Answer:
P-value = 0.009997
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 24
Nature of test: One-tailed test(Left tailed test)
[tex]t_{\text{statistic}} = -2.50[/tex]
We have to calculate the p-value.
Degree of freedom = n - 1 = 23
We calculate the p-value from the table corresponding to t score -2.50 and degree of freedom 23.
P-value = 0.009997
0.009997 is the probability of a value being to the left of a number that has a student-t statistic of –2.50.
