As we are testing here whether the variance is less than 521 or not, therefore the null and the alteranative hypothesis here are given as:
As we are testing for variance here, this is a chi square test here. Therefore chi square test statistic is to be used here.
The test statistic here is computed as:
Therefore 5.159 is the test statistic value here.
For 0.05 level of significance and 14 degrees of freedom, as this is a one tailed test, we have from chi square distribution tables here:
Therefore 23.685 is the required critical value here.
As the test statistic value here is less than the critical value, therefore it lies in the non rejection region here. Therefore we cannot support the claim here that the variance is less than 521. Therefore No is the correct answer here.
The cholesterol levels of adult males are approximately normally distributed with a mean of 224 mg/dl,...
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