Question

An article reported the following data on oxygen consumption (mL/kg/min) for a sample of ten firefighters performing a fire-suppression simulation: 29.6 49.3 30.1 28.3 28.7 26.3 33.5 29.8 23.7 31.0 Compute the following. (Round your answers to four decimal places.)
(a) the sample range(b) the sample variance s2 from the definition(c) the sample standard deviation(d) s2 using the shortcut method

Answer 1

Answer:

a) Range = 25.6

b) Variance = 48.1446

c) s = 6.9386

d) See below

Step-by-step explanation:

(a)  the sample range

To obtain the sample range, we must sort the data increasingly :

23.7, 26.3, 28.3, 28.7, 29.6, 29.8, 30.1, 31, 33.5, 49.3

Then, find the distance between the largest and the lowest

Range = 49.3 - 23.7 = 25.6

(b) the sample variance $s^2$ from the definition

In order to find the variance from the definition, we need first the mean. The mean is defined as the average

$\bar x=\displaystyle\frac{\displaystyle\sum_{i=1}^{n}x_i}{n}$

where the $x_i$ are the values of the data collected and n=10 the size of the sample.

So, the mean is

$\bar x=31.03$

Now, the variance of the sample is defined as  

$s^2=\displaystyle\frac{\displaystyle\sum_{i=1}^n(x_i-\bar x)^2}{n-1}$

and we have that the variance is

$s^2=48.1446$

(c) the sample standard deviation

The sample standard deviation is nothing but the square root of the variance

$s=\sqrt{48.1446}=6.9386$

d) $s^2$  using the shortcut method

The shortcut method figures out the variance without having to compute the mean. The formula is

$s^2=\frac{n\sum x_i^2-(\sum x_i)^2}{n(n-1)}$

where n=10 is the sample size .

So, using the shortcut method,

$s^2=\frac{10*10,061.91-96,286.09}{10*9}=48.1446$