Question 1025445
claim is that the mean family income is at most 22,450 per year.


sample taken.
mean of sample is 19769.
standard deviation of sample is 529.
sample size is 150.


standard error is equal to standard deviation divided by square root of sample size.


standard error is therefore equal to 529 / sqrt(150) = 43.19 rounded to 2 decimal places.


sample mean is 19769.
standard error of the mean, which is also referred to as the standard deviation of the distribution of sample means, is equal to 43.19.


at 5% level of significance, you would expect the z-score to be between -1.96 and 1.96 for 95% of the samples taken.


this means the the sample mean would be between 19769 - 1.96 * 43.19 and 19769 + 1.96 * 43.19 for 95% of the sample taken.


this would put the sample mean between 19684.35 and 19853.65 for 95% of the sample taken.


this is well below the claimed maximum mean of 22450.


the claim is therefore supported.


it's not even close.


each of the sample is assumed to have the same sample size of 150.