Q. 3.175 - Elementary Statistics

Question

In the below exercise:

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.

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.

Step-by-Step Solution

Answer

a) The quartiles are $26.685, 27.12, 28.87.$ The bottom $25\%$of the observations is less than or equal to the quartiles.

b) The interquartile range is $2.185.$ The diameter of $50\%$ of pizzas ranged over $2.185 cm.$

c) All observations are greater than or equal to the minimum. 

d) There are no potential outliers.

e) The boxplot is 

Step 1/5

a) Obtain the first quartile, second quartile (median) and third quartile by using MINITAB

MINITAB procedure:

Step 1: Choose Stat > Basic Statistics > Display Descriptive Statistics.

Step 2: In Variables enter the columns Diameter.

Step 3: Choose option statistics, and select First quartile, median, second quartile, minimum, and maximum.

Step 4: Click OK.

MINITAB output:

Descriptive Statistics: DIAMETER

$\begin{array}{lrrrrrrrr}\text{ Variable }& \text{ N }& N^{*}& \text{ Minimum }& Q1& \text{ Median }& \text{ Q3 }& \text{ Maximum }& \text{ IQR }\\ \text{ DIAMETER }& 41& 0& 26.090& 26.685& 27.120& 28.870& 29.400& 2.185\end{array}$

 From the MINITAB output, the first quartile is $26.685$, the second quartile is $27.12$and the third quartile is $28.87$.

From the results, it is clear that the bottom $25\%$of the observations is less than or equal to the quartiles.

Step 2/5

b) Obtain the interquartile range.

$\begin{array}{r}\mathrm{IQR}=Q_3-Q_1\\ =28.87-26.685\\ =2.185\end{array}$

The middle half of the diameter ranged over $2.185 cm.$

Step 3/5

c) Find the five-number summary.

$\begin{array}{|ccccc|}\hline \text{ Minimum }& \text{ First quartile }\left({\mathrm{Q}}_1\right)& \text{ Median }& \text{ Third quartile }\left({\mathrm{Q}}_3\right)& \text{ Maximum }\\ 26.09& 26.685& 27.12& 28.87& 29.4\\ \hline\end{array}$

From the result, it is clear that all observations are greater than or equal to the minimum.

Step 4/5

d) Identify the potential outliers.

Lower limit $={\mathrm{Q}}_1-(1.5\times 1\mathrm{QR})$

Upper limit $={\mathrm{Q}}_3+(1.5\times \mathrm{QQR})$

The lower limit is obtained below.

$\begin{array}{r}\text{ Lower limit }=26.685-(1.5\times 2.185)\\ =26.685-3.2775\\ =23.4075\end{array}$

The upper limit is

$\begin{array}{r}\text{ Upper limit }=28.87+(1.5\times 2.185)\\ =28.87+3.2775\\ =32.1475\end{array}$

All observations are within the limit. So there are no potential outliers.

Step 5/5

e) Using MINITAB boxplot is drawn.

The data is skewed to the right because the distance between the median and minimum value is less than the distance between the median and maximum value.