in a random sample of 32 gas grills, the mean price was $280.90 with a standard deviation of $123.70

Question

i need help on my homework.

Answer 1

Are you looking for confidence interval?  i will assume so    its usually 95 percent C.I which means Z of 1.96 

 

mean + or _(Z)  std/square root of the population(n)

 

280.90 +(1.96)123.70/sq root of 32

280-(1.96)123.70/sq root of 32

 

sq root of 32 is about 5.6

 

 

so 280.90+(1.96)123.70/5.6=324

 

280.90-(1.96)123.70/5.6=238

 

therefore my 95 percent confidence interval is in the range of 238-324 dollars   in other words, I am 95 percent certain of this range ;)