Question:

A pharmaceutical firm has surveyed a random sample of 120 persons suffering from Parkinson’s disease. Among the facts obtained is the following bivariate frequency distribution of disease duration and degree of self-reliance:

1. Test whether disease duration and degree of self-reliance are independent, controlling the risk at 0.01. State the alternatives, the decision rule, the value of the test statistic, and the conclusion.

   p -Value	0.006145
   Reject the null hypothesis

   
   The Chi-Square statistics is computed as
   \[{{\chi }^{2}}=\sum{\frac{{{\left( O-E \right)}^{2}}}{E}}=12.395\]
   The critical value is \(\chi _{C}^{2}=11.34\), which means that the reject the null hypothesis of independence.
2. Computer output gives a p–value of 0.006 for the contingency table test in part a). Is this value consistent with the test results in part a). Explain.

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