Question

In 1994, 6.4% of people used marijuana.

This year, a company wishes to use their employment drug screening to test a claim. They take a simple random sample of 2954 job applicants and find that 167 individuals fail the drug test for marijuana. They want to test the claim that the proportion of the population failing the test is lower than 6.4%. Use .10 for the significance level. Round to three decimal places where appropriate.

Hypotheses:

Ho:p=6.4%

H1:p<6.4%

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 6.4%
• There is NOT sufficient evidence to support the claim that the proportion of the population failing the test is lower than 6.4%
• There is sufficient evidence to warrant rejection of the claim that the proportion of the population failing the test is lower than 6.4%
• There is NOT sufficient evidence to warrant rejection of the claim that the proportion of the population failing the test is lower than 6.4%

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

Answer 1

given data and necessary calculation are:-

sample size (n) = 2954

sample proportion () = 167/2954 = 0.056534

hypothesis:-

[ claim ]

the test statistic :-

critical value :-

[ z critical value for alpha=0.10, left tailed test ]

p value is :-

[ in any blank cell of excel type =NORMSDIST(-1.658) ]

Conclusion About the Null: Reject the null hypothesis.

[ p value = 0.049 < 0.10 (alpha) ]

Conclusion About the Claim:

There is sufficient evidence to support the claim that the proportion of the population failing the test is lower than 6.4%

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