Q. 3.177

Question

Nicotine Patches. In the paper "The Smoking Cessation Efficacy of Varying Doses of Nicotine Patch Delivery Systems $4$ to $5$ Years Post-Quit Day" (Preventative Medicine, $28$ . pp. $113 - 118$ ) D. DAUGHTON CT al discussed the long-term effectiveness of transdermal nicotine patches on participants who had previously smoked at least $20$ cigarettes per day. A sample of $15$ participants in the Transdermal Nicotine Study Group (TNSG) reported that they now smoke the following number of cigarettes per day.

a. obtain and interpret the quartiles.

b. determine and interpret the interquartile range.

c. find and interpret the five number summary.

d. identify potential outliers, if any

e. construct and interpret a boxplot.

Step-by-Step Solution

Verified

Answer

(a) The quartiles are $Q_{1} = 9, Q_{2} = 8, Q_{3} = 10$ .

(b) The interquartile range is $1$ .

(d) The potential outlier is $6$ .

(e) See the boxplot in step $10$ .

1Part (a) Step 1: Given Information

We are given that A sample of $15$ participants in the Transdermal Nicotine Study Group (TNSG) reported and no of smoke they take is the table and we have to find out all the quartiles.

2Part (a) Step 2: Explanation

First, organize the frequencies in increasing order

we get, $6, 7, 8, 8, 8, 8, 9, 910, 10, 10, 10, 10, 10, 10$

then find the median of entire data as no. of observations is $15$ , so median will be at $\frac{15 + 1}{2} = 8 t h t e r m = 9$ and

We get $9$ as the median which is the $Q_{1}$ .

Divide it in half and include median in both as observations are odd.

.We get, Bottom Half is $6, 7, 8, 8, 8, 8, 9, 9$ and Top Half is $9, 10, 10, 10, 10, 10, 10, 10$ Then find the median of bottom half and median is at the position $\frac{8 + 1}{2} = 4.5$ and

Which gives median is between fourth and fifth position which is $\frac{8 + 8}{2} = 8$ and this is $Q_{2}$ and last find the median of the top half and median is at position between fourth and fifth position and that means median is $\frac{10 + 10}{2} = 10$ and that is $Q_{3}$ .

Hence, these are the all quartiles.

3Part (b) Step 1: Given Information

We are given that A sample of $15$ participants in the Transdermal Nicotine Study Group (TNSG) reported and no of smoke they take is the table and we have to find out the interquartile range.

4Part (b) Step 2: Explanation

The interquartile range is the difference between the first and the third quartiles,

By which we get, IQR $= Q_{3} - Q_{1}$

Putting the values,

We get, $I Q R = 10 - 9 = 1$

Hence this is the required interquartile range.

5Part (c) Step 1: Given Information

We are given that A sample of $15$ participants in the Transdermal Nicotine Study Group (TNSG) reported and no of smoke they take is the table and we have to find out the five number summary and interpret it.

6Part (c) Step 2: Explanation

The five-number summary of a data set is Min , $Q_{1}, Q_{2}, Q_{3}$ , Max

We get, Min $= 6, Q_{1} = 9, Q_{2} = 8, Q_{3} = 10$ , Max $= 10$

Now to get the variation subtract lowest from highest by which we get, $3, - 1, 2, 0$ and according to it first quarter has high variation but last quarter has less variation.

7Part (d) Step 1: Given Information

We are given that A sample of $15$ participants in the Transdermal Nicotine Study Group (TNSG) reported and no of smoke they take is the table and we have to find out the potential outliers.

8Part (d) Step 2: Explanation

First find the lower and upper limit which is,

Lower limit $Q_{1} - 1.5 \times I Q R$ and Upper Limit $Q_{3} + 1.5 \times I Q R$

Put the values of all variable we get,

Lower Limit $= 9 - 1.5 \times 1 = 7.5$ and Upper Limit $= 10 + 1.5 \times 1 = 11.5$ .

and potential outlier is defined as a value that is outside of lower and upper limit and according to this question only $6$ is outside the lower limit. So, this is the potential outlier.

9Part (e) Step 1: Given Information

We are given that A sample of $15$ participants in the Transdermal Nicotine Study Group (TNSG) reported and no of smoke they take is the table and we have to construct the boxplot.

10Part (e) Step 2: Explanation

First to make the boxplot we need quartile, potential outliers and adjacent values.

and from previous parts we have, $Q_{1} = 9, Q_{2} = 8, Q_{3} = 10$ and potential outlier is $6$ and adjacent values are $7$ and $9$ now plot these