Question

1. State the assumptions

2. State the appropriate hypothesis test

3. State the appropriate test statistics formula

4. Find the value for the test statistics via calculator 5. Find the p-value via calculator

6. Reject H0 if p-value is small

7. Explain your decision in a full sentence

Humerus bones from the same species of animal tend to have approximately the same length-to-width ratios. When fossils of humerus bones are discovered, archeologists can often determine the species of animal by examining the length-to-width ratios of the bones. It is known that species A has a mean ratio of 8.5. Suppose 41 fossils of humerus bones were un- earthed at an archeological site in East Africa, where species A is believed to have flourished. The length-to-width ratios of the bones were measured given below.

10.73 9.57 6.66 9.89 8.89 9.29 9.35 8.17 9.07 9.94 8.86 8.93 9.20 8.07 9.93 8.80 10.33 8.37 8.91 10.02 9.98 6.85 11.77 8.38 9.84 8.52 10.48 11.67 9.59 8.87 10.39 8.30 8.48 6.23 9.39 9.17 8.71 9.41 9.17 12.00 9.38

Do these data present sufficient evidence to indicate that the mean ratio of all bones of this species differs from 8.5 at 0.05 significance level?

please write out steps like shown above. Thank you !

Answer 1

1. Assumptions:
   It is assumed that the length-to-width ratio follows a Normal distribution.
   
   
2. Hypothesis
   Here, we are to test if the length-to-width ratio differs significantly from 8.5.
   That is, we are to test:
   
   
   
3. Test Statistic
   The test statistic is given by:
   
   
   where,
   
   
   
   
   Putting in the given values, we get,
   
   
   
4. p-value
   
   Under Null,
   
   The p-value is given by:
   
   
   Now,
   
   
   
   And,
   
   
   
   Hence,
   
   
   
   
   
5. Conclusion
   It is observed that, p-value<0.05 and hence we reject .
   
   
6. Decision
   The sample values indicate that the Null Hypothesis is rejected and hence it can be claimed that, there is sufficient evidence to indicate that the mean ratio of all bones of this species differs from 8.5 at 0.05 significance level.
   
   

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