Question

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. We want to find if these data provide convincing evidence to suggest that the percentage of students who smoke has increased over the last five years. What are the test statistic (Z) and p-value of the test

Answer 1

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, $H_0$ : p = 0.18 {means that the percentage of students who smoke has remained same at 18% over the last five years}

Alternate Hypothesis, $H_1$ : 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. = $\frac{\hat p -p}{\sqrt{\frac{\hat p (1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of students = 50/200 = 0.25

            n = sample size = 200

So, test statistics = $\frac{0.25 -0.18}{\sqrt{\frac{0.25 (1-0.25)}{200} } }$

                            = 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 .