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The variance value for an F-distribution with parameters = 5 and = 9 is :
Question

The variance value for an F-distribution with parameters = 5 and = 9 is :

A.

1.587

B.

1.378

C.

2.578

D.

1.498

Correct option is A

The correct answer is (a).

Introduction 

The F-distribution is a continuous probability distribution heavily utilized in statistical hypothesis testing, specifically in ANOVA.

The variance of an F-distribution depends strictly on its two degrees of freedom.

By plugging the given parameters into the standard mathematical formula for F-distribution variance, we can compute the precise statistical dispersion.

Information Booster
  • The parameters for an F-distribution are its degrees of freedom, denoted as d1d_1​ (numerator) and d2d_2​ (denominator).
  • In this problem, the parameters are given as d1=5andd2=9d_1 = 5 and d_2 = 9​​
  • The variance of an F-distribution is valid only when d2>4d_2 > 4​​
  • The formula for the variance is:
  • Var(X)=2d22(d1+d22)d1(d22)2(d24)Var(X) = \frac{2 d_2^2 (d_1 + d_2 - 2)}{d_1 (d_2 - 2)^2 (d_2 - 4)}
  • Calculating the numerator: 2×81×12=19442 \times 81 \times 12 = 1944​​
  • Calculating the denominator:5×49×5=12255 \times 49 \times 5 = 1225​.
  • Dividing the values yields 1944/12251.58691944 / 1225 \approx 1.5869​, which rounds to 1.587.
Additional Knowledge
  • Option B (1.378): This is an incorrect mathematical calculation and does not result from the standard variance formula of the F-distribution for the given parameters.
  • Option C (2.578): This value is substantially higher than the actual variance and implies a misapplication of the degrees of freedom in the formula.
  • Option D (1.498): This is a numerically close distractor but mathematically incorrect for degrees of freedom 5 and 9.

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