Question

construct one table that includes relative frequencies based on the frequency distributions shown below, then compare the amounts of tar in non - filtered and filtered cigarettes. do the cigarette filters appear to be effective? (hint: the filters reduce the amount of tar ingested by the smoker.) click the icon to view the frequency distributions. complete the following relative frequency table. (round to the nearest percent as needed.) do cigarette filters appear to be effective? a. yes, because the relative frequency of the higher tar classes is greater for non - filtered cigarettes. b. no, because the relative frequency of the higher tar classes is greater for filtered cigarettes. c. no, because the relative frequencies for each are not substantially different. d. this cannot be determined.

Explanation:

Step1: Recall relative - frequency formula

Relative frequency = $\frac{\text{Frequency}}{\text{Total frequency}}$

Let's assume we have the frequency distributions for non - filtered and filtered cigarettes. First, we need to find the total frequency for non - filtered and filtered cigarettes separately.

Let $f_{n1},f_{n2},\cdots,f_{n6}$ be the frequencies of non - filtered cigarettes in each class and $f_{f1},f_{f2},\cdots,f_{f6}$ be the frequencies of filtered cigarettes in each class.

The total frequency of non - filtered cigarettes $T_n=\sum_{i = 1}^{6}f_{ni}$ and the total frequency of filtered cigarettes $T_f=\sum_{i = 1}^{6}f_{fi}$

Step2: Calculate relative frequencies for non - filtered

For the class $3 - 8$ of non - filtered, relative frequency $r_{n1}=\frac{f_{n1}}{T_n}\times100\%$
For the class $9 - 14$ of non - filtered, relative frequency $r_{n2}=\frac{f_{n2}}{T_n}\times100\%$
For the class $15 - 20$ of non - filtered, relative frequency $r_{n3}=\frac{f_{n3}}{T_n}\times100\%$
For the class $21 - 26$ of non - filtered, relative frequency $r_{n4}=\frac{f_{n4}}{T_n}\times100\%$
For the class $27 - 32$ of non - filtered, relative frequency $r_{n5}=\frac{f_{n5}}{T_n}\times100\%$
For the class $33 - 38$ of non - filtered, relative frequency $r_{n6}=\frac{f_{n6}}{T_n}\times100\%$
For the class $39 - 44$ of non - filtered, relative frequency $r_{n7}=\frac{f_{n7}}{T_n}\times100\%$

Step3: Calculate relative frequencies for filtered

For the class $3 - 8$ of filtered, relative frequency $r_{f1}=\frac{f_{f1}}{T_f}\times100\%$
For the class $9 - 14$ of filtered, relative frequency $r_{f2}=\frac{f_{f2}}{T_f}\times100\%$
For the class $15 - 20$ of filtered, relative frequency $r_{f3}=\frac{f_{f3}}{T_f}\times100\%$
For the class $21 - 26$ of filtered, relative frequency $r_{f4}=\frac{f_{f4}}{T_f}\times100\%$
For the class $27 - 32$ of filtered, relative frequency $r_{f5}=\frac{f_{f5}}{T_f}\times100\%$
For the class $33 - 38$ of filtered, relative frequency $r_{f6}=\frac{f_{f6}}{T_f}\times100\%$
For the class $39 - 44$ of filtered, relative frequency $r_{f7}=\frac{f_{f7}}{T_f}\times100\%$

Step4: Analyze effectiveness

If the relative frequency of the higher tar classes (say, classes with tar levels $21$ and above) is greater for non - filtered cigarettes, it means that non - filtered cigarettes have a higher proportion of higher - tar amounts. So, if the relative frequency of the higher tar classes is greater for non - filtered cigarettes, the filters are effective.

If we assume that after calculating the relative frequencies, we find that the relative frequency of the higher tar classes is greater for non - filtered cigarettes.

Answer:

A. Yes, because the relative frequency of the higher tar classes is greater for nonfiltered cigarettes.