21 students took a exam. The sample variance is foundto be 80. The examiner claimsthat based on his past experience the true variance has been atleast 100. Does the sample result shows that the population variance significantly less than 100?
[Hint $${x^2} = \frac{{\left( {r - 1} \right)s}}{{{\sigma ^2}}} = \frac{{\left( {21 - 1} \right) \times 80}}{{100}} = 16$$
The value of x² for 20 df 5% level of significance is 31.4. x² value is less than table value.]

A. The variability is significant

B. Significant with + 1

C. The variability is not significant

D. All of the above

Answer: Option C