Hypothesis testing with TScore

Question

The question I am trying to solve is as follows;

A premium golf ball production line must produce all of its balls to 1.615 ounces in order to get the top rating (and therefore the top dollar). Samples are drawn hourly and checked. If the production line gets out of sync with a statistical significance of more than 1%, it must be shut down and repaired. This hour’s sample of 18 balls has a mean of 1.611 ounces and a standard deviation of 0.065 ounces. Do you shut down the line?

I understand the setting up of the hypotheses $$ H_0: \mu=1.615 $$ $$ H_A: \mu\neq 1.615 $$ and that we can calculate a T-Score as follows;

\begin{eqnarray*} t_{\hat\mu} &=& \frac{\bar{\mu}-\mu_0}{\hat{\sigma}/\sqrt{n}}\\ &=& \frac{1.611-1.615}{0.065/\sqrt{18}}\\ &=& -0.261 \end{eqnarray*}

I also understand to look up the T table with degrees of freedom = n-1 where n = 18, n-1 =17 and get a critical t-score of 2.898 for 2 tails with p=0.01

What I don't understand is how to use these two results to determine whether I can reject the null hypothesis.

Answer 1

In this case, the absolute t-value you derived is smaller than the t-value associated with p=0.01 (two-tailed). Thus, you can not reject the null hypothesis of the two means being the same.

The intuition behind it is basically: The more extrem your difference, the higher the t-value. The assoicated p-value then indicates the probability with which you would expect the given sample under the assumption that the true mean is given by the null hypothesis.

Also notice that the t-value is calculated with respect to the variance within the sample. The lower the variance, the more "precise" the estimated mean, and thus, the more weight is given to the actual difference between means.