To determine the organic material in a dried lake bed, the percent carbon by mas

Question

To determine the organic material in a dried lake bed, the percent carbon by mass is measured at two Map different locations. In order 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 with two significant figures. Location 1 Number 0.055 Location 1 Location 2 (% C) |(% C) 60.40 60.30 60.30 60.40 60.30 Location 2 60.10 60.90 60.20 60.70 60.40 Number 2 0.34 4 What is the calculated F value for comparing the standard deviations? Number 37.68 Are the two standard deviations for the locations significantly different at the 95% confidence level? yes O no Scroll down.

Explanation / Answer

Two-Sample T-Test and CI: Location 1, Location 2

Two-sample T for Location 1 vs Location 2

N Mean StDev SE Mean
Location 1 5 60.3400 0.0548 0.024
Location 2 5 60.460 0.336 0.15

Difference = mu (Location 1) - mu (Location 2)
Estimate for difference: -0.120
95% CI for difference: (-0.543, 0.303)
T-Test of difference = 0 (vs not =): T-Value = -0.79 P-Value = 0.475 DF = 4

t-value=-0.79

Ans. No. Since p-value= 0.475>0.05 and 95% confidence interval, (-0.543, 0.303) contains zero.

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