1. ## Normally distribuded

A manufacturer makes ball bearings that are supposed to have a mean weight of 30g. A retailer suspects that the mean weight is not 30 g and wants to verify that the product is meeting specifications. The mean weight for a random sample of 16 ball bearings is 29.5 g with a standard deviation of 4.1 g. At the 0.05 significance level, test the claim that the mean is not 30 g. You may assume the weights are normally distributed.

2. Originally Posted by harry
A manufacturer makes ball bearings that are supposed to have a mean weight of 30g. A retailer suspects that the mean weight is not 30 g and wants to verify that the product is meeting specifications. The mean weight for a random sample of 16 ball bearings is 29.5 g with a standard deviation of 4.1 g. At the 0.05 significance level, test the claim that the mean is not 30 g. You may assume the weights are normally distributed.
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