Question

The Occupational Safety and Health Administration (OSHA) has identified the gaseous chemical benzene as a possible cancer-causing substance. A federal regulation requires factories to keep the amount of benzene in the air at or below 1 part per million (ppm). OSHA sends an investigator to inspect the air quality at a certain factory. The factory’s manager insists that their air quality is safe. In OSHA’s investigation, they collected 20 random air samples from the factory and found a mean benzene content of 1.4 ppm with a standard deviation of 1.25 ppm.

At a significance level of 0.01, conduct a hypothesis test for the claim, “The air in this factory has a mean benzene content that is less than or equal to 1 ppm.”

1. Select the pair of hypotheses that are appropriate for testing this claim.

		H0: µ > 1 (claim) H1: µ ≤ 1
		H0: µ ≥ 1 H1: µ < 1 (claim)
		H0: µ < 1 H1: µ ≥ 1 (claim)
		H0: µ ≤ 1 H1: µ > 1 (claim)
		H0: µ ≥ 1 (claim) H1: µ < 1
		H0: µ ≤ 1 (claim) H1: µ > 1
		H0: µ < 1 (claim) H1: µ ≥ 1
		H0: µ > 1 H1: µ ≤ 1 (claim)

2. Select the choice that best describes the nature and direction of a hypothesis test for this claim.

		This is a right-tail t-test for µ.
		This is a left-tail t-test for µ.
		This is a two-tail z-test for µ.
		This is a left-tail z-test for µ.
		This is a two-tail t-test for µ.
		This is a right-tail z-test for µ.

3. Find the standardized test statistic for this hypothesis test. Round your answer to 2 decimal places.

4. Find the P-value for this hypothesis test. Round your answer to 4 decimal places.

5. Using your previous calculations, select the correct decision for this hypothesis test.

6. Consider the following statements related to the claim. Based on the results of your hypothesis test, which of these statements is true? Select the best choice.

Answer 1

1)

H0: µ ≤ 1 (claim)
H1: µ > 1

2)
This is a right-tail t-test for µ.

3)

sample std dev ,    s =    1.2500                  
Sample Size ,   n =    20                  
Sample Mean,    x̅ =   1.4000                  
                          
degree of freedom=   DF=n-1=   19                  
                          
Standard Error , SE = s/√n =   1.2500   / √    20   =   0.2795      
t-test statistic= (x̅ - µ )/SE = (   1.400   -   1   ) /    0.2795   =   1.43

4) p-Value   =   0.0843   [Excel formula =t.dist.rt(t-stat,df) ]

5) Decision:   p-value>α, Do not reject null hypothesis   

6) Conclusion: There is not enough evidence to reject the claim