A tobacco company advertises that the average nicotine content of its cigarettes

Question

A tobacco company advertises that the average nicotine content of its cigarettes is at most 14 milligrams. A consumer protection agency wants to determine whether the average nicotine content is in fact greater than 14. A random sample of 300 cigarettes of the company's brand yield an average nicotine content of 14.6 and a standard deviation of 3.8 milligrams. Determine the level of significance of the statistical lest of the agency's claim that mu is greater than 14. If alpha = 0.1 is there significant evidence that the agency's claim has been supported by the data? Additional exercise: Use the information in exercise 5.26 (page 204) to answer these questions. State the null and alternative hypotheses Identify which type of test this is (choose one): right-tailed, left-tailed, or two-tailed test. Identify the value for each of the following: (a) alpha (b) mu_0 (c) Y (d) s (e) n Calculate the test statistic. Identify the critical value. Identify the rejection region. Do we reject HO? (yes or no) What is the p-value for this test? Is there evidence that the consumer protection agency's claim is correct? Is there evidence that the tobacco company's advertisement is misleading? In the context of this problem what is a type I error? In the context of this problem, what is a Type II error? Calculate the power of this test if the actual mean nicotine content is 14.7 milligrams.

Explanation / Answer

1. Ans> The null and alternative hypotheses are as follows:

H0:mu=14 (the average nicotine content is 14 miligrams)

H1:mu>14 (the average nicotine content is greater than 14 miligrams)

2. Ans> The alternative hypothesis states that the test is right-tailed (the greater than sign indicates that).

3. Ans> alpha=0.01, mu0=14, ybar=14.6, s=3.8, n=300.

4. Ans> For large sample, and unknown population standard deviation, use student's t model to compute 1-sample t test statistic.

t=(ybar-mu0)/(s/sqrt n)

=(14.6-14)/(3.8/sqrt 300)

=2.73

5. The t critical at 299 degrees of freedom [df=n-1] and alpha=0.01 is 2.59.

6. Reject H0, if observed test statistic is greater than 2.59.

7. Reject H0, observed t>critical t.

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