1. Blood cocaine concentration (milligrams per liter) was determined for a set o

Question

1. Blood cocaine concentration (milligrams per liter) was determined for a set of individuals who had died from a cocaine-induced delirium (went crazy) versus individuals who dies of a cocaine overdose with no delirium (simply fell asleep). The data is displayed below. Answer the questions on a separate sheet of paper and turn in at the end of the hour. You may work together with a single partner and you may use your notes, text, computer, etc. Delirium: 0 0 0 0 0.1 0.1 0.1 0.2 0.2 0.3 0.3 0.5 0.7 0.7 1.0 5 2.7 2.8 3.5 4.0 8.9 9.2 .7 21.0 No Delirium: 0 0 o 0 00.1 0.1 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.3 0.4 0.5 0.5 0.60.8 0.9 1.0 1.2 14 15 1.7 2.03.2 3.54.1 4.3 4.8 5.05.6 5.9 6.06.4 7.9 8.38.7 9.1 9.6 9.9 0 11.5 12.2 12.7 14.0 6.6 17.8 a) Describe the population of interest, the samples, the variable(s) and the subject:s (individuals) in the experiment. b) Which clssification is appropriate for the data Quantitative Categorical c) Present the two data sets in a back-to-back stem-and-leaf plot (show the key) d) Describe briefly the main features of the distribution for each data set (symmetry, skewness, outliers, unimodal/bimodal as applicable) e) Give the 5-number summary for each data set. (you may not use your calculator's statistical functions on this question and you need to show your work). f) Identify any mild and extreme outliers and construct a (modified) boxplot for both samples. Use one real axis and stack the boxplots. f) Calculate the sample mean and standard deviation for each data set (you may use your calculator's statistical functions)

Explanation / Answer

a) Population of interest is Blood cocaine concentration was determined for a set of individuals

Sample: two types of sample is there

individuals who had died from a cocaine-induced delirium

individuals who had died from a cocaine-overdose with no delirium

variable: cocaine-induced delirium and with no delirium

b) Categorical variable

c)

D=c(0,0,0,0,0.1,0.1,0.1,0.1,0.2,0.2,0.3,0.3,0.5,0.7,0.7,1.0,1.5,2.7,
2.8,3.5,4.0,8.9,9.2,11.7,21.0)
ND=c(0,0,0,0,0,0.1,0.1,0.1,0.1,0.2,0.2,0.2,0.3,0.3,0.3,0.4,0.5,0.5,0.6,
0.8,0.9,1.0,1.2,1.4,1.5,1.7,2.0,3.2,3.5,4.1,4.3,4.8,5.0,5.6,5.9,6.0,6.4,
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)
stem(D)
stem(ND)

stem(D)

The decimal point is 1 digit(s) to the right of the |

0 | 000000000000111123344
0 | 99
1 | 2
1 |
2 | 1

> stem(ND)

The decimal point is at the |

   0 | 00000111122233345568902457
   2 | 025
   4 | 138069
   6 | 049
   8 | 37169
10 | 05
12 | 27
14 | 0
16 | 68

d) Data one is unimodel with outliers and it is right skewed

Data second is unimodel and it is right skewed

e)

summary(D)
   Min. 1st Qu. Median    Mean 3rd Qu.    Max.
0.000   0.100   0.500   2.784   2.800 21.000
summary(ND)
   Min. 1st Qu. Median    Mean 3rd Qu.    Max.
0.000   0.300   1.600   4.250   7.525 17.800

f)

par(mfrow=c(1,2))
boxplot(D)
boxplot(ND)

f)

DD=c(mean(D), sd(D));DD
[1] 2.784000 4.983864
NDD=c(mean(ND), sd(ND));NDD
[1] 4.250000 4.936381

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