To determine the organic material in a dried lake bed, you have determined the percent carbon...

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Answer 1

Answer #1

From the given data

Descriptive Statistics: Location 1, Location 2 Total Variable Count Mean StDev Variance Location 1 Location 2 5 60.320 0.0837 0.0070 5 60.460 0.329 0.108

Test Statistic F = 0.0837^2 / 0.329^2 = 0.064723

Test Statistic, F: 0.0648
Lower Critical F: 0.1041176
Upper Critical F: 9.60452
P-Value: 0.0213

Here P-value < alpha 0.05 so we reject H0

Correct answer: YES

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Answer 2

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