Dating Stonehenge (Case Study Sample)

Running head: DATING STONEHENGE. 1
Statistics: Case Study 3
Name
Institution
QUESTION 1
* Step 1. Set up hypotheses and determine level of significance
Null hypothesis: H0: μ = 2950 B.C
Alternative hypothesis: H1: μ > 2950 B.C               α =0.05
n = 9 , X= 3033.1 s = 66.9 , μ = 2950 B.C
*  Step 2. Select the appropriate test statistic.  
Because the sample size is small (n<30) the appropriate test statistic is
 .
* Step 3. Set up decision rule.  
This is a lower tailed test, using a t statistic and a 5% level of significance.
In order to determine the critical value of t, we need degrees of freedom, df, defined as df=n-1. Here df=9-1=8.
The critical value for a lower tailed test with df=9 and a =0.05 is  2.3060 (from t-test tables) and the decision rule is as follows:  
Reject H0 if t < 2.3060.
*  Step 4. Compute the test statistic.  
We now substitute the sample data into the formula for the test statistic identified in Step 2.  
9525001854200t = 3033.1 – 2950
66.9 / ‚ 9
=3.7265
* Step 5. Conclusion.  
We do not reject H0 because 3.7265 >  2.3060. We do not have statistically significant evidence at α=0.05 to show that the mean date for the construction of the ditch of 2950 B.C is lower than the estimates from unshed antlers excavated from the ditch produced a mean of 3033.1 B.C.
QUESTION 2
* Step 1. Set up hypotheses and determine level of significance
Null hypothesis : H0: μ = 2950 B.C
Alternative hypothesis: H1: μ > 2950 B.C               α =0.05
n = 3 , X= 2193.3 s = 104.1 , μ = 2950 B.C
* Step 2. Select the appropriate test statistic.  
Because the sample size is small (n<30) the appropriate test statistic is
 .
* Step 3. Set up decision rule.  
This is a lower tailed test, using a t statistic and a 5% level of significance.
In order to determine the critical value of t, we need degrees of freedom, df, defined as df=n-1. Here df=3-1=2.
The critical value for a lower tailed test with df=3 and a =0.05 is 4.3027 (from t-test tables) and the decision rule is as follows:  
Reject H0 if t < 4.3027.
*  Step 4. Compute the test statistic.  
We now substitute the sample data into the formula for the test statistic identified in Step 2.  
9525001854200t = 2193.3 – 2950
104.1 / ‚ 3
= -12.5902
* Step 5. Conclusion.  
We reject H0 because -12.5902˂ 4.3027. We have statistically significant evidence at α=0.05 to claim that the population mean age of the Bluestone formations is not different from Corbin’s declared mean age of the ditch, that is, 2950 B.C.
QUESTION 3
* Step 1. Set up hypotheses and determine level of significance
Null hypothesis : H0: μ = 2950 B.C
Alternative hypothesis: H1: μ > 2950 B.C               α =0.05
n = 3 , X= 1671.7 s = 99.7 , μ = 2950 B.C
* Step 2. Select the appropriate test statistic.  
Because the sample size is small (n<30) the appropriate test statistic is
 .
* Step 3. Set up decision rule.  
This is a lower tailed test, using a t statistic and a 5% level of significance.
In order to determine the critical value of t, we need degrees of freedom, df, defined as df=n-1. Here df=3-1=2.
The critical value for a lower tailed test with df=2 and a =0.05 is 4.3027 (from t-test tables) and the decision rule is as follows:  
Reject H0 if t < 4.3027.
*  Step 4. Compute the test statistic.  
We now substitute the sample data into the formula for the test statistic identified in Step 2.  
9525001854200t = 1671.7 – 2950
99.7/ ‚ 3
= -22.2074
* Step 5. Conclusion.  
We reject H0 because -22.2074˂ 4.3027. We have statistically significant evidence at α=0.05 to claim that the population mean age of the Y and Z holes is not different from Corbin’s stated mean age of the ditch—that is, 2950 B.C.
QUESTION 4
1st site:
n = 9 , X= 3033.1 s = 66.9
We arrive at z = -1.96.  Now we solve for x:
                      x – 3033.1         x – 3033.1        -1.96  =                    =                                                                             66.9/ ‚9               22.3                                                
Hence
        x – 3033.1  =  -51.514
We say that  51.514 is the margin of error.
We have that a 95% confidence interval for the mean clarity is
        (2981.587, 3084.613)
In other words there is a 95% chance that the mean years is between 2981.587 and 3084.613 B.C
2nd site:
n = 3 , X= 2193.3 s = 104.1
We arrive at z = -1.96.  Now we solve for x:
                      x –2193.3 x – 3033.1        -1.96  =                    =                                                                             104.1/ ‚3            60.102                                                
<...