Question 1202043
population mean is assumed to be 50
population standard deviation is assumed to be 10.
sample mean is 56.
two tail confidence interval of 95% has critical z-score of plus and minus 1.96.
sample size is 15.
standard error = population standard deviation / square root of sample size = 10 / sqrt(15) = 2.58199 rounded to 5 decimal places.
test z-score = (x-m)/s
x is the sample mean
m is the population mean
s is the standard error.
formula becomes z = (56 - 50) / 2.58199 = 2.32379 rounded to 5 decimal places.
that is considerably more than the critical z-score of 1.96, indicating that the results are significant.
this means that the sample indicates that the differencre in the mean sample versus the population mean indicates that the children sampled most probably don't come from the population of normal children.