Respuesta :
Using the z-distribution, it is found that since the test statistic is more than the critical value for the left-tailed test, it is found that there is not enough evidence to conclude that illegal drug use in her school is below the current national average.
At the null hypothesis, it is tested if drug use in her school is not below the national average, that is:
[tex]H_0: p \geq 0.098[/tex]
At the alternative hypothesis, it is tested if it is below, that is:
[tex]H_1: p < 0.098[/tex]
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.
93 out of 1046 reported using illegal drugs, hence the parameters are:
[tex]n = 1046, \overline{p} = \frac{93}{1046} = 0.0889, p = 0.098[/tex]
Hence, the value of the test statistic is:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.0889 - 0.098}{\sqrt{\frac{0.098(0.902)}{1046}}}[/tex]
[tex]z = -0.99[/tex]
The critical value for a left-tailed test, as we are testing if the proportion is less than a value, is [tex]z^{\ast} = -1.645[/tex]
Since the test statistic is more than the critical value for the left-tailed test, it is found that there is not enough evidence to conclude that illegal drug use in her school is below the current national average.
To learn more about the use of the z-distribution to test an hypothesis, you can take a look at https://brainly.com/question/25584945
