A regional retailer would like to determine if the variation in average monthly
ID: 3320963 • Letter: A
Question
A regional retailer would like to determine if the variation in average monthly store sales can, in part, be explained by the size of the store measures in square feet. A random sample of 2121 stores was selected and the store size and average monthly sales were computed, with the accompanying results. Complete parts a through d below. Use a 95% confidence level and Xp =15,000 where needed.
C. Produce a 95% confidence interval for the average value of monthly sales when xp = 15,000. The 95% confidence interval is Round to the nearest integer as needed.) d. Produce a 95% prediction interval for a particular month's average sales when the store size is 15,000 square feet. The 95% prediction interval is (DD Round to the nearest integer as needed.)Explanation / Answer
Solutionc:
Code in R:
storesize <- c(17400,15910,17430,17320,15750,20200,15270,17010,11930,12410,15630,12550,21680,14130,16670,14920,18370,18440,16730,19890,17880)
averagemonthlysales <- c(581240,538276,636060,574478,558043,689256,552570,584737,470551,520797,619704,465416,730864,501500,624254,567043,612975,618123,691403,719275,536593)
mod1.lm = lm(averagemonthlysales ~ storesize)
newdata = data.frame(storesize=15000)
predict(mod1.lm, newdata, interval="confidence")
output:
fit lwr upr
550940.9 529119.4 572762.3
95% confidence interval is 529119, 572762
Solutiond:
storesize <- c(17400,15910,17430,17320,15750,20200,15270,17010,11930,12410,15630,12550,21680,14130,16670,14920,18370,18440,16730,19890,17880)
averagemonthlysales <- c(581240,538276,636060,574478,558043,689256,552570,584737,470551,520797,619704,465416,730864,501500,624254,567043,612975,618123,691403,719275,536593)
mod1.lm = lm(averagemonthlysales ~ storesize)
newdata = data.frame(storesize=15000)
predict(mod1.lm, newdata, interval="predict")
output:
fit lwr upr
550940.9 463432.9 638448.8
95% prediction interval is
463433 and 638449(rounding to nearest integer)
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