From a sample of 41 orders for an on-line bookseller, the average order size is $75, and the sample standard deviation is $18. Assume the distribution of orders is normal. For which interval can one be exactly 90 percent confident that the population mean is contained in that interval? A. $62.00 to $88.00. B. $71.29 to 78.71. C. $70.27 to $79.73. The answer says that we don’t know the population variance therefore we should used T distribution… But as we do have the std deviation, can’t we just deduct the variance ??? What am I missing ?

Don’t quite understand your question. The formula to use is “sample mean ± Critical Value X SE”

We hqve to use a T test as we don’t know the variance, but as they give the sdt dev why can’t I just square it and use a Z test ?

How did they get those numbers?? Shouldnt it be: 75± 1.65(18/ sq root of 41)… This comes out to 70.36 - 79.64 Is that right?

PatBateman You need a T test, the 1,65 comes from a Z test

D’Artagnan Wrote: ------------------------------------------------------- > We hqve to use a T test as we don’t know the > variance, but as they give the sdt dev why can’t I > just square it and use a Z test ? You need the population variance/sd in order to use Z-test. Question gave the sample’s sd, not the population’s.

N is over 30. You could use z-test if you wanted.

job71188 Wrote: ------------------------------------------------------- > N is over 30. You could use z-test if you wanted. QBank’s answer tend to use t-test even if over 30. It would be safer to use Z only when n is over 120. But of course, Schweser != CFAI.

The question gives sample std dev. So if you square it, it would be sample variance and not population variance. CFA book says that it would be rarely possible to get variance for population. So generally, we will have sample std dev on problems.

What’s the answer anyway?