Answer:
We can conclude that the result is significant and production differ in cost variance.
Explanation:
Given :
n1 = 16
n2 = 16
s1² = 5.7
s2² = 2.8
α = 0.10
H0 : σ1² = σ2²
H1 : σ1² ≠ σ2²
The test statistic :
Ftest = s1² / s2² =
Ftest = 5.7 / 2.8
Ftest = 2.036
Using the Pvalue from Fratio calculator :
df numerator = 16 - 1 = 15
df denominator = 16 - 1 = 15
Pvalue(2.036, 15, 15) = 0.0898
Pvalue = 0.0898
Since the Pvalue is < α ; We can conclude that the result is significant and production differ in cost variance.