Question

To determine the organic material in a dried lake bed, the percent carbon by mass is measured at two different locations. To compare the means of the two different locations, it must first be determined whether the standard deviations of the two locations are different. For each location, calculate the standard deviation and report it with two significant figures.

Answer 1

Answer:

$\sigma_1 = 0.08$ --- Location 1

$\sigma_2 = 0.34$ --- Location 2

Step-by-step explanation:

Given

See attachment for the given data

Required

The standard deviation of each location

For location 1

First, calculate the mean

$\bar x_1 =\frac{\sum x}{n}$

So, we have:

$\bar x_1 =\frac{30.40+30.20+30.30+30.40+30.30}{5}$

$\bar x_1 =\frac{151.60}{5}$

$\bar x_1 =30.32$

The standard deviation is calculated as:

$\sigma_1 = \sqrt{\frac{\sum(x - \bar x_1)^2}{n-1}}$

$\sigma_1 = \sqrt{\frac{(30.40 - 30.32)^2+(30.20 - 30.32)^2+(30.30 - 30.32)^2+(30.40 - 30.32)^2+(30.30 - 30.32)^2}{5-1}}$

$\sigma_1 = \sqrt{\frac{0.028}{4}}$

$\sigma_1 = \sqrt{0.007}$

$\sigma_1 = 0.08$

For location 2

First, calculate the mean

$\bar x_2 =\frac{\sum x}{n}$

So, we have:

$\bar x_2 =\frac{30.10+30.90+30.20+30.70+30.30}{5}$

$\bar x_2 =\frac{152.2}{5}$

$\bar x_2 =30.44$

The standard deviation is calculated as:

$\sigma_2 = \sqrt{\frac{\sum(x - \bar x_2)^2}{n-1}}$

$\sigma_2 = \sqrt{\frac{(30.10-30.44)^2+(30.90-30.44)^2+(30.20-30.44)^2+(30.70-30.44)^2+(30.30-30.44)^2}{5-1}}$

$\sigma_2 = \sqrt{\frac{0.472}{4}}$

$\sigma_2 = \sqrt{0.118}$

$\sigma_2 = 0.34$