Answer:
Test statistics = 2.29
P-value = 0.01101
Step-by-step explanation:
We are given that a survey conducted five years ago by the health center at a university showed that 18% of the students smoked at the time. This year a new survey was conducted on a random sample of 200 students from this university, and it was found that 50 of them smoke.
Let Null Hypothesis, [tex]H_0[/tex] : p = 0.18 {means that the percentage of students who smoke has remained same at 18% over the last five years}
Alternate Hypothesis, [tex]H_1[/tex] : p > 0.18 { means that the percentage of students who smoke has increased over the last five years}
The test statistics used here will be ;
T.S. = [tex]\frac{\hat p -p}{\sqrt{\frac{\hat p (1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of students = 50/200 = 0.25
n = sample size = 200
So, test statistics = [tex]\frac{0.25 -0.18}{\sqrt{\frac{0.25 (1-0.25)}{200} } }[/tex]
= 2.29
Now, P-value is given by = P(Z > 2.29) = 1 - P(Z <= 2.29)
= 1 - 0.98899 = 0.01101 or 1.101 %
Therefore, P-value of the test is 0.01101 .