In an investigation of environmental causes of disease, data were collected on t

Question

In an investigation of environmental causes of disease, data were collected on the annual mortality rate (deaths per 100 000) for males in 61 large towns in England and Wales. In addition, the water hardness was recorded as the calcium concentration (parts per million, ppm) in the drinking water. Below, we provide some descriptive statistics for both variables, i.e. mortality and calcium concentration.

> summary(calcium)

Min. 1st Qu. Median Mean 3rd Qu. Max.

5.00 14.00 39.00 47.18 75.00 138.00

> summary(mortality)

Min. 1st Qu. Median Mean 3rd Qu. Max.

1096 1379 1555 1524 1668 1987

Here are histograms for both variables.

(a) For each variable (i.e. calcium concentration and mortality) describe its distribution by describing

(i) its shape;

(ii) its central tendencies;

(iii) its dispersion.

?(b) For each variable, are there any outliers in the sample? (Explain.)

1. In an investigation of environmental causes of disease, data were collected on the annual mortality rate (deaths per 100000) for males in 61 large towns in England and Wales In addition, the water hardness was recorded as the calcium concentration (parts per million, ppm) in the drinking water. Below, we provide some descriptive statistics for both variables, i.e. mortality and calcium concentration summary (calcium,) Min. 1st Qu. Median Mean 3rd Qu 5.00 14.00 39.00 47.18 75.00 138.00 ?? summary (mortality) Min. 1st Qu. Median Mean 3rd Qu. 1096 1379 1555 1524 1668 Max 1987 Here are histograms for both variables Histogram of calcium 20 40 80 80 100 140 caldum Histogram of mortality 1000 1600 2000 (a) For each variable (i.e. calcium concentration and mortality) describe its distribu tion by describing (i) its shape; (ii) its central tendencies; (ii) its dispersion b) For each variable, are there any outliers in the sample? (Explain.)

Explanation / Answer

a) i) Calcium curve is positively skewed but not symmetrical

Mortality curve is approximately bell shaped and normal

ii) Calcium :

Mean = 47.18

Median = 39

Mortality :

Mean = 1524

Median = 1555

iii) Dispersion

Calcium :

Range = 138 - 5 = 133

Interquartile range = 75 - 14 = 61

Quartile deviation = 61 / 2 = 30.5

Mortality :

Range = 1987 - 1096 = 891

Interquartile range = 1668 - 1379 = 144.5

Quartile deviation = 144.5 / 2 = 72.25

b) Calcium :

Q1–1.5 × IQR = 14 - 1.5 * 61 = -77.5

Q1+1.5 × IQR = 75 + 1.5 * 61 = 166.5

Yes there is no outlier , since 165.5 > 138

Mortality :

Q1–1.5 × IQR = 1379 - 1.5 * 144.5 = 1162.25

Q3 +1.5 × IQR = 1668 - 1.5 * 144.5 = 1884.75

Yes there is an outlier , since 1096 < 1162.5

One of the outlier is 1096.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.