Math Problem Statement

For the family of Neyman–Pearson test, show that the larger the α, the smaller the β (P(type II error)).

Solution

To address the question regarding the relationship between α\alpha (Type I error probability) and β\beta (Type II error probability) in the Neyman-Pearson testing framework:

Explanation:

The Neyman-Pearson Lemma outlines how hypothesis tests are designed to maximize power (i.e., minimize β\beta) for a fixed level of α\alpha. However, the trade-off between α\alpha and β\beta is intrinsic to hypothesis testing. Increasing α\alpha (the probability of rejecting H0H_0 when H0H_0 is true) typically results in a decrease in β\beta (the probability of failing to reject H0H_0 when H1H_1 is true). Here's why:

  1. Critical Region Expansion:

    • α\alpha represents the size of the critical region (the region where H0H_0 is rejected).
    • Increasing α\alpha expands the critical region, making it more likely to reject H0H_0.
  2. Impact on β\beta:

    • A larger critical region means that for a fixed H1H_1, fewer observations fall into the acceptance region of H0H_0. This reduces the probability of a Type II error (β\beta).
  3. Mathematical Intuition:

    • Consider the test statistic and the distributions of the null (H0H_0) and alternative (H1H_1) hypotheses.
    • Increasing α\alpha shifts the threshold closer to the center of the null distribution. This decreases overlap with H1H_1, reducing β\beta.

Proof (Conceptual Overview):

Let the test statistic be TT, and denote the critical value for rejecting H0H_0 by tct_c. Then:

  • α=P(T>tcH0)\alpha = P(T > t_c \mid H_0),
  • β=P(TtcH1)\beta = P(T \leq t_c \mid H_1).

If tct_c is reduced to increase α\alpha:

  1. The rejection region expands.
  2. The probability of correctly rejecting H0H_0 under H1H_1 increases, thus lowering β\beta.

Conclusion:

This relationship underscores the trade-off between sensitivity (avoiding β\beta) and specificity (avoiding α\alpha) in hypothesis testing.


Would you like me to:

  1. Provide a more detailed mathematical derivation using distributions?
  2. Explore an example of this relationship using normal distributions?
  3. Clarify the Neyman-Pearson Lemma in more depth?
  4. Discuss practical implications of this trade-off in experimental design?
  5. Illustrate this with a visual diagram of Type I and Type II errors?

Tip:

When designing a test, consider the context and consequences of both errors to balance α\alpha and β\beta effectively.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Type I and Type II Errors
Neyman–Pearson Lemma

Formulas

α = P(Rejecting H₀ | H₀ is true)
β = P(Failing to Reject H₀ | H₁ is true)

Theorems

Neyman–Pearson Lemma

Suitable Grade Level

Undergraduate