Question

Consider a drug testing company that provides a test for marijuana usage. Among 310 tested​ subjects, results from 28 subjects were wrong​ (either a false positive or a false​ negative). Use a 0.05 significance level to test the claim that less than 10 percent of the test results are wrong.

Answer 1

Answer:

$z=\frac{0.0903 -0.1}{\sqrt{\frac{0.1(1-0.1)}{310}}}=-0.569$  

The p value for this case can be calculated with this probability:

$p_v =P(z$  

Since the p value is higher than significance level we don't have enough evidence to conclude that the true proportion is significantly less than 0.1

Step-by-step explanation:

Information given

n=310 represent th sample selected

X=28 represent the subjects wrong

$\hat p=\frac{28}{310}=0.0903$ estimated proportion of subjects wrong

$p_o=0.1$ is the value to verify

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to test the claim that less than 10 percent of the test results are wrong ,and the hypothesis are:  

Null hypothesis:$p\geq 0.1$  

Alternative hypothesis:$p < 0.1$  

The statistic is given by:

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

Replacing the info we got:

$z=\frac{0.0903 -0.1}{\sqrt{\frac{0.1(1-0.1)}{310}}}=-0.569$  

The p value for this case can be calculated with this probability:

$p_v =P(z$  

Since the p value is higher than significance level we don't have enough evidence to conclude that the true proportion is significantly less than 0.1