$H_0: \mu = \mu_0 \text{ against } H_1: \mu \ne \mu_0$
Understanding Confidence Intervals and Hypothesis Tests
The core principle connecting confidence intervals (CI) and hypothesis testing is that a $(1-\alpha)$ confidence interval provides a range of plausible values for a population parameter. For a two-sided hypothesis test of the form $H_0: \mu = \mu_{value}$ versus $H_1: \mu \ne \mu_{value}$, the null hypothesis $H_0$ is rejected at the significance level $\alpha$ if and only if the hypothesized value $\mu_{value}$ falls outside the $(1-\alpha)$ confidence interval.
Analyzing the Given Confidence Interval
We are given a 95% confidence interval for the population mean $\mu$, which is $(2.5, 3.6)$.
- This 95% CI corresponds to a significance level of $\alpha = 1 - 0.95 = 0.05$.
- The hypothesis test is $H_0: \mu = \mu_0$ versus $H_1: \mu \ne \mu_0$.
- We know the hypothesized value $\mu_0$ satisfies $\mu_0 < 2.5$.
Testing Hypothesis at $\alpha = 0.1$
A significance level of $\alpha = 0.1$ corresponds to a $(1-0.1) = 0.90$ or 90% confidence interval.
Key points:
- The 90% confidence interval is wider than the 95% confidence interval.
- We established that $\mu_0 < 2.5$. Since the 95% CI is $(2.5, 3.6)$, the value $2.5$ is the lower boundary. As $\mu_0$ is strictly less than $2.5$, it lies outside the 95% CI.
- Because $\mu_0$ lies outside the 95% CI, it must also lie outside the wider 90% CI.
- Therefore, we must reject $H_0$ at the $\alpha = 0.1$ significance level. This aligns with Option A.
Testing Hypothesis at $\alpha = 0.025$
A significance level of $\alpha = 0.025$ corresponds to a $(1-0.025) = 0.975$ or 97.5% confidence interval.
Key points:
- The 97.5% confidence interval is narrower than the 95% confidence interval.
- We know $\mu_0 < 2.5$. The 95% CI starts at $2.5$. The 97.5% CI will have a lower bound that is greater than $2.5$.
- We only know $\mu_0$ is below $2.5$. We don't know if $\mu_0$ falls above or below the lower bound of the narrower 97.5% CI. It could be inside or outside this interval.
- Therefore, the information provided (the 95% CI) is not sufficient to determine whether $H_0$ would be rejected at the $\alpha = 0.025$ significance level. This aligns with Option D.
Conclusion
Based on the analysis:
- $H_0$ would be necessarily rejected at $\alpha = 0.1$.
- For $\alpha = 0.025$, the information is not enough to draw a conclusion.
Important Questions from Elementary Bayesian Inference
- Suppose the distribution of $X$ given $\theta$ is normal with mean $\theta$ and variance $15$. Further, let the prior (improper) distribution of $\theta$ be proportional to $1, \ -\infty<\theta<\infty$. If the observed value of $X$ is $13$, then which of the following statements is true?
- Let $X_1, X_2, . . ., X_n$ be a random sample from $N(\theta, 1)$, $\theta \in R$. If $\hat{\theta}$ is the Bayes estimator of $\theta$ with respect to some prior $\pi(\theta)$ and loss function $L(\theta, d)$. Then, which of the following statements are true?
- Let $X|\theta \sim \text{Uniform}[0, \theta]$ and $\theta$ has an Exponential distribution with mean $\lambda$, where $\lambda > 3$ is known. If the realized value of $X$ is $2025$, then the posterior mode equals
- $X_1, X_2, \cdots, X_n$ are independent and identically distributed $N(\theta, 1)$ random variables, where $\theta$ takes only integer values i.e.
$\theta \in \{\cdots, -2, -1, 0, 1, 2, \cdots\}$.
Which of the following is the maximum likelihood estimator of $\theta$? - Suppose the probability mass function of a random variable X under the parameter $\theta = \theta_0$ and $\theta = \theta_1 (\ne \theta_0)$ are given by
x 0 1 2 3 $p_{\theta_0}(x)$ 0.01 0.04 0.5 0.45 $p_{\theta_1}(x)$ 0.02 0.08 0.4 0.5
Define a test $\phi$ such that $\phi(x) = 1$ if $x = 0, 1$, and $0$ if $x = 2, 3$.
For testing $H_0: \theta = \theta_0$ against $H_1: \theta = \theta_1$, the test $\phi$ is
