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=\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.