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

i need help on my homework.

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You need to state the problem completely.
 
Jim

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OLUWASEUN O. | Tutor in Math, Pre-Medical/Medical courses and Fitness/Martial Arts,Tutor in Math, Pre-Medical/Medical cours...
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 ;)