Question

In 1993, 5.6% 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 2926 job applicants and find that 125 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 5.6%. Use .01 for the significance level. Round to three decimal places where appropriate.

Hypotheses:

Ho:p=5.6%

H1:p<5.6%

Test Statistic: z =

Critical Value: z =

p-value:

Conclusion About the Null:

• Reject the null hypothesis
• Fail to reject the null hypothesis

Conclusion About the Claim:

• There is sufficient evidence to support the claim that the proportion of the population failing the test is lower than 5.6%
• There is NOT sufficient evidence to support the claim that the proportion of the population failing the test is lower than 5.6%
• There is sufficient evidence to warrant rejection of the claim that the proportion of the population failing the test is lower than 5.6%
• There is NOT sufficient evidence to warrant rejection of the claim that the proportion of the population failing the test is lower than 5.6%

Do the results of this hypothesis test suggest that fewer people use illegal opioids? Why or why not?

Please explain how to solve on a TI-84 calculator.  Thanks!

Answer 1