Question

Please help explain this to me. I have tried using an example but i still dont understand it.

Answer 1

The mean of the $n=15$ estimates is $\bar x=62.2$ seconds, and the standard deviation of the sample is about $s=18.1$. Then the test statistic is

$t=\dfrac{\bar x-\mu_0}{\frac s{\sqrt n}}$

where $\mu_0$ is the mean under the null hypothesis. So

$t=\dfrac{62.2-60}{\frac{18.1}{\sqrt{15}}}\approx0.47$

We go on to find

$P(T>0.47)\approx0.32$

so the difference is statistically significant and you reject the null hypothesis.