Question

A study of the relationship between age and various visual functions (such as acuity and depth perception) reported the following observations on the area of scleral lamina (mm2) from human optic nerve heads: 2.79 2.57 2.73 3.75 2.27 2.75 4.00 4.22 3.88 4.35 3.41 4.57 2.38 3.73 2.75 3.47 3.06(a) Calculate ?xi and ?xi2. (Round ?xi2 to two decimal places.)(b) Use the values calculated in part (a) to compute the sample variance s2 and then the sample standard deviation s. (Round your answers to three decimal places.)

Answer 1

Answer:

(a) $\sum x_i=56.68$ and $\sum x_i^2=197.46$

(b) The sample variance is $s^2=0.530$ and the sample standard deviation is $s=0.728$

Step-by-step explanation:

(a)

The sum of these 17 sample observations is

$\sum x_i=2.79\:+2.57+\:2.73\:+3.75\:+2.27+\:2.75+\:4.00+\:4.22+\:3.88+\:4.35+\:3.41+\:4.57+\:2.38+\:3.73\:+2.75+\:3.47+\:3.06\\\\\sum x_i=56.68$

and the sum of their squares is

$\sum x_i^2=2.79^2\:+2.57^2+\:2.73^2\:+3.75^2\:+2.27^2+\:2.75^2+\:4.00^2+\:4.22^2+\:3.88^2+\:4.35^2+\:3.41^2+\:4.57^2+\:2.38^2+\:3.73^2\:+2.75^2+\:3.47^2+\:3.06^2\\\\\sum x_i^2=197.46$

(b)

The sample variance, denoted by $s^2$, is given by

$s^2=\frac{S_{xx}}{n-1}$

where $S_{xx} =\sum x_i^2-\frac{(\sum x_i)^2}{n}$

Applying the above formula we get that

$S_{xx}=197.46-\frac{(56.68)^2}{17}\\\\S_{xx} =8.482$

$s^2=\frac{8.482}{17-1}=0.530$

The sample standard deviation, denoted by s, is the (positive) square root of the variance:

$s=\sqrt{s^2}$

Applying the above formula we get that

$s=\sqrt{0.530}=0.728$