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?Select one:a. Z=2.58, p-value=0.0049b. Z=2.29, p-value=0.0220c. Z=2.29, p-value=0.0110d. Z=2.58, p-value=0.0098

Answer 1

Answer: Z=2.58, p-value=0.00988

Step-by-step explanation:

Null hypothesis : $H_0: p=0.18$

Alternative hypothesis : $H_0: p\neq0.18$

Since alternative hypothesis is two-tailed so the test is a two tailed test.

Given : Population proportion : $p=0.18$

Sample size : n=200

Number of people smoke in sample = 50

then the sample proportion of smokers : $\hat{p}=\dfrac{50}{200}=0.25$

Test statistic for proportion :

$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

i.e. $z=\dfrac{0.25-0.18}{\sqrt{\dfrac{0.18(1-0.18)}{200}}}$

i.e. $z=\dfrac{0.07}{\sqrt{\dfrac{0.1476}{200}}}$

i.e. $z=2.57673560841\approx2.58$

P-value for two-tailed test = $2(P>|z|)=2(P(z>2.58))=2(1-P(z<2.58))$

$=2(1- 0.99506)=2(0.00494)=0.00988$

Hence, the test statistic test statistic (Z) and p-value of the test are 2.58 and 0.00988 respectively.

Answer 2

Using the z-distribution, as we are working with a proportion, it is found that the test statistic and the p-value of the test are given by:

d. Z=2.58, p-value=0.0098.

What are the hypothesis tested?

At the null hypothesis, it is tested if the proportion is of 18%, that is:

$H_0: p = 0.18$

At the alternative hypothesis, it is tested if the proportion has changed, that is:

$H_1: p \neq 0.18$

What is the test statistic?

The test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

In which:

• $\overline{p}$ is the sample proportion.

• p is the proportion tested at the null hypothesis.

• n is the sample size.

In this problem, the parameters are:

$p = 0.18, n = 200, \pi = \frac{50}{200} = 0.25$

Hence, the value of the test statistic is:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

$z = \frac{0.25 - 0.18}{\sqrt{\frac{0.18(0.82)}{200}}}$

$z = 2.58$

What is the p-value of the test?

Using a z-distribution calculator, with a two-tailed test, as we are testing if the proportion is different of a value, and z = 2.58, it is found that the p-value is of 0.0098.

Hence, option D is correct.

More can be learned about the z-distribution at https://brainly.com/question/26454209