$H_0: \mu = \mu_0 \text{ against } H_1: \mu \ne \mu_0$
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
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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 $X_2$?
Let $(X,Y)$ have the joint discrete distribution such that $X \mid Y=y \sim \text{Binomial} \quad (y, 0.5)$ and $Y \sim \text{Poisson}(\lambda), \lambda > 0$, where $\lambda$ is an unknown parameter. Let $T = T(X,Y)$ be any unbiased estimator of $\lambda$. Then
Consider the pdf
$f(x;\theta,\sigma) = \frac{0.9}{\sigma}\phi\left(\frac{x-\theta}{\sigma}\right) + 0.1\ \phi(x-\theta)$, where $-\infty < \theta < \infty$ and $\sigma > 0$ are unknown parameters and $\phi$ denotes the pdf of $N(0,1)$. Let $X_1, X_2, \cdots, X_n$ be a random sample from this probability distribution. Then which of the following is (are) correct?
