1 Answer

MatG · Answer 1 · 2020-05-11T23:53:59+0000

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

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

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

$t=(62.2-60)/((18.1)/(√(15)))\approx0.47$

We go on to find

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

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

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