You need the t-distributing for this problem, as we have a small sample and

are using the sample SD.

z = (29.5-30)/(4.1/sqrt(16)) ~= -0.456

So the probability of getting a mean this small or less if the true mean

were 30g is (looking up -0,456 in a table of the cumulative t-distribution

with 15 degrees of freedom):

0.3275 (or 32.75%)

which is very far from being significant (on a one sided test we would usually

want this to be <= 0.05 to reject the hypothesis that there is no difference)

(If you have not done the t-distribution you are probably supposed to

look the z score up in a table of the normal distribution when you will find

the p-value is 0.3242)

RonL