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?
$H_0$ is accepted both at $5\%$ and $1\%$ levels of significance.
Hypothesis Test Setup
This problem concerns a hypothesis test for the mean ($\mu$) of a normally distributed population where both the mean and variance ($\sigma^2$) are unknown. The test is conducted using a Uniformly Most Powerful (UMP) approach.
- Sample size: $n=7$
- Observed sample values: $x_1, \dots, x_7 = \{1.2, 1.3, 1.7, 1.8, 2.1, 2.3, 2.7\}$
- Hypotheses: Null hypothesis $H_0: \mu = 2$, Alternative hypothesis $H_1: \mu > 2$.
- Significance levels ($\alpha$): $0.05$ and $0.01$.
Calculation of Sample Statistics
To perform the t-test, we first compute the sample mean ($\bar{x}$) and sample standard deviation ($s$).
- Sample Mean ($\bar{x}$): $ \bar{x} = \frac{\sum_{i=1}^{7} x_i}{n} = \frac{1.2 + 1.3 + 1.7 + 1.8 + 2.1 + 2.3 + 2.7}{7} = \frac{13.1}{7} \approx 1.8714 $
- Sample Standard Deviation ($s$): First, calculate the sample variance ($s^2$). $ s^2 = \frac{1}{n-1} \sum_{i=1}^{7} (x_i - \bar{x})^2 $ Sum of squared deviations: $\sum (x_i - \bar{x})^2 \approx 1.7343$. $ s^2 = \frac{1.7343}{7-1} = \frac{1.7343}{6} \approx 0.2891 $ $ s = \sqrt{s^2} \approx \sqrt{0.2891} \approx 0.5377 $
Applying the UMP (t-test)
Since the population variance is unknown, we use a t-statistic. The UMP test for $H_0: \mu = \mu_0$ against $H_1: \mu > \mu_0$ rejects $H_0$ if the t-statistic is greater than the critical value.
The test statistic is calculated as: $ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} $ Substituting the values ($\bar{x} \approx 1.8714$, $\mu_0 = 2$, $s \approx 0.5377$, $n=7$): $ t = \frac{1.8714 - 2}{0.5377/\sqrt{7}} \approx \frac{-0.1286}{0.5377 / 2.6458} \approx \frac{-0.1286}{0.2032} \approx -0.6328 $
The degrees of freedom are $df = n-1 = 6$. The critical region is $t > t_{\alpha, 6}$.
- For $\alpha = 0.05$: The critical value $t_{0.05, 6} \approx 1.943$. The calculated $t \approx -0.6328$. Since $-0.6328$ is not greater than $1.943$, $H_0$ is accepted.
- For $\alpha = 0.01$: The critical value $t_{0.01, 6} \approx 2.968$. The calculated $t \approx -0.6328$. Since $-0.6328$ is not greater than $2.968$, $H_0$ is accepted.
Final Conclusion
Based on the UMP t-test, the null hypothesis $H_0: \mu = 2$ is accepted for both significance levels, $\alpha=0.05$ and $\alpha=0.01$. The sample mean is below the hypothesized mean, and the test statistic is significantly negative, falling far short of the critical values needed for rejection in this right-tailed test.
Important Questions from Elementary Bayesian Inference
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- 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
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