Dating Stonehenge Case Study - 275 Words

Dating Stonehenge (Case Study Sample) Content: Running head: DATING STONEHENGE. 1Statistics: Case Study 3NameInstitutionQUESTION 1 * Step 1.ÂSet up hypotheses and determine level of significanceNull hypothesis: H0: ÃŽÂ ¼ = 2950 B.CAlternative hypothesis: H1: ÃŽÂ ¼ 2950 B.C        Î =0.05n = 9 , X= 3033.1 s = 66.9 , ÃŽÂ ¼ = 2950 B.C * ÂStep 2.ÂSelect the appropriate test statistic. ÂBecause the sample size is small (n30) the appropriate test statistic isÂ. * Step 3.ÂSet up decision rule. ÂThis 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 Â2.3060 (from t-test tables) and the decision rule is as follows: ÂReject H0Âif tÂÂ2.3060. * ÂStep 4.ÂCompute the test statistic. ÆšWe now substitute the sample data into the formula for the test statistic identified in Step 2. Â9525001854200t = 3033.1 à ¢Ã¢â€š ¬ 295066.9 / à ¢Ã… ¡ 9=3.7265 * Step 5.ÂConclusion. ÂWe do not reject H0Âbecause 3.7265 Â2.3060. We do not have statistically significant evidence at ÃŽ=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.ÂSet up hypotheses and determine level of significanceNull hypothesis : H0: ÃŽÂ ¼ = 2950 B.CAlternative hypothesis: H1: ÃŽÂ ¼ 2950 B.C        Î =0.05n = 3 , X= 2193.3 s = 104.1 , ÃŽÂ ¼ = 2950 B.C * Step 2.ÂSelect the appropriate test statistic. ÂBecause the sample size is small (n30) the appropriate test statistic isÂ. * Step 3.ÂSet up decision rule. ÂThis 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: ÂReject H0Âif tÂÂ4.3027. * ÂStep 4.ÂCompute the test statistic. ÂWe now substitute the sample data into the formula for the test statistic identified in Step 2. Â9525001854200t = 2193.3 à ¢Ã¢â€š ¬ 2950104.1 / à ¢Ã… ¡ 3= -12.5902 * Step 5.ÂConclusion. ÂWe reject H0Âbecause -12.5902Ë 4.3027. We have statistically significant evidence at ÃŽ=0.05 to claim that the population mean age of the Bluestone formations is not different from Corbinà ¢Ã¢â€š ¬s declared mean age of the ditch, that is, 2950 B.C.QUESTION 3 * Step 1.ÂSet up hypotheses and determine level of significanceNull hypothesis : H0: ÃŽÂ ¼ = 2950 B.CAlternative hypothesis: H1: ÃŽÂ ¼ 2950 B.C        Î =0.05n = 3 , X= 1671.7 s = 99.7 , ÃŽÂ ¼ = 2950 B.C * Step 2.ÂSelect the appropriate test statistic. ÂBecause the sample size is small (n30) the appropriate test statistic isÂ. * Step 3.ÂSet up decision rule. ÂThis 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: ÂReject H0Âif tÂÂ4.3027. * ÂStep 4.ÂCompute the test statistic. ÂWe now substitute the sample data into the formula for the test statistic identified in Step 2. Â9525001854200t = 1671.7 à ¢Ã¢â€š ¬ 295099.7/ à ¢Ã… ¡ 3= -22.2074 * Step 5.ÂConclusion. ÂWe reject H0Âbecause -22.2074Ë 4.3027. We have statistically significant evidence at ÃŽ=0.05 to claim that the population mean age of the Y and Z holes is not different from Corbinà ¢Ã¢â€š ¬s stated mean age of the ditchà ¢Ã¢â€š ¬that is, 2950 B.C.QUESTION 41st site:n = 9 , X= 3033.1 s = 66.9We arrive atÂz = -1.96. ÂNow we solve forÂx: x à ¢Ã¢â€š ¬ 3033.1  x à ¢Ã¢â€š ¬ 3033.1ÂÂÂÂÂÂÂÂ-1.96 = = 66.9/Âà ƒ ¢Ã… ¡9  22.3ÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂHence x à ¢Ã¢â€š ¬ 3033.1 = -51.514We say thatÂÂ51.514Âis theÂmargin of error.We have that aÂ95% confidence interval for the meanÂclarity is (2981.587, 3084.613)In other words there is aÂ95%Âchance that the mean years is betweenÂ2981.587 and 3084.613 B.C2nd site:n = 3 , X= 2193.3 s = 104.1We arrive atÂz = -1.96. ÂNow we solve forÂx: x à ¢Ã¢â€š ¬2193.3 x à ¢Ã¢â€š ¬ 3033.1ÂÂÂÂÂÂÂÂ-1.96 = = 104.1/Âà ¢Ã… ¡3 60.102ÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂ