In: Statistics and Probability
In 2003, 5.9% of people used illegal opioids.
This year, a company wishes to use their employment drug screening to test a claim. They take a simple random sample of 2880 job applicants and find that 136 individuals fail the drug test for illegal opioids. They want to test the claim that the proportion of the population failing the test is lower than 5.9%. Use .10 for the significance level. Round to three decimal places where appropriate.
Hypotheses:
Ho:p=5.9%Ho:p=5.9%
H1:p<5.9%H1:p<5.9%
Test Statistic: z =
Critical Value: z =
p-value:
Conclusion About the Null:
Conclusion About the Claim:
Do the results of this hypothesis test suggest that fewer people use illegal opioids? Why or why not?
H0: Population proportion P = 0.059
H1: Population proportion P < 0.059
Significance level alpha = 0.10
Sample proportion = 0.0472
Test statistic z:
= -0.0118/0.00439 = -2.6879
Critical value of z = 1.28
For the test statistic value, p-value = .003605
Since .003605 < 0.10, p-value < alpha. Hence we reject the null hypothesis.
There is sufficient evidence to support the claim that the proportion of the population failing the test is lower than 5.9%.
Do the results of this hypothesis test suggest that fewer people use illegal opioids? Why or why not?
This test suggests that few people (< 5.9%) fail the test. So it can be inferred that few people use the illegal opioids. However, it should also be considered that the test itself may not necessarily be accurate, so even if a larger number of people use the opioids, they may still pass the test.