In: Statistics and Probability
The types of raw materials used to construct stone tools found at an archaeological site are shown below. A random sample of 1486 stone tools were obtained from a current excavation site.
Raw Material | Regional Percent of Stone Tools | Observed Number of Tools as Current excavation Site |
Basalt | 61.3% | 898 |
Obsidian | 10.6% | 163 |
Welded Tuff | 11.4% | 165 |
Pedernal chert | 13.1% | 197 |
Other | 3.6% | 63 |
Use a 1% level of significance to test the claim that the regional distribution of raw materials fits the distribution at the current excavation site.
(a) Find the value of the chi-square statistic for the sample.
(Round the expected frequencies to at least three decimal places.
Round the test statistic to three decimal places.)
=
Solution:
Given:
Claim: the regional distribution of raw materials fits the distribution at the current excavation site.
Thus hypothesis are:
H0: the regional distribution of raw materials fits the distribution at the current excavation site.
H0: The distributions are the
same.
Vs
H1: the regional distribution of raw materials does NOT fit the distribution at the current excavation site.
H1: The distributions are different.
Level of significance = 0.01
Part a) Find the value of the chi-square statistic for the sample
Chi square test statistic for goodness of fit
Where
Oi = Observed Counts
Ei =Expected Counts = N * Regional Percent of Stone Tools
Thus we need to make following table:
Raw Material | Regional Percent of Stone Tools | Oi: Observed Number of Tools as Current excavation Site | Ei: Expected frequencies | Oi^2/Ei |
---|---|---|---|---|
Basalt | 61.30% | 898 | 910.918 | 885.265 |
Obsidian | 10.60% | 163 | 157.516 | 168.675 |
Welded Tuff | 11.40% | 165 | 169.404 | 160.710 |
Pedernal chert | 13.10% | 197 | 194.666 | 199.362 |
Other | 3.60% | 63 | 53.496 | 74.192 |
N = 1486 |
Thus
Are all the expected frequencies greater than 5?
Yes
Sampling distribution:
Chi-square
the degrees of freedom
df = k - 1 = 5 -1 = 4
Part c) estimate the P-value
Use following Excel command:
=CHISQ.DIST.RT(2.205,4)
=0.698
P-value = 0.698
Part d) will you reject or fail to reject the null hypothesis of independence?
Since P-value = 0.698 > 0.01 level of significance, we fail to reject H0.
Since the P-value > α, we fail to reject the null hypothesis
Part e) Interpret your conclusion in the context of the application.
At the 0.01 level of significance, there is insufficient evidence to conclude that the regional distribution of raw materials and the current excavation site distribution are not independent.