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Blood cocaine concentration (mg/L) was determined both for a sample of individuals who had died from cocaine-induced excited delirium (ED) and for a sample of those who had died from a cocaine overdose without excited delirium; survival time for people in both groups was at most 6 hours. The accompanying data was read from a comparative boxplot in the article 鈥淔atal Excited Delirium Following Cocaine Use鈥 (J.

of Forensic Sciences, 1997: 25鈥31).

ED0 0 0 0 .1 .1 .1 .1 .2 .2 .3 .3

.3 .4 .5 .7 .8 1.0 1.5 2.7 2.8

3.5 4.0 8.9 9.2 11.7 21.0

Non-ED0 0 0 0 0 .1 .1 .1 .1 .2 .2 .2

.3 .3 .3 .4 .5 .5 .6 .8 .9 1.0

1.2 1.4 1.5 1.7 2.0 3.2 3.5 4.1

4.3 4.8 5.0 5.6 5.9 6.0 6.4 7.9

8.3 8.7 9.1 9.6 9.9 11.0 11.5

12.2 12.7 14.0 16.6 17.8

a. Determine the medians, fourths, and fourth spreads for the two samples.

b. Are there any outliers in either sample? Any extreme outliers?

c. Construct a comparative boxplot, and use it as a basis for comparing and contrasting the ED and non-ED samples.

Short Answer

Expert verified

a. For ED,

The median value is 0.4.

The lower fourth is 0.1.

The Upper fourth is 2.8.

The fourth spread is 2.7.

For non-ED,

The median value is 1.6.

The lower fourth is 0.3.

The Upper fourth is 7.9.

The fourth spread is 7.6.

b. For ED,

The outliers are 8.9, 9.2,11.7 and 21.

The extreme outliers are 11.7 and 21.

For non-ED,

There are no outliers.

c. The boxplot is represented as,

Step by step solution

01

Computing the median

Let x represents the ED samples.

Let y represent the non-ED samples.

a.

The data is arranged in ascending order.

For x,

For the odd number of observations, the median value is computed as,

\(\begin{aligned}\tilde x = {\left( {\frac{{n + 1}}{2}} \right)^{th}}ordered\;value\\ = {\left( {\frac{{27 + 1}}{2}} \right)^{th}}ordered\;value\\ = {\left( {14} \right)^{th}}ordered\;value\end{aligned}\)

Thus, the median value of x is 0.4.

For y,

For the even number of observations, the median value is computed as,

\(\begin{aligned}\tilde y = average\;of\;{\left( {\frac{n}{2}} \right)^{th}}and\;{\left( {\frac{n}{2} + 1} \right)^{th}}\;ordered\;values\\ = average\;of\;{\left( {\frac{{50}}{2}} \right)^{th}}and\;{\left( {\frac{{50}}{2} + 1} \right)^{th}}\;ordered\;values\\ = average\;of\;{\left( {25} \right)^{th}}and\;{\left( {26} \right)^{th}}\;ordered\;values\\ = \frac{{1.5 + 1.7}}{2}\\ = 1.6\end{aligned}\)

Thus, the median value is 1.6.

02

Computing the fourths and fourth spread

For ED,

\(\begin{array}{l}{n_{smallest}} = 13\\{n_{{\rm{largest}}}} = 13\end{array}\)

The lower fourth is computed as,

\(\begin{aligned}\frac{{{n_{smallest}} + 1}}{2} = \frac{{13 + 1}}{2}\\ = {7^{th}}value\end{aligned}\)

The 7th value of the smallest half is 0.1.

This implies that the lower fourth is 0.1.

The upper fourth is computed as,

\(\begin{aligned}\frac{{{n_{{\rm{largest}}}} + 1}}{2} = \frac{{13 + 1}}{2}\\ = {7^{th}}value\end{aligned}\)

The 7th value of the largest half is 2.8.

This implies that the upper fourth is 2.8.

The fourth spread is computed as,

\(\begin{aligned}{f_s} = {\rm{Upper}}\,{\rm{fourth}} - {\rm{Lower}}\;{\rm{fourth}}\\ = 2.8 - 0.1\\ = 2.7\end{aligned}\)

Therefore, the fourth spread is 2.7.

For non-ED,

\(\begin{aligned}{l}{n_{smallest}} = 25\\{n_{{\rm{largest}}}} = 25\end{aligned}\)

The lower fourth is computed as,

\(\begin{aligned}\frac{{{n_{smallest}} + 1}}{2} = \frac{{25 + 1}}{2}\\ = {13^{th}}value\end{aligned}\)

The 13th value of the smallest half is 0.3.

This implies that the lower fourth is 0.3.

The upper fourth is computed as,

\(\begin{aligned}\frac{{{n_{{\rm{largest}}}} + 1}}{2} = \frac{{25 + 1}}{2}\\ = {13^{th}}value\end{aligned}\)

The 13th value of the largest half is 7.9.

This implies that the upper fourth is 7.9.

The fourth spread is computed as,

\(\begin{aligned}{f_s} = {\rm{Upper}}\,{\rm{fourth}} - {\rm{Lower}}\;{\rm{fourth}}\\ = 7.9 - 0.3\\ = 7.6\end{aligned}\)

Therefore, the fourth spread is 7.6.

03

Checking the outliers

b.

Referring to part a,

For ED,

The median value is 0.4.

The lower fourth is 0.1.

The Upper fourth is 2.8.

The fourth spread is 2.7.

For non-ED,

The median value is 1.6.

The lower fourth is 0.3.

The Upper fourth is 7.9.

The fourth spread is 7.6.

Any observation farther than 1.5\({f_s}\)from the closest fourth is an outlier.

An outlier is extreme if it is more than 3\({f_s}\)from the nearest fourth, otherwise, it is mild.

For x,

The calculations are as follows,

\(\begin{aligned}Lower\;fourth - 1.5{f_s} = 0.1 - \left( {1.5 \times 2.7} \right)\\ = 0.1 - 4.05\\ = - 3.95\end{aligned}\)

\(\begin{aligned}Upper\;fourth + 1.5{f_s} = 2.8 + \left( {1.5 \times 2.7} \right)\\ = 2.8 + 4.05\\ = 6.85\end{aligned}\)

It can be observed that there are no values that are less than -3.95.but there are few values that exceeds 6.85; they are, 8.9, 9.2,11.7 and 21.

For extreme outlier,

\(\begin{aligned}Upper\;fourth + 3{f_s} = 2.8 + \left( {3 \times 2.7} \right)\\ = 2.8 + 8.1\\ = 10.9\end{aligned}\)

Therefore, the extreme outliers are 11.7 and 21.

For y,

The calculations are as follows,

\(\begin{aligned}Lower\;fourth - 1.5{f_s} = 0.3 - \left( {1.5 \times 7.6} \right)\\ < 0\end{aligned}\)

\(\begin{aligned}Upper\;fourth + 1.5{f_s} = 7.9 + \left( {1.5 \times 7.6} \right)\\ = 7.9 + 11.4\\ = 19.3\end{aligned}\)

It can be observed that there are no values that are negative and no values are greater than 19.3.

This implies that are no outliers.

04

Construction of a comparative boxplot

Referring to parts a and b,

The five-number summary is,

For ED,

The smallest observation is 0.

The median value is 0.4.

The lower fourth is 0.1.

The Upper fourth is 2.8.

The fourth spread is 2.7.

The largest observation is 21.

For non-ED,

The smallest observation is 0.

The median value is 1.6.

The lower fourth is 0.3.

The Upper fourth is 7.9.

The fourth spread is 7.6.

The largest observation is 17.8.

The steps to construct a boxplot are as follows,

1. Compute the values of the five-number summary (smallest value, lower fourth, median, upper fourth, largest value).

2. Construct a line segment from the smallest value to the largest value of the dataset.

3. Construct a rectangular box from the lower fourth to the upper fourth and draw a line in the box at the median value.

The boxplot is represented as,

The red-colored boxplot represents ED, and the green-colored boxplot represents non-ED.

The comparison is as follows,

1. There are four outliers for ED and no outliers for non-ED samples.

2. The distribution is positively skewed for both samples.

3. There is less variability in ED than in the non-ED sample.

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Most popular questions from this chapter

The National Health and Nutrition Examination Survey (NHANES) collects demographic, socioeconomic, dietary, and health related information on an annual basis. Here is a sample of \({\rm{20}}\) observations on HDL cholesterol level \({\rm{(mg/dl)}}\) obtained from the \({\rm{2009 - 2010}}\) survey (HDL is 鈥済ood鈥 cholesterol; the higher its value, the lower the risk for heart disease):

\(\begin{array}{l}{\rm{35 49 52 54 65 51 51}}\\{\rm{47 86 36 46 33 39 45}}\\{\rm{39 63 95 35 30 48}}\end{array}\)

a. Calculate a point estimate of the population mean HDL cholesterol level.

b. Making no assumptions about the shape of the population distribution, calculate a point estimate of the value that separates the largest \({\rm{50\% }}\) of HDL levels from the smallest \({\rm{50\% }}\).

c. Calculate a point estimate of the population standard deviation.

d. An HDL level of at least \({\rm{60}}\) is considered desirable as it corresponds to a significantly lower risk of heart disease. Making no assumptions about the shape of the population distribution, estimate the proportion \({\rm{p}}\) of the population having an HDL level of at least \({\rm{60}}\).

The accompanying frequency distribution of fracture strength (MPa) observations for ceramic bars fired in aparticular kiln appeared in the article 鈥淓valuating TunnelKiln Performance鈥 (Amer. Ceramic Soc. Bull., Aug.1997: 59鈥63).

Class81-<83 83-<85 85-<87 87-<89 89-<91

Frequency6 7 17 30 43

Class91-<93 93-<95 95-<97 97-<99

Frequency28 22 13 3

  1. Construct a histogram based on relative frequencies, and comment on any interesting features.
  2. What proportion of the strength observations are at least 85? Less than 95?

c. Roughly what proportion of the observations are less than 90?

A sample of 26 offshore oil workers took part in a simulated escape exercise, resulting in the accompanying data on time (sec) to complete the escape (鈥淥xygen Consumption and Ventilation During Escape from an Offshore Platform,鈥 Ergonomics, 1997: 281鈥292):

389 356 359 363 375 424 325 394 402

373 373 370 364 366 364 325 339 393

392 369 374 359 356 403 334 397

a. Construct a stem-and-leaf display of the data. How does it suggest that the sample mean and median will compare?

b. Calculate the values of the sample mean and median.(Hint:\(\sum {{x_i} = } \)9638.)

c. By how much could the largest time, currently 424, be increased without affecting the value of the sample median? By how much could this value be decreased without affecting the value of the sample median?

d. What are the values of \(\bar x\)and \(\tilde x\), when the observations are re expressed in minutes?

Let \({\rm{X}}\) be the total medical expenses (in \({\rm{1000}}\) s of dollars) incurred by a particular individual during a given year. Although \({\rm{X}}\) is a discrete random variable, suppose its distribution is quite well approximated by a continuous distribution with pdf \({\rm{f(x) = k(1 + x/2}}{\rm{.5}}{{\rm{)}}^{{\rm{ - 7}}}}\) for.

a. What is the value of\({\rm{k}}\)?

b. Graph the pdf of \({\rm{X}}\).

c. What are the expected value and standard deviation of total medical expenses?

d. This individual is covered by an insurance plan that entails a \({\rm{\$ 500}}\) deductible provision (so the first \({\rm{\$ 500}}\) worth of expenses are paid by the individual). Then the plan will pay \({\rm{80\% }}\) of any additional expenses exceeding \({\rm{\$ 500}}\), and the maximum payment by the individual (including the deductible amount) is\({\rm{\$ 2500}}\). Let \({\rm{Y}}\) denote the amount of this individual's medical expenses paid by the insurance company. What is the expected value of\({\rm{Y}}\)?

(Hint: First figure out what value of \({\rm{X}}\) corresponds to the maximum out-of-pocket expense of \({\rm{\$ 2500}}\). Then write an expression for \({\rm{Y}}\) as a function of \({\rm{X}}\) (which involves several different pieces) and calculate the expected value of this function.)

In a famous experiment carried out in 1882, Michelson,and Newcomb obtained 66 observations on the time it took for light to travel between two locations in Washington, D.C. A few of the measurements(coded in a certain manner) were 31, 23, 32, 36, 22, 26, 27, and 31.

a. Why are these measurements not identical?

b. Is this an enumerative study? Why or why not?

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