Question

Sanitation inspection of cruise ships. Refer to Exercise 2.81 (p. 111) and the data on the sanitation levels of passenger cruise ships.

1. Use the box plot method to detect any outliers in the data set.
2. Use the z-score method to detect any outliers in the data set.
3. Do the two methods agree? If not, explain why

Short answer

1. 105.5

2. This approach considers any point with a z-score more than 2 or less than -2 an outlier.

3. As a result, no outliers are discovered using the z-score approach.

Step-by-step solution

(a) Box plot method

We find the 5-point summary, as given below, and afterward calculate the IQR as well as outlier boundaries.

The five-point summary:

$$\begin{gathered} \min =69 \\ 1\mathrm{st}\mathrm{quartile}=93 \\ \mathrm{median}=96 \\ 3\mathrm{rd}\mathrm{quartile}=98 \\ \max =100 \end{gathered}$$

$$\begin{gathered} \mathrm{skewness}=3\left(\frac{\mathrm{mean}-\mathrm{median}}{\mathrm{\sigma }}\right) \\ =3\left(\frac{94.4409-96}{5.3352}\right) \\ =0.8766869 \end{gathered}$$

IQR=Q3-Q1

=98-93

=5

To find the outlier:

$$\begin{gathered} \mathrm{lower}\mathrm{limit}=\mathrm{Q}1-1.5\mathrm{IQR} \\ =93-(1.5\times 5) \\ =85.5 \\ \mathrm{Higher}\mathrm{limit}=\mathrm{Q}1+1.5\mathrm{IQR} \\ =93+(1.5\times 5) \\ =105.5 \end{gathered}$$

Thus, any worth less than 85.5 or more than 105.5 is an outlier.

As we can see, some points are less than 85.5 and thus are considered outliers.

(b) z-score method

As illustrated, we compute the z-score for every observation.

This approach considers any point with a z-score more than 2 or less than -2 an outlier.

However, no outlier is discovered using this approach because no point has a z-score more than or even less than 2.

(c) Two methods

In the box plot approach, they only examine the centre 50% of the information to calculate the IQR, which is used as a measurement of the information's dispersion. When we plot the data as a histogram, we see that it is skewed to the left. As a result, employing the median as a central tendency measurement is more suitable in this circumstance.

One utilized the standard deviation and mean from the complete dataset in the z-score approach. Both are heavily influenced by outlier estimates and thus tend to distort the standard deviation and mean readings. As a result, no outliers are discovered using the z-score approach.