Chapter 4, Problem 15PT1

(a)

To determine

To compare: The scores of the Machine A and Machine B

Explanation of Solution

Given information:

The stem and leaf diagram for the two machines is given below

Machine A		Machine B
	5	3, 5, 9
6,1	4	2, 5, 7
9,7,4,1,1	3	2, 4, 8, 9
8,7,6,3,2,0	2	1, 5, 9
5,4	1	0
	0	2

Formula used:

  $\begin{array}{l}\text{Range = Maximum - Minimum}\\ \text{Median = }{\left(\frac{n+1}{2}\right)}^{th}term\end{array}$

Calculation:

For machine A

  $\begin{array}{l}\text{Minimum value = 14}\\ \text{Maximum value = 46}\\ \text{Range = 46 -14 }\\ \text{Range =}32\end{array}$

Median

  $\begin{array}{l}n=15\\ \text{Median = }{\left(\frac{15+1}{2}\right)}^{th}term\\ Q_2=8^{th}term=28\end{array}$

Skewness

Most of the data values are between $20-29$

It seems to be symmetric roughly.

For machine B

  $\begin{array}{l}\text{Minimum value = 2}\\ \text{Maximum value = 59}\\ \text{Range = 59 -2 }\\ \text{Range =57}\end{array}$

Median

  $\begin{array}{l}n=15\\ \text{Median = }{\left(\frac{15+1}{2}\right)}^{th}term\\ Q_2=8^{th}term=38\end{array}$

Skewness

Most of the data values are between $30-39$

It seems to be negatively skewed as it is skewed towards left.

Conclusion:

Center

Center can be compared on the basis of median.

Here, $\text{Machine A }\left(=28\right)<\text{Machine B }\left(=38\right)$

Spread

Here, machine B is more spread than machine A as the range of machine B is greater than machine A.

Skewness

Machine A is roughly symmetrical while Machine B is negatively skewed.

(b)

To determine

To find: The machine achieving the overall highest gain.

Answer to Problem 15PT1

Machine B should be advertised.

Explanation of Solution

Formula used:

  $\overline{x}=\frac{\Sigma x}{n}$

Calculation:

The machine achieving overall highest gain can be decided on the basis of the average gain in cardiovascular fitness.

For machine A

  $\begin{array}{l}\overline{A}=\frac{\Sigma A}{n}\\ \overline{A}=\frac{14+15+\mathrm{..................................46}}{15}\\ \overline{A}=\frac{434}{15}\\ \overline{A}=28.9333\end{array}$

For machine B

  $\begin{array}{l}\overline{B}=\frac{\Sigma B}{n}\\ \overline{B}=\frac{2+10+\mathrm{.............................}+59}{15}\\ \overline{B}=\frac{531}{15}\\ \overline{B}=35.4\end{array}$

Conclusion:

Mean overall gain is higher in machine B.

(c)

To determine

To find: The machine with highest consistency.

Answer to Problem 15PT1

Machine A should be advertised for the most consistent gain in cardiovascular fitness.

Explanation of Solution

Formula used:

  $s^2=\frac{1}{n-1}\times \left(\Sigma x^2-n\times {\overline{x}}^2\right)$

Calculation:

The consistency can be measured by variance.

More variance is an indication of less consistency.

For machine A

  $\begin{array}{l}{s}_A^2=\frac{1}{15-1}\times \left(\Sigma A^2-n\times {\overline{A}}^2\right)\\ \Sigma A^2={14}^2+{15}^2+{\mathrm{...............................46}}^2=13788\\ s_A^2=\frac{1}{15-1}\times \left(13788-15\times {28.9333}^2\right)\\ s_A^2=87.9259\end{array}$

For machine B

  $\begin{array}{l}{s}_B^2=\frac{1}{15-1}\times \left(\Sigma B^2-n\times {\overline{B}}^2\right)\\ \Sigma B^2=2^2+{10}^2+{\mathrm{...............................59}}^2=22469\\ s_B^2=\frac{1}{15-1}\times \left(22469-15\times {35.4}^2\right)\\ s_B^2=262.2571\end{array}$

Conclusion:

The variance of machine B is greater than machine A.

(d)

To determine

To find: The reason for limitedness of the experiment.

Explanation of Solution

Here, the volunteers are chosen from the area near the company’s headquarters.

This may not represent the entire population of the cardiovascular fitness gain. Other centres may vary in results.