Question

Sample 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. 2 T Location 1 | Location 2 (% C) (% C) 30.40 30.10 30.20 30.90 30.30 30.20 30.40 30.70 30.30 30.30 For each location, calculate the standard deviation and report it with two significant figures. 3 4. 5 location 1 standard deviation: location 2 standard deviation: What is the calculated F value for comparing the standard deviations?

Answer 1

Location 1	Location 2
30.4	30.1
30.2	30.9
30.3	30.2
30.4	30.7
30.3	30.3
0.08	0.34	Stdev

Use Excel function =STDEV.S(select the data)

$F_{stats}= \frac{s_{2}^2}{s_{1}^2}= \frac{0.34^2}{0.08^2} = \boldsymbol{18.06}$

_________________________

P-value = F.DIST.RT(18.06,4,4) = 0.007

p-value is less than 0.05 so, reject Ho we can conclude that the standard dev are not equal

YES

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t-Test: Two-Sample Assuming Unequal Variances
	Location 1	Location 2
Mean	30.32	30.44
Variance	0.007	0.118
Observations	5	5
Hypothesized Mean Difference	0
df	4
t Stat	-0.76
P(T<=t) one-tail	0.2450741
t Critical one-tail	2.1318468
P(T<=t) two-tail	0.4901481
t Critical two-tail	2.7764451

tcalc = -0.76

$\boldsymbol{No}$

Since P-value is greater than 0.05