Question 1137774
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.