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:
0.20
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.
\(f(x,\theta ) = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{\theta };0 \le x \le \theta }\\ {0;elsewhere} \end{array}} \right.\)
This is the PDF of a uniform distribution on the interval \([0, \theta]\).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.
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: \(X < 0.4\)
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 \(P(X < 0.4 | \theta = 2)\).
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.
| 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 \(H_1\) | \(\int_0^{0.4} \frac{1}{2} dx = 0.2\) |
| 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. |
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.
For frequency distribution presentation, which option is wrong?
If the mean and variance of a binomial distribution are 5 and 4, respectively, then the value of n is:
A discrete random variable X has the following probability distribution.
| X | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X) | K | 2K | 2K | 3K | K 2 | 2K 2 | 7K 2 + K |
What is the value of K?
Which of the following are the properties of Binomial distribution?
A. May be symmetrical or skewed
B. Uni-modal, bell-shaped and symmetrical
C. Asymptotic to the x-axis
D. n and p are the two parameters
E. μ and σ are the two parameters
Choose thecorrectanswer from the options given below:
The following two statements relate to probability distributions. Choose the correct code for the statements being correct or incorrect.
Statement I: When ‘p' and 'q' are equal in the binomial distribution, the shape of the distribution is perfectly symmetrical irrespective of the size of 'n'.
Statement II: The mean and the variance of Poisson distribution are not equal.
Which of the following are the properties of normal distribution?
(A) May be symmetrical or skewed
(B) Uni-modal, bell-shaped and symmetrical
(C) Asymptotic to the x-axis
(D) m and p are the two parameters
(E) μ and σ are the two parameters
Choose the correct answer from the options given below:
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: