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 changed over the last five years. What are the test statistic (Z) and p-value of the test?

Answer 1

Answer:

test statistic (Z) is 2.5767 and p-value of the test is .009975

Step-by-step explanation:

$H_{0}$: percentage of students who smoke did not change

$H_{a}$: percentage of students who smoke has changed

z-statistic for the sample proportion can be calculated as follows:

z=$\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } }$ where

• p(s) is the sample proportion of smoking students ($\frac{50}{200}$ =0.25)

• p is the proportion of smoking students in the survey conducted five years ago (18% or 0.18)
• N is the sample size (200)

Then, z=$\frac{0.25-0.18}{\sqrt{\frac{0.18*0.82}{200} } }$ ≈ 2.5767

What is being surveyed is if the percentage of students who smoke has changed over the last five years, therefore we need to seek two tailed p-value, which is .009975.

This p value is significant at 99% confidence level. Since .009975 <α/2=0.005, there is significant evidence that the percentage of students who smoke has changed over the last five years