Question

The article “Solid-Phase Chemical Fractionation of Selected Trace Metals in Some Northern Kentucky Soils” (A. Karathanasis and J. Pils, Soil and Sediment Contamination, 2005:293-308) reports that in a sample of 26 soil specimens taken in a region of northern Kentucky, the average concentration of chromium (Cr) in mg/kg was 20.75 with a standard deviation of 3.93.

a. Can you conclude that the mean concentration of Cr is greater than 20 mg/kg?

b. Can you conclude that the mean concentration of Cr is less than 25 mg/kg?

Answer 1

Solution:

Given:

Sample size = n = 26

the average concentration of chromium (Cr) in mg/kg = = 20.75 and a standard deviation = s = 3.93.

Part a) Can you conclude that the mean concentration of Cr is greater than 20 mg/kg?

Step 1) State H0 and H1:

Step 2) Test statistic:

Step 3) t critical value:

df = n - 1= 26 - 1 = 25

Right tail area = 0.05

From t table t critical value for df = 25 and right tail area ( one tail area) = 0.05 is t = 1.708

Step 4) Decision rule : Reject H0, if t test statistic value > 1.708, otherwise we fail to reject H0.

Since t test statistic value = 0.973 is not greater than 1.708 , hence we fail to reject H0.

Step 5) Conclusion: Since we failed to reject H0, there is not sufficient evidence to conclude that: the mean concentration of Cr is greater than 20 mg/kg.

Part b) Can you conclude that the mean concentration of Cr is less than 25 mg/kg?

Step 1) State H0 and H1:

Step 2) Test statistic:

Step 3) t critical value:

df = n - 1= 26 - 1 = 25

Left tail area = 0.05

From t table t critical value for df = 25 and left tail area ( one tail area) = 0.05 is t = -1.708

Step 4) Decision rule : Reject H0, if t test statistic value < -1.708, otherwise we fail to reject H0.

since t test statistic value = t = -5.514 < t critical value = -1.708, we reject H0.

Step 5) Conclusion: Since we rejected H0, there is sufficient evidence to support the claim that: the mean concentration of Cr is less than 25 mg/kg