Q 3.175

Question

Pizza Diameters. In the article, "Assessing Claims Made by a Pizza Chain" by P. Dunn, (Journal of Statistics Education, Vol. 20, No. 1. Pp. 1-19), a sample of pizzas from Domino's was collected. The intention of the study was to test the claim of a local pizza shop that its pizzas are bigger than Domino's. The following table gives the diameters, in centimeters (cm), of 41 large Supreme pizzas from Domino's.

$\begin{array}{l} 29.40 & 27.45 & 26.38 & 26.50 & 26.55 & 26.64 & 26.87 \\ 26.61 & 29.33 & 26.67 & 26.09 & 29.32 & 27.12 & 26.83 \\ 27.04 & 29.17 & 27.14 & 27.04 & 29.19 & 26.26 & 26.16 \\ 26.70 & 28.78 & 29.33 & 27.27 & 28.90 & 29.14 & 29.27 \\ 28.32 & 27.12 & 27.05 & 28.44 & 26.97 & 27.52 & 26.79 \\ 26.54 & 26.70 & 28.79 & 28.97 & 28.84 & 26.71 \end{array}$

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 am:

e. construct and interpret a boxplot.

Step-by-Step Solution

Verified

Answer

a)Interpretation: from the above result clear that the bottom $25 %$ of the observations are less than or equal to

$Q_{1} (= 26.685)$ , the next $25 %$ of the observation are lies between $Q_{1} (= 26.685)$ and

$Q_{2} (= 27.12)$ , the next $25 %$ of the observation are lies between $Q_{2} (= 27.12)$ and

$Q_{3} (= 28.87)$ And the top $25 %$ of the observation are greater than or equal to $Q_{3} (= 28.87) .$

b)The interquartile range is therefore $2.185 .$

Interpretation: The middle $50 %$ , or the middle half of the diameter, was around $2.185 cm$

c)Interpretation: It is clear from the above result that all observations are greater than or equal to the minimum, the bottom 25% of observations are less than or equal to $Q 1,$ the next 25% of observations are between $Q 1$ and , the next 25% of observations are between $Q 2$ and $Q 3$ , the top 25% of observations are greater than or equal to $Q 3$ , and all observations are less than or equal to $Q 3$ .

d)As a result, there are no outliers because all data fall between the lower and upper limits.

e)The data is skewed to the right in the box plot above because the distance between the median and highest value is greater than the distance between the median and minimum value. Furthermore, the right-hand whisker is larger than the left-hand whisker.

1Part (a) Step 1: Given information

Given in the question that,

2Part (a) Step 2: Explanation

$Q_{1} (= 26.685)$ MINITAB MINITAB procedure: Calculate the first quartile, second quartile (9median), and third quartile.

Step 1: Select Start > Basic Statistics > Display Descriptive Statistics from the drop-down menu.

Step 2: In the Variables section, type Diameter in the column Diameter.

Step 3: Select first quartile, median, second quartile, minimum, and maximum from the choice statistics menu. Step 4: Select OK from the drop-down menu.

MINITAB RESULTS:

DIAMETER is a descriptive statistic.

Variable ${NN}^{*}$ minimum $Q_{1}$ median $Q_{3}$ maximum IQR

DIAMETER $41026.09026 . 68527 . 12028 . 87029 . 4002 . 185$

From the MINITAB output, the first quartile $(Q_{1})$ is $26.685,$ the second quartile $(Q_{2})$ is $27.12$

The third is

And the third quartile is $28.87$ $(Q_{3})$

Interpretation: From the above result clear that the bottom $25 %$ of the observations are less than or equal to, the next $25 %$ of the observation are lies between $Q_{1} (= 26.685)$ and $Q_{2} (= 27.12)$ , the next $25 %$ of the observation are lies between $Q_{2} (= 27.12)$ and $Q_{3} (= 28.87)$ And the top $25 %$ of the observation are greater than or equal to $Q_{3} (= 28.87) .$

3Part (b) Step 1: Given information

Given in the question that,

4Part b Step 2 Explanation

Calculate the range between the quartiles.

The difference between the first and third quartiles is the inter quartile range (IQR)

$I Q R = Q_{3} - Q_{1}$

= 28.87 -26.685

= 2.185

The interquartile range is therefore 2.185.

Interpretation: The middle $50 %$ , or the middle half of the diameter, was around $2.185 cm$

5Part (c) Step 1: Given information

We have to find and interpret the five number summary.

6Part (c) Step 2: Explanation

Calculate the five numbers summery

The part's MINITAB output (a). The following are the five numerical summaries:

$\begin{matrix} Minimum & \begin{matrix} First quartile \\ (Q_{1}) \end{matrix} & Median & \begin{matrix} Third quartile \\ (Q_{3}) \end{matrix} & Maximum \\ 26.09 & 26.685 & 27.12 & 28.87 & 29.4 \end{matrix}$

Interpretation: It is clear from the above result that all observations are greater than or equal to the minimum, the bottom 25% of observations are less than or equal to $Q 1,$ the next 25% of observations are between $Q 1$ and $Q 2$ , the next 25% of observations are between $Q 2$ and $Q 3$ , the top 25% of observations are greater than or equal to $Q 3,$ and all observations are less than or equal to $Q 3$ .

7Part (d) Step 1: Given information

We have to identify the potential outliers

8Part d Step 2 Explanation

Calculate the lower and upper limit of data set.

Lower limit

$= Q_{1} - 1.5 (I Q R)$

= 26.685 - 1.5(2.185)

= 23.4075

Upper limit

$= Q_{3} + 1.5 (I Q R)$

= 28.28 + 1.5(2.185)

= 31.1475

As a result, there are no outliers because all data fall between the lower and upper limits.

9Part (e) Step 1: Given information

Given in the question that,

10Part (e) Step 2: Explanation

Construct the box plot by MINITAB

MINITAB procedure:

Step 1: Choose Graph > Box plot or start > EDA > Box plot.

Step 2: Under one Y, choose Simple. Click OK

Step 3: In Graph Variables, enter the data of Diameter.

Step 4: Click OK.

MINITAB output:

The data is skewed to the right in the box plot above because the distance between the median and highest value is greater than the distance between the median and minimum value. Furthermore, the right-hand whisker is larger than the left-hand whisker.

Q 3.173

Q.3.170

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