$H_0: \mu = \mu_0 \text{ against } H_1: \mu \ne \mu_0$
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
- 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
- Suppose $X|\theta \sim \text{Binomial}(7,\theta)$, $0 < \theta < 1$, and the prior distribution of $\theta$ is $\text{Beta}(\alpha, \beta)$ where $\alpha > 0$ and $\beta > 0$ are known. Then which of the following statements MAY NOT be true?
Let $X$ be a random sample from an exponential distribution with mean $1/\lambda$. If $\lambda$ has a prior distribution with probability density function
$g(\lambda) = \begin{cases} \lambda e^{-\lambda} & ; \quad \lambda > 0 \\ 0 & ; \quad \lambda \leq 0 \end{cases}$
then the Bayes estimator of $1/\lambda$ with respect to the squared error loss function is
Let $X_1, X_2, \dots, X_7$ be a random sample from $N(\mu, \sigma^2)$ where $\mu$ and $\sigma^2$ are unknown. Consider the problem of testing $H_0: \mu = 2$ against $H_1: \mu > 2$. Suppose the observed values of $x_1, x_2, \dots, x_7$ are $1.2, 1.3, 1.7, 1.8, 2.1, 2.3, 2.7$. If we use the Uniformly Most Powerful test, which of the following is true?
Suppose $X_i \mid \theta_i \sim N(\theta_i, \sigma^2), i = 1, 2$ are independently distributed. Under the prior distribution, $\theta_1$ and $\theta_2$ are i.i.d $N(\mu, \tau^2)$, where $\sigma^2, \mu$ and $\tau^2$ are known. Then which of the following is true about the marginal distributions of $X_1$ and $X_2$?
