Question

Refer to the accompanying data​ table, which shows the amounts of nicotine​ (mg per​ cigarette) in​ king-size cigarettes,​ 100-mm menthol​ cigarettes, and​ 100-mm nonmenthol cigarettes. The​ king-size cigarettes are​ nonfiltered, while the​ 100-mm menthol cigarettes and the​ 100-mm nonmenthol cigarettes are filtered. Use a 0.05 significance level to test the claim that the three categories of cigarettes yield the same mean amount of nicotine. Given that only the​ king-size cigarettes are not​ filtered, do the filters appear to make a​ difference?

	King Size Nicotine (mg)	100 mm Menthol Nicotine (mg)	100 mm Non Menthol Nicotine (mg)
1	1.1	1.1	.8
2	1.0	.9	1.2
3	1.0	1.2	.4
4	1.1	.8	1.2
5	1.4	1.3	1.0
6	1.2	1.4	.8
7	1.1	1.0	1.1
8	1.0	1.2	1.2
9	1.2	.8	.9
10	1.2	.9	.9

A. Determine the null and alternative hypotheses.

B. Find F test statistic

C. Find P value using F test statistic

D. What is the conclusion for this hypothesis test?

E. Do the filters appear to make a difference?

Answer 1

(A) The Hypothesis:

H0: There is no difference between the mean nicotine content of the 3 types of cigarettes.

Ha: The mean nicotine content of at least one cigarette is different from the others.

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(B) The ANOVA Table is below

Source	SS	DF	Mean Square	F
Between	0.17	2	0.08	2.01
Within/Error	1.11	27	0.041
Total	1.28	29

Calculations For the ANOVA Table:

Overall Mean : Since n is equal , the overall mean = mean of means

= (1.13 + 1.06 + 0.95) / 3 = 1.05

SS treatment = SUM [n* ( Individual Mean - overall mean)²] = 10 * (1.13 - 1.05)² + 10 * (1.13 1.05)² + 10 * (1.13 - 1.05)² = 0.17

df1 = k - 1 = 3 - 1 = 2

MSTR = SS treatment / df1 = 0.17 / 2 = 0.08

SS error = SUM (Sum of Squares) = 0.14 + 0.4 + 0.57 = 1.11

df2 = N - k = 30 - 3 = 27

Therefore MS error = SS error / df2 = 1.11 / 27 = 0.041

F = MSTR / MSE = 0.08 / 0.041 = 2.01

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(C) The p value is calculated for F = 2.01 for df1 = 2 and df2 = 27

p value = 0.1541

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(D) Since p value is > , We fail to reject H0.

The Conclusion: There isn't sufficient evidence at the 95% level of significance to conclude that the

mean nicotine content of at least one cigarette is different from the others.

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(E) No, since the null hypothesis that the mean nicotine content of at least one cigarette is different from the others has been failed to be rejected, we dont have sufficient evidence that the filters are effective.

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