Question 1137774: The treasurer of a municipality claims that the average net worth of families in this municipality is 730,000.00php. a random sample of 50 families selected from this area produced a mean net worth of 860,000.00php. with standard deviation of 65,000.00php. using 0.1 significance level, can we conclude that the claim is true?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! population mean is 730,000.
sample size is 50.
sample mean is 860,000.
sample standard deviation is 65,000.
since you do not have the population standard deviation, i believe you need to find the t-score rather than the z-score.
since the sample size is quite large, the z-score would probably be close.
using the z-score, you would get z = (860,000 - 730,000) / s
s is the stnadard error which is equal to standard deviation of the sample in this case divided by the square root of the sample size.
that makes s = 65,000 / sqrt(50) = 9192.388155.
the z-score is therefore (860,000 - 730,000) / 9192.388155 = 14.14213562.
the critical z-score at 1% significance level is either 1.2i8... if it's a one tail distributuion, or 1.645... if it's a two tail distributuion.
in either case, a z-score of 14 + is so far behyond this that there is no way the population average could be 730,000 if the sample average of 50 families is 860,000.
if you do this using t-scores, you'll get a similar answer because the sample size is quite large, making the t-score get pretty close to the z-score.
the critical t-score with 49 degrees of freedom and .1 significance level is somewhere between 1.303 and 1.296 for a tone tail distribution and somewhere between 1.684 and 1.671 for a two tail distribution.
a t-score of 14 + is way beyond this as well.
you would have to reject the claim that the population average is 730,000 based on the results of the sample taken.
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