Answer:
There is enough evidence, at a significance level of 0.05, that the population mean age of the Bluestone formations is different from 2950 BCE.
Explanation:
We have a sample and we want to perform a hypothesis test on the mean.
The null hypothesis is the Corbin's declared age (2950 BCE). The alternative hypothesis states that the age differ from that value.
They can be expressed as:
![\H_0:\mu=2950\\\\H_a:\mu\\eq2950](https://img.qammunity.org/2021/formulas/mathematics/college/so917wiz5urbmx7fq02kpzwhv7mdxanjgr.png)
The significance level is 0.05.
The sample has a size of n=3, a mean of 2193.3 BCE and a standard deviation of 104.1 years.
As the standard deviation is estimated from the sample, we have to calculate the t-statistic.
![t=(\bar x-\mu)/(s/√(n))=(2193.3-2950)/(104.1/√(3))=(-756.7)/(60.1)=-12.59](https://img.qammunity.org/2021/formulas/mathematics/college/ni8j4xeu5uqcu5h8y8c036x1exinxh7pcl.png)
The degrees of freedom for this test are:
![df=n-1=3-1=2](https://img.qammunity.org/2021/formulas/mathematics/college/lesqq8r28c6iird0ikezrrob86ej3jzusv.png)
The critical value for a two side test with level of significance α=0.05 and 2 degrees of freedom is t=±4.271.
As the statistic t=-12.59 lies outside of the acceptance region, the null hypothesis is rejected.
There is enough evidence, at a significance level of 0.05, that the population mean age of the Bluestone formations is different from 2950 BCE.