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Question

For the distribution with unknown θ

\(f(x,\theta ) = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{\theta };0 \le x \le \theta }\\ {0;elsewhere} \end{array}} \right.\)

We set the testing of hypothesis H 0 ∶ θ = 1 vs H 1 ∶ θ = 2. When the critical region X ≥ 0.4, the value of probability of type-II error is:

The correct answer is

0.20

Understanding the Hypothesis Test and Type II Error

This question asks us to calculate the probability of a Type II error for a specific hypothesis test involving a uniform distribution. Let's first define the problem and the concepts involved.

  • The distribution is given by the probability density function (PDF):

    \(f(x,\theta ) = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{\theta };0 \le x \le \theta }\\ {0;elsewhere} \end{array}} \right.\)

    This is the PDF of a uniform distribution on the interval \([0, \theta]\).
  • The null hypothesis is \(H_0: \theta = 1\).
  • The alternative hypothesis is \(H_1: \theta = 2\).
  • The critical region for a single observation \(X\) is given as \(X \ge 0.4\).
  • We need to find the probability of a Type II error.

What is a Type II Error in Hypothesis Testing?

A Type II error occurs when we fail to reject the null hypothesis (\(H_0\)) when the alternative hypothesis (\(H_1\)) is actually true. The probability of a Type II error is often denoted by \(\beta\).

To calculate the probability of a Type II error, we need to determine the acceptance region and then find the probability that the observed data falls into this acceptance region, assuming the alternative hypothesis (\(H_1\)) is true.

Determining the Acceptance Region

The critical region is the set of values of the test statistic (in this case, the single observation \(X\)) that lead to the rejection of the null hypothesis. Given the critical region \(X \ge 0.4\), the acceptance region is the complement of the critical region.

Acceptance Region: \(X < 0.4\)

Calculating the Probability of Type II Error (\(\beta\))

The probability of Type II error is \(P(\text{Accept } H_0 | H_1 \text{ is true})\).

In this case, accepting \(H_0\) means the observation \(X\) falls into the acceptance region \(X < 0.4\). The condition "$H_1$ is true" means \(\theta = 2\).

So, we need to calculate \(P(X < 0.4 | \theta = 2)\).

Using the PDF under the Alternative Hypothesis

When \(H_1\) is true, \(\theta = 2\). The distribution is uniform on \([0, 2]\). The PDF becomes:

\(f(x, 2) = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{2};0 \le x \le 2}\\ {0;elsewhere} \end{array}} \right.\)

To find \(P(X < 0.4 | \theta = 2)\), we integrate the PDF \(f(x, 2)\) over the region \(0 \le x < 0.4\).

\(\beta = P(X < 0.4 | \theta = 2) = \int_0^{0.4} f(x, 2) dx\)

Since \(f(x, 2) = \frac{1}{2}\) for \(0 \le x \le 2\), the integral is:

\(\beta = \int_0^{0.4} \frac{1}{2} dx\)

\(\beta = \left[ \frac{1}{2}x \right]_0^{0.4}\)

\(\beta = \frac{1}{2}(0.4) - \frac{1}{2}(0)\)

\(\beta = 0.2 - 0\)

\(\beta = 0.20\)

The probability of the Type II error for this hypothesis test with the given critical region is 0.20.

Summary of Calculation

Concept Description Value/Region
Null Hypothesis (\(H_0\)) Parameter value under \(H_0\) \(\theta = 1\)
Alternative Hypothesis (\(H_1\)) Parameter value under \(H_1\) \(\theta = 2\)
Distribution PDF Uniform distribution \(f(x,\theta) = 1/\theta\) for \(0 \le x \le \theta\)
Critical Region Region to reject \(H_0\) \(X \ge 0.4\)
Acceptance Region Region to accept \(H_0\) \(X < 0.4\)
Type II Error (\(\beta\)) \(P(\text{Accept } H_0 | H_1 \text{ true})\) \(P(X < 0.4 | \theta = 2)\)
PDF under \(H_1\) Uniform distribution on \([0, 2]\) \(f(x, 2) = 1/2\) for \(0 \le x \le 2\)
\(\beta\) Calculation Integral of PDF over acceptance region under \(H_1\) \(\int_0^{0.4} \frac{1}{2} dx = 0.2\)

Revision Table: Key Concepts in Hypothesis Testing

Concept Definition Impact on Error
Null Hypothesis (\(H_0\)) The statement being tested; often represents no effect or no change. Basis for calculating Type I error.
Alternative Hypothesis (\(H_1\)) The statement accepted if \(H_0\) is rejected; often represents an effect or change. Basis for calculating Type II error and power.
Critical Region Set of outcomes where \(H_0\) is rejected. Determines the probability of Type I error (\(\alpha\)) and influences Type II error (\(\beta\)).
Acceptance Region Set of outcomes where \(H_0\) is not rejected. Complement of the critical region. Used to calculate \(\beta\).
Type I Error (\(\alpha\)) Rejecting \(H_0\) when it is true. Probability depends on critical region and distribution under \(H_0\).
Type II Error (\(\beta\)) Failing to reject \(H_0\) when \(H_1\) is true. Probability depends on critical region and distribution under \(H_1\).
Power of the Test \(1 - \beta\); Probability of correctly rejecting \(H_0\) when \(H_1\) is true. Higher power is desirable.

Additional Information on Probability Distributions and Errors

The uniform distribution is a simple continuous probability distribution. It has a constant probability density over its support interval.

In hypothesis testing, the choice of the critical region directly affects the probabilities of Type I and Type II errors. There is often a trade-off between these two types of errors. Decreasing the probability of one type of error typically increases the probability of the other for a fixed sample size. The Neyman-Pearson lemma provides a framework for finding the "most powerful" test for a simple null versus a simple alternative hypothesis, which minimizes \(\beta\) for a given \(\alpha\).

Understanding Type I and Type II errors is crucial for interpreting the results of any statistical hypothesis test. The significance level (\(\alpha\)) is the maximum acceptable probability of a Type I error, chosen before conducting the test.

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Important Questions from Probability Distribution

  1. If the mean and variance of a binomial distribution are 5 and 4, respectively, then the value of n is:

  2. For the cumulative distribution function \(F(x) = \left\{ {\begin{array}{*{20}{c}} {0;x < - 1}\\ {\frac{1}{2}{{(x + 1)}^2}; - 1 \le x < 0}\\ {1 - \frac{{{{(1 - x)}^2}}}{2};0 \le x < 1}\\ {1.1 < x < \infty } \end{array}} \right.\)

    the upper quartile point is

  3. Let the joint probability density function of \( (X, Y) \) be

    \[f(x, y) = \begin{cases} 6xy^2 & \text{if } 0 < x < 1, 0 < y < 1 \\ 0, & \text{otherwise} \end{cases}\]

     

    Then \( P\left(\frac{1}{2} < X < \frac{3}{4}\right) \) is:

  4. Let X and Y have the joint p.m.f. f(x, y) = x + y / 21, where x = 1, 2, 3 and y = 1, 2. The marginal p.m.f. of X is:

  5. For a random variable X following Poisson distribution with parameter 5, the variance of X is:

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