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9. To determine the organic material in a dried lake bed, you have Location 1 (w

ID: 566012 • Letter: 9

Question

9. To determine the organic material in a dried lake bed, you have Location 1 (wt% OC) Location 2 (wt%OC) 70.10 | Sample # determined the weight percent of organic carbon (wt% OC) at two different locations. Five samples from each location were analyzed. The results are shown in the table. Is the wt% of orgaric carbon at the two locations significantly different at the 95% confidence level? Show all work below (except if you reed to calculate means or standard deviations, for which 70.40 70.30 70.90 70.30 70.20 70.40 70.70 0.20 70.80 you can use Excel), then circle "YES" or "NO" at the bottom of the page.

Explanation / Answer

Solution:

The given data is-

Location 1

Location 2

1

70.40

70.10

2

70.30

70.90

3

70.30

70.20

4

70.40

70.70

5

70.20

70.80

To calculate the standard deviation we need the mean of the observations;

Mean of set 1 = (70.40+70.20+70.30+70.40+70.30)/5 = 70.32 = x1

Mean of set 2 = (70.20+70.80+70.30+70.80+70.30)/5 = 70.48 = x2

The formula for standard deviation is;

s =

Where, x =value of each observation,

x = Mean of all values

n = number of observations = 5 for both sets

Now we calculate the standard deviation using the following table for convenience-

Set 1-

Sr. No

x

x

x- x

(x- x)2

1

70.40

70.32

0.08

0.0064

2

70.20

70.32

-0.12

0.0144

3

70.30

70.32

-0.02

0.0004

4

70.40

70.32

  0.10

0.0100

5

70.30

70.32

-0.02

0.0004

Sum of all (x- x) 2= 0.0316

Thus the standard deviation s = (0.0316/4) = 0.0079

Set 2-

Sr. No

x

x

x- x

(x- x)2

1

70.20

70.48

-0.28

0.0784

2

70.80

70.48

0.32

0.1024

3

70.30

70.48

-0.18

0.0324

4

70.80

70.48

0.32

0.1024

5

70.30

70.48

-0.18

0.0324

Sum of all (x- x)2 =0.348

Thus the standard deviation s = (0.348/4) = 0.2949

F-testè For the F-test we need the variances
We know that variance f2 = deviation2

Thus for set 1- variance = 6.241*10-5

For set 2- variance = 0.0869

To calculate the F-value, f1 > f2 , so that we get a positive F-value.

Here 0.0869 > 6.241*10-5

Hence F = 0.0869 / 6.241*10-5 =1398.4

Location 1

Location 2

1

70.40

70.10

2

70.30

70.90

3

70.30

70.20

4

70.40

70.70

5

70.20

70.80