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Dating Stonehenge (Case Study Sample)

Instructions:

Case Study at the end of chapter 10. Complete the Case Study at the end of chapter 10, "How Old Is Stonehenge?" Make sure you follow directions, use good writing skills, and correctly apply the concepts of the chapter.

source..
Content:
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                                                
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