In: Statistics and Probability
In 2005, 5.9% 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 2488 job applicants and find that 120 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 5.9%. Use .05 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 marijuana? Why or why not?
Null and Alternative Hypotheses
Ho:p=5.9%
H1:p<5.9%
Test Statistics
The following information is provided: The sample size is N=120, the number of favorable cases is X=2488, and the sample proportion is , and the significance level is α=0.05
The z-statistic is computed as follows:
z=−2.28
Based on the information provided, the significance level is α=0.05, and the critical value for a left-tailed test is \
zc=−1.64.
The rejection region for this left-tailed test is R={z:z<−1.64}
p value = 0.0113
Decision about the null hypothesis
Since it is observed that z=−2.28<zc=−1.64, it is then concluded that the null hypothesis is rejected.
Using the P-value approach: The p-value is p=0.0113, and since p=0.0113<0.05, it is concluded that the null hypothesis is rejected.
Conclusion
It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion p is less than p0, at the α=0.05 significance level.
There is sufficient evidence to support the claim that the proportion of the population failing the test is lower than 5.9%
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