A large high school has been waging a campaign against drug use, particularly marijuana. Before the campaign began in
2007, a random sample of 100 students had their lockers checked for drugs. In 27 of the lockers, marijuana was found.
To assess if the program has been successful, another random locker check was conducted in 2017. This time, 30 out of
175 students had marijuana in their lockers. Using a = 0.05, is there evidence to suggest that the proportion of students
who are found with marijuana in their lockers has decreased?

Respuesta :

Using the z-distribution, as we are working with a proportion, it is found that there is evidence that the proportion has decreased.

What are the hypothesis tested?

At the null hypothesis, it is tested if the proportion is still the same, of 27/100 = 0.27, hence:

[tex]H_0: p = 0.27[/tex]

At the alternative hypothesis, it is tested if the proportion has decreased, hence:

[tex]H_1: p < 0.27[/tex].

What is the test statistic?

The test statistic is given by:

[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]

In which:

  • [tex]\overline{p}[/tex] is the sample proportion.
  • p is the proportion tested at the null hypothesis.
  • n is the sample size.

In this problem, the parameters are as follows:

[tex]n = 175, \overline{p} = \frac{30}{175} = 0.1714[/tex]

Hence, the value of the test statistic is given by:

[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]

[tex]z = \frac{0.1714 - 0.27}{\sqrt{\frac{0.27(0.73)}{175}}}[/tex]

z = -2.94.

What is the decision?

Considering a left-tailed test, as we are testing if the propoortion is less than a value, the critical value is of [tex]z^{\ast} = -1.645[/tex].

Since the test statistic is less than the critical value for the left-tailed test, it is found that there is evidence that the proportion has decreased.

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