Question

Consider the demand for Fresh Detergent in a future sales period when Enterprise Industries' price for Fresh will be x₁ = 3.70, the average price of competitors’ similar detergents will be x₂ = 3.90, and Enterprise Industries' advertising expenditure for Fresh will be x₃ = 6.50, y = the demand in hundreds of thousands of bottles. A 95 percent prediction interval for this demand is given on the following Excel add-in (MegaStat) output:

	95% Confidence Interval			95% Prediction Interval
Predicted	lower	upper		lower	upper	Leverage
8.51504	8.11805	8.91203		6.83864	10.19143	.059

(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 ($3.70), we can be very confident that the resulting dollar amount will be the minimal 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 = .792. Optional technical note needed. The distance value equals Leverage. (Round your answers to 5 decimal places.)

99% PI; [, ]

Answer 1

a)

• Since we are using the mean demand, then from the table we see that the 95 percent confidence interval is given as

→[8.11805,8.91203]

• In order to have an ample supply, we must round up any decimal, so we would have

→8.91203≈9 bottles in supply.

• The minimum value we would have in revenue is given as

→8.11805*3.70=$30.036785

b)

We know that n=30 and s=0.792

From the output, we also know that μ=8.51504

Now, for a 99 percent interval, the z value is given as

Zα/2=Z.01/2=Z.005=2.575

So the confidence interval is given as

[ 8.142370 , 8.88738 ]