Let X1, ..., Xn be a random sample from N(μ, 1) distribution, where μ ∈ ℝ is unknown. In order to test H0 : μ = μ0 against H1 : μ > μ0, where μ0 ∈ ℝ is some specified constant, consider the following two tests: (A) Reject H0 if and only if X̅n > c1, where c1 is such that \(P_{μ_0}\) (X̅n > c1) = α ∈ (0, 1) and X̅n = \(\frac{1}{n} \sum_{i=1}^n X_i\). (B) Reject H0 if and only if Median {X1, ..., Xn} > c2, where c2 is such that \(P_{μ_0}\)(Median{X1, ..., Xn} > c2) = α ∈ (0, 1). Then which of the following statements are true?
Hypothesis Testing for Normal Mean
This problem deals with hypothesis testing for the mean (\(\mu\)) of a Normal distribution \(N(\mu, 1)\) with known variance (1). The specific test is a one-sided test: \(H_0: \mu = \mu_0\) against \(H_1: \mu > \mu_0\).
We are asked to evaluate the properties of two proposed tests:
- Test (A) based on the sample mean (\(\bar{X}_n\)).
- Test (B) based on the sample median (Median\(\{X_1, ..., X_n\}\)).
Analysis of Test (A) based on Sample Mean
Test (A) rejects \(H_0\) if \(\bar{X}_n > c_1\), where \(c_1\) is chosen such that the size of the test is \(\alpha\), i.e., \(P_{\mu_0}(\bar{X}_n > c_1) = \alpha\). The sample mean \(\bar{X}_n\) is the maximum likelihood estimator (MLE) and a sufficient statistic for \(\mu\) in a Normal distribution with known variance.
For testing \(H_0: \mu = \mu_0\) against \(H_1: \mu > \mu_0\) in a \(N(\mu, \sigma^2)\) distribution with known \(\sigma^2\), the Uniformly Most Powerful (UMP) test of size \(\alpha\) is based on the sample mean \(\bar{X}_n\). The critical region for this UMP test is of the form \(\bar{X}_n > c\) for some constant \(c\).
Analysis of Test (B) based on Sample Median
Test (B) rejects \(H_0\) if Median\(\{X_1, ..., X_n\} > c_2\), where \(c_2\) is chosen such that \(P_{\mu_0}(\text{Median}\{X_1, ..., X_n\} > c_2) = \alpha\). While the sample median is also a statistic, for a Normal distribution, the sample mean is more efficient than the sample median as an estimator of the population mean/median.
Since the sample mean is a sufficient statistic and the UMP test exists and is based on the sample mean for this problem, a test based on the sample median is generally less powerful than the UMP test based on the sample mean, especially for small sample sizes. For large sample sizes, the sample median is asymptotically Normal, but the sample mean still provides a more powerful test due to its smaller variance.
Evaluation of Statements
Statement 1: The test described in (A) is the uniformly most powerful test of size \(\alpha\).
For a random sample from \(N(\mu, \sigma^2)\) with known \(\sigma^2\), the UMP test for \(H_0: \mu = \mu_0\) against \(H_1: \mu > \mu_0\) is based on the sample mean \(\bar{X}_n\). The test that rejects \(H_0\) when \(\bar{X}_n\) is large is indeed the UMP test. Test (A) is exactly this test with the critical value \(c_1\) set to achieve size \(\alpha\).
Conclusion: Statement 1 is true.
Statement 2: The test described in (B) is the uniformly most powerful test of size \(\alpha\).
As discussed, the UMP test for this problem is based on the sample mean, not the sample median. The sample median is less efficient than the sample mean for Normal data, and a test based on the median will be less powerful than the UMP test based on the mean.
Conclusion: Statement 2 is false.
Statement 3: \(P_{\mu}(X̅_n > c_1) \to 1\) as \(n \to \infty\) for all \(\mu > \mu_0\).
This statement describes the power of Test (A). The power function of Test (A) is \(\beta(\mu) = P_{\mu}(\bar{X}_n > c_1)\). Under \(P_{\mu}\), \(\bar{X}_n \sim N(\mu, 1/n)\). We standardized \(\bar{X}_n\) as \(\sqrt{n}(\bar{X}_n - \mu) \sim N(0, 1)\). The critical value \(c_1\) is determined by the size \(\alpha\):
\(P_{\mu_0}(\bar{X}_n > c_1) = \alpha\)
\(P\left(\frac{\bar{X}_n - \mu_0}{1/\sqrt{n}} > \frac{c_1 - \mu_0}{1/\sqrt{n}}\right) = \alpha\)
Let \(Z \sim N(0, 1)\). Then \(\frac{c_1 - \mu_0}{1/\sqrt{n}} = z_\alpha\), where \(P(Z > z_\alpha) = \alpha\). Thus, \(c_1 = \mu_0 + \frac{z_\alpha}{\sqrt{n}}\).
Now, consider the power for \(\mu > \mu_0\):
\(\beta(\mu) = P_{\mu}(\bar{X}_n > c_1) = P\left(\frac{\bar{X}_n - \mu}{1/\sqrt{n}} > \frac{c_1 - \mu}{1/\sqrt{n}}\right) = P\left(Z > \sqrt{n}(c_1 - \mu)\right)\)
Substitute \(c_1\):
\(\beta(\mu) = P\left(Z > \sqrt{n}\left(\mu_0 + \frac{z_\alpha}{\sqrt{n}} - \mu\right)\right) = P\left(Z > \sqrt{n}(\mu_0 - \mu) + z_\alpha\right)\)
For \(\mu > \mu_0\), we have \(\mu_0 - \mu < 0\). As \(n \to \infty\), \(\sqrt{n}(\mu_0 - \mu) \to -\infty\). Therefore, \(\sqrt{n}(\mu_0 - \mu) + z_\alpha \to -\infty\).
So, \(\lim_{n \to \infty} \beta(\mu) = \lim_{n \to \infty} P\left(Z > \sqrt{n}(\mu_0 - \mu) + z_\alpha\right) = P(Z > -\infty) = 1\).
This means the power approaches 1 for any fixed alternative \(\mu > \mu_0\) as the sample size increases. This is a desirable property of consistent tests.
Conclusion: Statement 3 is true.
Statement 4: \(P_{\mu_0}(\text{Median}\{X_1, ..., X_n\} > \mu_0) = \frac{1}{2}\).
Under \(H_0\), the data \(X_1, ..., X_n\) are drawn from \(N(\mu_0, 1)\). The Normal distribution is symmetric about its mean (which is \(\mu_0\)). The sample median of a random sample from any continuous distribution which is symmetric about a point \(\theta\) has a distribution that is also symmetric about \(\theta\). In this case, the distribution of the sample median is symmetric about \(\mu_0\).
For any random variable \(Y\) with a distribution symmetric about \(\theta\), \(P(Y > \theta) = P(Y < \theta)\). Since the distribution is continuous, \(P(Y = \theta) = 0\). Therefore, \(P(Y > \theta) + P(Y < \theta) + P(Y = \theta) = 1\), which implies \(2 P(Y > \theta) = 1\), so \(P(Y > \theta) = 1/2\).
Applying this to the sample median from \(N(\mu_0, 1)\), its distribution is symmetric about \(\mu_0\). Thus, \(P_{\mu_0}(\text{Median}\{X_1, ..., X_n\} > \mu_0) = 1/2\).
Conclusion: Statement 4 is true.
Summary of Results
Based on our analysis:
- Statement 1 is true (Test A is UMP).
- Statement 2 is false (Test B is not UMP).
- Statement 3 is true (Power of Test A approaches 1 as n grows for alternatives).
- Statement 4 is true (Probability of median exceeding population median under H0 is 1/2 due to symmetry).
The statements found to be true are 1, 3, and 4.
Important Questions from Sampling Distributions
Let X1, X2, ..., X6 be a random sample from a gamma distribution with the probability density function
\(f(x \mid \lambda)=\left\{\begin{array}{cl} \frac{\lambda^4}{6} e^{-\lambda x} x^3, & \text { if } x>0 \\ 0, & \text { if } x \leq 0 \end{array},\right.\)
where λ > 0 is unknown. Let \(T=\sum_{i=1}^6 X_i\) and ψ be the uniformly most powerful test of size α = 0.05 for testing null hypothesis H0 : λ = 1 against alternative hypothesis H1 : λ > 1. For any positive integer v, let \(\chi_{v, α}^2\) denote the (1 - α)th quantile of \(\chi_v^2\) distribution. Then the test ψ rejects H0 if and only if
For n ≥ 2, let X1, X2, ..., Xn be a random sample from a distribution with the probability density function
\(f(x \mid θ)=\left\{\begin{array}{cc} θ x^{θ-1}, & 0<x<1 \\ 0, & \text { otherwise } \end{array},\right.\)
where θ > 0 is an unknown parameter. Then which of the following is the uniformly minimum variance unbiased estimator for \(\frac{1}{\theta}\) ?
Let X1, X2, ..., Xn be a random sample from an unknown distribution with absolutely continuous cumulative distribution function (cdf) F. Let F0 be a specified absolutely continuous cdf. For testing H0 : F(x) = F0(x) for all x against H1 : F(x) ≠ F0(x) for some x, consider the following two test statistics:
\(\displaystyle T_{1, n}=\sup _{x \in \mathbb{R}}\left|\frac{1}{n} \sum_{i=1}^n I_{\left\{X_i \leq x\right\}}-F_0(x)\right| \), and \(\displaystyle T_{2, n}=\sup _{x \in \mathbb{R}} n\left|\frac{1}{n} \sum_{i=1}^n I_{\left\{X_i \leq x\right\}}-F_0(x)\right|\), where \(I_{\left\{X_i \leq x\right\}}=\left\{\begin{array}{ll}1, & \text { if } X_i \leq x \\ 0, & \text { if } X_i>x\end{array}\right.\) for i = 1, 2, ..., n.
Then which of the following statements are true?
- Which one of the following is the formula to calculate standard error of mean differences of two sample groups ?
- Standard deviation of a sampling distribution of a statistic is termed as,
