Question

In 2004, 6.5% 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 1957 job applicants and find that 104 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 6.5%. Use .10 for the significance level. Round to three decimal places where appropriate.

Hypotheses:

Ho:p=6.5%Ho:p=6.5%

H1:p<6.5%H1:p<6.5%

Test Statistic: z =

Critical Value: z =

p-value:

Conclusion About the Null:

• Reject the null hypothesis or 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 6.5%
• There is NOT sufficient evidence to support the claim that the proportion of the population failing the test is lower than 6.5%
• There is sufficient evidence to warrant rejection of the claim that the proportion of the population failing the test is lower than 6.5%
• There is NOT sufficient evidence to warrant rejection of the claim that the proportion of the population failing the test is lower than 6.5%

(Statements above are multiple choice ^^)

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

Answer 1

Answer)

Ho : P = 6.5%

H1 : P < 6.5%

Sample size (n) = 1957

As the sample size is large enough, we can use standard normal z table to conduct the test

Test statistics z = (obtained p - claimed p)/standard error

Standard error = √claimed p*(1-claimed p)/√n

Obtained p = 104/1957

N = 1957

Claimed p = 0.065 (6.5%)

Z = -2.128

Critical value z for 0.1 alpha is -1.28

P-value from z table is = p(z<-2.128) = 0.0166

Aa the obtained p-value is less than the given significance level of 0.1

We reject the null hypothesis

There is sufficient evidence to support the claim