H0:σ=0.25.
Ha:σ>0.25.
Test statistic: χ2=0.25(20−1)0.32=24.32.
Degrees of freedom: df=20−1=19.
P-value: p=P(χ2>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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