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2. A random sample of 45 door-to-door encyclopedia salespersons was asked how long on average they were able to talk to the potential customer. Their answers revealed a mean of 8.5 minutes with a variance of 9 minutes. Compute a point estimate for the average conversation length and the margin of error.
The point estimate of the mean conversation length is the sample mean,Quote:
Originally Posted by bombo31
so in this case it is 8.5 minutes. The margin of error of this estimate is
usually the standard error (SE) of the mean which is the population standard
deviation (SD) divided by the square root of the sample size. But we usually
use the sample SD as we don't know the population SD.
Now the problem you are given a variance (which is the square of the SD)
but with units of minutes - its units should be minutes squared. But let's
assume that this is a typo and the variance is 9 minutes squared, so the
SD is 3 minutes. Then the SE of the mean is 3/sqrt(45)~=0.447 minutes.
So your point estimate of the mean conversation length is 8.5 minutes with
a standard error of ~=0.477 minutes.
RonL
Thanks. IN the scenario don't you use the estimator +or - 1.96 multiplied by the 3/ the sqrt of 45 in the calculation of the margin of error.
There is no indication that margin of error means that, though you mayQuote:
Originally Posted by bombo31
mave been taught that as a convention. I would have said it does not
matter how you express the margin of error so long as you say what
it is you are quoting.
RonL
Ok. Thanks.