Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Identify the null and alternative? hypotheses, test? statistic, P-value, and state the final conclusion that addresses the original claim.
A simple random sample of 25 filtered 100 mm cigarettes is obtained, and the tar content of each cigarette is measured. The sample has a mean of 19.6 mg and a standard deviation of 3.24 mg. Use a 0.05 significance level to test the claim that the mean tar content of filtered 100 mm cigarettes is less than 21.1 mg, which is the mean for unfiltered king size cigarettes. What do the results suggest, if anything, about the effectiveness of the filters?
What are the? hypotheses? Identify the test statistic. t=? The P value is? State the final conclusion that addresses the original claim.
Choose the correct answer below.
A. Fail to reject H0. There is insufficient evidence to support the claim that the mean tar content of filtered 100 mm cigarettes is less than 21.1 mg.
B. Reject H0. There is sufficient evidence to support the claim that the mean tar content of filtered 100 mm cigarettes is less than 21.1 mg.
C. Fail to reject H0. There is sufficient evidence to support the claim that the mean tar content of filtered 100 mm cigarettes is less than 21.1 mg.
D. Reject H0. There is insufficient evidence to support the claim that the mean tar content of filtered 100 mm cigarettes is less than 21.1 mg. What do the results suggest, if anything, about the effectiveness of the filters?
A. The results are inconclusive because the sample size is less than 30.
B. The results suggest that the filtered cigarettes have the same tar content as unfiltered king size cigarettes.
C. The results do not suggest that the filters are effective.
D. The results suggest that the filters are effective.
E. The results suggest that the filters increase the tar content.
x̅ = 19.6
s = 3.24
n = 25
α = 0.05
Null and Alternative hypothesis:
Ho : µ = 21.1
H1 : µ < 21.1
Test statistic:
t = (x̅- µ)/(s/√n) = (19.6 - 21.1)/(3.24/√25) =
-2.3148
df = n-1 = 24
p-value :
Left tailed p-value = T.DIST(-2.3148, 24, 1) =
0.0147
Decision:
p-value < α, Reject the null hypothesis.
Conclusion:
Answer B. Reject H0. There is sufficient evidence
to support the claim that the mean tar content of filtered 100 mm
cigarettes is less than 21.1 mg.
Answer A. The results are inconclusive because the sample size is less than 30.
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