Answer:
The P-value is between 2.5% and 5% from the t-table.
Step-by-step explanation:
We are given that a random sample of 16 students selected from the student body of a large university had an average age of 25 years and a standard deviation of 2 years.
Let [tex]\mu[/tex] = true average age of all the students at the university.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex] 24 years {means that the average age of all the students at the university is less than or equal to 24}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 24 years {means that the average age of all the students at the university is significantly more than 24}
The test statistics that will be used here is One-sample t-test statistics because we don't know about the population standard deviation;
T.S. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample average age = 25 years
s = sample standard deviation = 2 years
n = sample of students = 16
So, the test statistics = [tex]\frac{25-24}{\frac{2}{\sqrt{16} } }[/tex] ~ [tex]t_1_5[/tex]
= 2
The value of t-test statistics is 2.
Also, the P-value of test-statistics is given by;
P-value = P( [tex]t_1_5[/tex] > 2) = 0.034 {from the t-table}
The P-value is between 2.5% and 5% from the t-table.