Answer:
[tex] t_{observed}= -7.44[/tex]
[tex] p_v <0.0001[/tex]
For this case since the [tex]p_v <\alpha[/tex] we have enough evidence to reject the null hypothesis at a significance of 5%. So then the mean prediction error seems to be different from 0.
Reject H0. There is enough evidence to conclude that the mean prediction error is different from 0 minutes.
Step-by-step explanation:
For this case they want to test this:
H0: [tex] \mu =0[/tex]
H1: [tex] \mu \neq 0[/tex]
So the correct option is:
B. Upper H0: μ=0 minutes
Upper H1: μ≠0 minutes
The output for the test is on the figure attached.
The significance level its [tex]\alpha=0.05[/tex]
The statistic on this case is calculated with the following formula:
[tex] t = \frac{x -\mu}{\frac{s}{\sqrt{n}}}[/tex]
The degreed of freedom for the statistic are given by the output [tex] df=98[/tex]
The statistic obtained for this case is given by the output:
[tex] t_{observed}= -7.44[/tex]
The p value from the output is given by:
[tex] p_v <0.0001[/tex]
We can calculate it like this manually:
[tex] p_v = P(t_{98} <-7.44) = 1.94x10^{-11}[/tex]
For this case since the [tex]p_v <\alpha[/tex] we have enough evidence to reject the null hypothesis at a significance of 5%. So then the mean prediction error seems to be different from 0.
The final conclusion for this case would be:
Reject H0. There is enough evidence to conclude that the mean prediction error is different from 0 minutes.
