Question

An article in the Transactions of the Institution of Chemical Engineers (Vol. 34, 1956, pp. 280–293) reported data from an experiment investigating the effect of several process variables on the vapor phase oxidation of naphthalene. A sample of the percentage mole conversion of naphthalene to maleic anhydride follows: 4.2, 4.7, 4.7, 5.0, 3.8, 3.6, 3.0, 5.1, 3.1, 3.8, 4.8, 4.0, 5.2, 4.3, 2.8, 2.0, 2.8, 3.3, 4.8, and 5.0. (a) Calculate the sample mean. (b) Calculate the sample variance and sample standard deviation. (c) Construct a box plot of the data

Answer 1

a) Mean = 4

b) Standard deviation = 0.9308 ; Variance = 0.8663

c) attached in the file

Explanation:

Given:

4.2, 4.7, 4.7, 5.0, 3.8, 3.6, 3.0, 5.1, 3.1, 3.8, 4.8, 4.0, 5.2, 4.3, 2.8, 2.0, 2.8, 3.3, 4.8, and 5.0

n = 20

a) Mean =
`\frac{\textup{sum of all observations}}{\textup{Total number of observations}}`

thus,

Mean =
`\frac{\textup{80}}{\textup{20}}`

or

Mean = 4

b)

data data-mean (data - mean)²

4.2 0.2 0.04

4.7 0.7 0.49

4.7 0.7 0.49

5.0 1 1

3.8 -0.2 0.04

3.6 -0.4 0.16

3.0 -1 1

5.1 1.1 1.21

3.1 -0.9 0.81

3.8 -0.2 0.04

4.8 0.8 0.64

4.0 0 0

5.2 1.2 1.44

4.3 0.3 0.09

2.8 -1.2 1.44

2.0 -2 4

2.8 -1.2 1.44

3.3 -0.7 0.49

4.8 0.8 0.64

5.0 1 1

====================================

Now, ∑(Data - Mean)² = 16.46

and,

Standard deviation =
`\sqrt{(\sum(Data - Mean)^2)/(n-1)}`

or

Standard deviation =
`\sqrt{(16.46)/(20-1)}`

or

Standard deviation = 0.9308

variance = ( Standard deviation )²

or

Variance = 0.9308² = 0.8663

c) the plot is attached in the file