Question 1202092
population mean = 60.
population standad deviation = 10.
sample size is 15.
sample mean is 56.
95% two tailed confidence interval has 2.5% on each tail with a critical z-score of plus or minus 1.96.
z = (x-m)/s
z is the z-score
x is the raw score
m is the mean
s is the standard eror, since you are looking at the mean of a sample rather than an individual score.
standard error = standard deviation / square root of sample size = 10 / sqrt(15) = 2.58199.
the formula becomes z = (56 - 50) / 2.58199 = 2.32379.
area to the right of that z-scorfe is equal to .01007.
since this is less than the critical p-value of .025, the results are significant and the conclusion is that the children are not a random sample from a population of normal children.
the critical z-score confirms this as well since the test z-score is greater than 1.96.