Question 1183061
population mean in 2012 is 2505 square feet in 2012.
15 home sample taken in 2013 has a mean area of 2627 with a standard deviation of 248 feet.


test is whether mean in the 2013 sample shows that the population mean in 2013 is greater than the population mean in 2012.


test is one tailed t-test.
degrees of freedom is 15 - 1 = 14.
standard error is equal to 248 / sqrt(15) = 64.0333 rounded to 4 decimal places.
t-score is (2627 - 2505) / 64.0333 =  1.9053 rounded to 4 decimal places.
area to the right of the t-scores with 14 degrees of freedom is .0387 rounded to 4 decimal places.
since that's less tha .05 critical alpha, the test is considered significant and the assumption that the mean area of home built in 2015 is greater than the mean area of homes built in 2012 is accepted.



standard error = standard deviation / square root of sample size


t-score formula is (x - m) / s for degrees of freedom indicated.
x is the raw score
m is the raw mean
s is the standard error.


degrees of freedom = sample size minus 1


the test statistics is t = 1.90 53
p-value is .0387


one tailed critical p-value is .05