       10.013361479*100000=   1001336.148
lower dollar amount=   7.0139632575*3.81*100000=   2672320.0

b)

CI=   99%          
α=   1%          
DF=n-p-1=30-3-1=   26          
t critical value=   2.779          
              
  
99% prediction interval = point estimate ±t*std error   
=8.5136623684±2.7787*0.729592621

(   6.48633   ,   10.54099   )