The manufacturer of exercise machines for fitness centers has designed two new elliptical machines that are meant to increase cardiovascular fitness. The two machines are being tested on $30$ volunteers at a fitness center near the company’s headquarters. The volunteers are randomly assigned to one of the machines and use it daily for two months. A measure of cardiovascular

fitness is administered at the start of the experiment and again at the end. The following table contains the differences in the two scores (After – Before)

for the two machines. Note that higher scores indicate larger gains in fitness.

(a) Write a few sentences comparing the distributions of cardiovascular

fitness gains from the two elliptical machines.

(b) Which machine should be chosen if the company wants to advertise it as achieving the highest overall gain in cardiovascular fitness? Explain your reasoning.

(c) Which machine should be chosen if the company wants to advertise it as achieving the most consistent gain in cardiovascular fitness? Explain your reasoning.

(d) Give one reason why the advertising claims of the company (the scope of inference) for this experiment would be limited. Explain how the company could broaden that scope of inference.

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

Range = Maximum - Minimum

Median $={(\frac{n+1}{2})}^{th}term$

For machine $A$

Minimum value $=14$

Maximum value $=46$

Range $=46-14$

Range $=32$

Median

$n=15$

Median$={(\frac{15+1}{2})}^{th}term$

$Q^2=8^{th}term=28$

Skewness

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

It seems to be symmetric roughly.

For machine $B$

Minimum value = $2$

Maximum value = $59$

Range = $59-2$

Range $=57$

$n=15$

$Q_2=8^{th}term=38$

Skewness

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

It appears to be negatively skewed, as it leans to the left.

Center

Center can be compared on the basis of the median.

Here, Machine $A(=28)<MachineB(=38)$

Spread

Machine $B$is more spread than machine $A$in this case because its range is bigger than machine $A\text{'}s$

Skewness

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

$\overline{x}=\frac{\Sigma x}{n}$

The machine with the greatest overall gain can be determined based on the average increase in cardiovascular fitness.

For machine A

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

For machine $B$

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

Mean overall gain is higher in machine $B$

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

Variance can be used to assess consistency.

More volatility indicates a lack of stability.

For machine $A$

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

For machine $B$

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

The variance of machine $B$is greater than machine $A$

Volunteers are picked from the surrounding area of the company's headquarters. This may not be representative of the total cardiovascular fitness gain population. The outcomes of other centers may differ.