A simple random sample of 29 filtered 100-mm cigarettes is obtained from a nonma

Question

A simple random sample of 29 filtered 100-mm cigarettes is obtained from a nonmally distributed population, and the tar content of each cigarette is measured. The sample has a standard deviation of 0.18 mg Use a 0.05 significance level to test the claim that the tar content of filtered 100-mm cigarettes has a standard deviation different from 0.25 mg, which is the standard deviation for unfitered king-size cigarettes. Complete parts (a) through (d) below. a. What are the null and alternative hypotheses? O A. Ho -0.25 mg O B. Hoe: 0.25 mg H1: o 0.25 mg H1: o 0.25 mg O D. Ho: o -0.25 mg C. Ho- o 0.25 mg H1: os0.25 mg H1: 0.25 mg

Explanation / Answer

Back-up Theory

X ~ N(µ, 2)

H0: 2 = 02   Vs HA: 2 02

Test Statistic: 2 = (n - 1)s2/02

where n = sample size and s = sample standard deviation.

Under H0, 2 ~ 2n - 1

H0 is rejected/accepted at % level of significance if p-value of 2cal </>

Given Data and Computations

n = 29, s = 0.18, = 0.05

Part (a)

Since the purpose of study is to test if the tar content of 100 mm filtered cigarettes has a standard deviation different from 0.25 mg, the alternative is: HA: 0.25 and hence null hypothesis is: H0: = 0.25. ANSWER option A

Part (b)

Test Statistic: 2 = (n - 1)s2/02 = 28x(0.18/0.25)2 = 14.5152 = 14.515 ANSWER

Part (c)

Under H0, the test statistic has chi-square distribution with 28 degrees of freedom. Hence, p-value = P(28> 14.5152) = 0.9831 ANSWER [probability is obtained using Exel Function]

Part (d)

Conclusion

Reject H0 because p-value is greater than the specified level of significance. There is sufficient evidence to reject the claim that the tar content of 100 mm filtered cigarettes has a standard deviation 0.25 mg.

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