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Chapter 7: Problem 35

Refer to Exercise \(7.34 .\) Suppose that \(F\) has an \(F\) distribution with \(\nu_{1}=50\) numerator degrees of freedom and \(\nu_{2}=70\) denominator degrees of freedom. Notice that Table 7 , Appendix 3 , does not contain entries for 50 numerator degrees of freedom and 70 denominator degrees of freedom. a. What is \(E(F) ?\) b. Give \(V(F)\) c. Is it likely that \(F\) will exceed \(3 ?[\text { Hint: Use Tchebysheff's theorem. }]\)

Short Answer

Expert verified
a. \(E(F) \approx 1.029\); b. \(V(F) \approx 0.076\); c. Unlikely, \(F\) exceeds 3.

Step by step solution

01

Calculate the Expected Value of F

The expected value of an F-distribution with numerator degrees of freedom \(u_1\) and denominator degrees of freedom \(u_2\) is given by\[ E(F) = \frac{u_2}{u_2 - 2} \]so long as \(u_2 > 2\). For this example,\[ u_2 = 70 \text{ and therefore } E(F) = \frac{70}{70 - 2} = \frac{70}{68} \approx 1.029 \].
02

Compute the Variance of F

The variance of an F-distribution is given by\[ V(F) = \frac{2u_2^2(u_1 + u_2 - 2)}{u_1(u_2 - 2)^2(u_2 - 4)} \]for \(u_2 > 4\). For this problem,\[ V(F) = \frac{2\times 70^2\times (50 + 70 - 2)}{50\times 68^2\times 66} \]Calculate:- Numerator: \(2 \times 4900 \times 118 = 1,156,400 \)- Denominator: \(50 \times 4624 \times 66 = 15,235,200 \)Thus, the variance is:\[ V(F) \approx \frac{1,156,400}{15,235,200} \approx 0.076 \].
03

Use Tchebysheff's Theorem to Estimate Probability

Tchebysheff's theorem gives a method to bound probabilities. It states that for any distribution, the probability that a random variable deviates from its mean \(\mu\) by more than \(k\sqrt{V(F)}\) is at most \(\frac{1}{k^2}\).Here, \(\mu = 1.029\), and we want to check \(P(F > 3)\).- Deviation from mean: \(3 - 1.029 = 1.971\)- \(k = \frac{1.971}{\sqrt{0.076}} \approx 7.14\)Using Tchebysheff's theorem,\[ P(|F - \mu| \geq 1.971) \leq \frac{1}{7.14^2} \approx 0.020 \].This indicates it is unlikely \(F\) will exceed \(3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Degrees of Freedom
In an F-distribution, degrees of freedom are vital parameters that define its shape. Here, we have two types:
  • Numerator degrees of freedom (\(u_1\)): These specify the variation due to different group means or conditions in your data. In this exercise, \(u_1 = 50\).
  • Denominator degrees of freedom (\(u_2\)): These relate to the variation within the groups, measured as the 'error' or residuals. In this case, \(u_2 = 70\).
Degrees of freedom relate to the amount of information available from a set of data. More degrees typically imply more reliable estimates for the distribution's parameters, underpinning the expected value and variance calculations in the F-distribution.
Expected Value
The expected value in an F-distribution refers to the average value or mean that the F-statistic is expected to have. The formula for calculating this is:\[E(F) = \frac{u_2}{u_2 - 2}\]provided that \(u_2 > 2\).
For this exercise, we find:
  • Numerator for \(E(F)\): \(u_2 = 70\)
  • Denominator for \(E(F)\): \(u_2 - 2 = 68\)
So, \[E(F) = \frac{70}{68} \approx 1.029\]
The expected value serves as a reference point for understanding typical outcomes. It is important as it centralizes the distribution, giving insights into the statistical behavior of the F-distribution.
Variance
Variance measures the spread or variability around the expected value in an F-distribution. It helps quantify how much the values differ from the mean. The formula for variance is:\[V(F) = \frac{2u_2^2(u_1 + u_2 - 2)}{u_1(u_2 - 2)^2(u_2 - 4)}\]if \(u_2 > 4\).
For this exercise, we substitute:
  • Numerator: \(2 \times 70^2 \times (50 + 70 - 2) = 1,156,400\)
  • Denominator: \(50 \times 68^2 \times 66 = 15,235,200\)
So, \[V(F) \approx \frac{1,156,400}{15,235,200} \approx 0.076\]
This small variance indicates the F-statistic values are generally close to the mean, suggesting less variability.
Tchebysheff's Theorem
Tchebysheff's theorem provides a way to determine the probability of deviations from the mean, irrespective of the distribution. It sets an upper bound on the probability that a random variable differs from its mean by more than a certain distance, using:\[P(|X - \mu| \geq k\sqrt{V(F)}) \leq \frac{1}{k^2}\]
In the context of our exercise:
  • We check if \(F\) exceeds 3, yielding a deviation of \(3 - 1.029 = 1.971\).
  • Calculate \(k\) using the deviation and variance: \(k = \frac{1.971}{\sqrt{0.076}} \approx 7.14\).
  • Apply the theorem: \[P(|F - 1.029| \geq 1.971) \leq \frac{1}{7.14^2} \approx 0.020\]
This calculation suggests there's only about a 2% chance that \(F\) will exceed 3, demonstrating Tchebysheff's usefulness for probability estimation.

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Most popular questions from this chapter

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