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Preliminary data analyses indicate that it is reasonable to use as to carry out the specified hypothesis test. Perform the be
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We are testing here whether the population mean is more than 2.85. The test statistic here is computed as:

t^* = \frac{\bar X - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{3.12 - 2.85}{\frac{0.52}{\sqrt{9}}} = 1.56

For n - 1 = 8 degrees of freedom, and 0.01 level of significance, we have from t distribution tables here:

P( t8 < 2.896 ) = 0.99

Therefore P( t8 > 2.896) = 1 - 0.99 = 0.01

Therefore 2.896 is the required critical value here.

As the test statistic value here is 1.56 < 2.896, therefore it lies in the non rejection region here. Therefore we cannot reject the null hypothesis here. We have insufficient evidence that the mean is more than 2.85 here.

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