"H_0:\\sigma=0.25."
"H_a:\\sigma>0.25."
Test statistic: "\\chi^2=\\frac{(20-1)0.32}{0.25}=24.32."
Degrees of freedom: "df=20-1=19."
P-value: "p=P(\\chi^2>24.32)=0.1842."
Since the p-value is greater than 0.05, fail to reject the null hypothesis.
There is no apparent increase in variability significant at the 0.05 level of significance.
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