Question

Background:

Excavations at Stonehenge uncovered a number of unshed antlers, antler tines, and animal bones. Carbon-14 dating methods were used to estimate the ages of the Stonehenge artifacts. Carbon-14 is one of three carbon isotopes found in Earth’s atmosphere. Carbon-12 makes up 99% of all of the carbon dioxide in the air. Virtually all of the remaining 1% is composed of carbon-13. By far, the rarest form of carbon isotope found in the atmosphere is carbon-14.

The ratio of carbon-14 to carbon-12 remains constant in living organisms. However, once the organism dies, the amount of carbon-14 in the remains of the organism begins to decline, because it is radioactive, with a half-life of 5730 years (the “Cambridge half-life”). So the decay of carbon-14 into ordinary nitrogen makes possible a reliable estimate about the time of death of the organism. The counted carbon-14 decay events can be modeled by the normal distribution.

The team used two different carbon-14 dating methods to arrive at age estimates for the numerous Stonehenge artifacts. The liquid scintillation counting (LSC) method utilizes benzene, acetylene, ethanol, methanol, or a similar chemical. Unlike the LSC method, the accelerator mass spectrometry (AMS) technique offers direct carbon-14 isotope counting. The AMS method's greatest advantage is that it requires only milligram-sized samples for testing. The AMS method was used only on recovered artifacts that were of extremely small size.

Question:

the center of the monument are two concentric circles of igneous rock pillars, called bluestones. The construction of these circles was never completed. These circles are known as the Bluestone Circle and the Bluestone Horseshoe. The stones in these two formations were transported to the site from the Prescelly Mountains in Pembrokeshire, Southwest Wales. Excavation at the center of the monument revealed an antler, an antler tine, and an animal bone. Each artifact was submitted for dating. It was determined that this sample of three artifacts had a mean age of 2193.3 BCE, with a standard deviation of 104.1 years. Assume that the ages are normally distributed with no obvious outliers. Use an α = 0.05 significance level to test the claim that the population mean age of the Bluestone formations is different from Corbin’s declared mean age of the ditch, that is, 2950 BCE.

Answer 1

The hypothesis being tested is:

H0: µ = 2950

Ha: µ ≠ 2950

x = 2193.3

µ = 2950

s = 104.1

n = 3

The test statistic, t = (x - µ)/s/√n

t = (2193.3 - 2950)/104.1/√3

t = -12.590

The p-value for t = -12.590 and df = 3 - 1 = 2 is 0.0062.

Since the p-value (0.0062) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that the population mean age of the Bluestone formations is different from Corbin’s declared mean age of the ditch, that is, 2950 BCE.

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