A 2-column table with 5 rows. The first column is labeled time (weeks) with entries 1, 2, 3, 4, 5. The second column is labeled
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Answer:
The P-value of this sample is 0 and is less than the significance level (0.05), so the effect is significant and the null hypothesis is rejected.
As there is significant evidence to reject , we can say that there is significant evidence to claim that the mean time delay for the hypothetical population of all persons who may be subjected to the stimulus differs from 1.6 seconds.
The significance level for this test is 0.05.
Step-by-step explanation:
In this case, we want to prove if there is significant evidence that the mean differs from 1.6 seconds. That is the same as having evidence to reject the the hypothesis (the null hypothesis always have the equal sign).
We have to test the null hypothesis
The significance level for this test is 0.05
Calculation of the t-statistic:
If we look up in a t-table for t=9.47 and df=(36-1)=35, we get this value appears with a probability of zero. A large number like that is very unlikely to happen.
The P-value of this sample is 0 and is less than the significance level (0.05), so the effect is significant and the null hypothesis is rejected.
As there is significant evidence to reject , we can say that there is significant evidence to claim that the mean time delay for the hypothetical population of all persons who may be subjected to the stimulus differs from 1.6 seconds.